If $$\int_{a}^{x} ty (t) dt = x^{2} + y(x)$$ then y as a function of x is:

$$y = 2 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$

$$y = 1 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$

$$y = 2 - (1 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$

MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 8

The solution of

$ye^{-x/y}dx-(xe^{(-x/y)} + y^{3}) dy=0$ is

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$e^{-x/y} + y^{2} = C$
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$xe^{-x/y} + y = C$
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$2e^{-x/y} + y^{2} = C$
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$e^{-x/y} + 2y^{2} = C$

Explanation

$ye^{-x/y}dx-(xe^{-x/y} + y^{3})dy = 0$

$\implies (ydx-xdy)e^{-x/y} - y^{3}dy = 0$

$\implies \dfrac{ydx-xdy}{y^{2}}e^{-x/y} = ydy$

$\implies d(x/y)e^{-x/y} = ydy$

$-e^{-x/y} = \dfrac{y^{2}}{2} + C$

$2 e^{-x/y}+ y^{2} = C$

The solution of the following differential equation

$[1+x\sqrt{(x^{2}+y^{2})}]dx+[\sqrt{(x^{2}+y^{2})-1}]ydy=0$ is equal to

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$x^{2}+\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c$
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$x-\dfrac{y^{3}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c$
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$x-\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c$
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None of these

Explanation

$[1+x\sqrt{(x^{2}+y^{2})}]dx+[\sqrt{(x^{2}+y^{2})-1}]ydy=0$

$\implies dx-ydy+\sqrt{(x^{2}+y^{2}})(xdx+ydy)=0$

$\implies dx-ydy+\dfrac{1}{2}\sqrt{(x^{2}+y^{2})}d(x^{2}+y^{2})=0$

Integrating, we have

$x-\dfrac{y^{2}}{2}+\dfrac{1}{2} \int \sqrt{t} dt=c, (t=\sqrt{(x^{2}+y^{2})})$

$x-\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2)}=c$

The solution of the differential equation

$\dfrac{dy}{dx}=\dfrac{3x^{2}y^{4}+2xy}{x^{2}-2x^{3}y^{3}}$ is

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$\dfrac{y^{2}}{x}-x^{3}y^{2}=c$
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$\dfrac{x^{2}}{y^{2}}+x^{3}y^{3}=c$
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$\dfrac{x^{2}}{y}+x^{3}y^{2}=c$
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$\dfrac{x^{2}}{3y}-2x^{3}y^{2}=c$

Explanation

Re-write the D.E. as

$(2xydx-x^{2}dy)+y^{2}(3x^{2}y^{2}dx+2x^{3}ydy)=0$

Dividing by

$y^{2}$ , we get

$\dfrac{y2xdx-x^{2}dy}{y^{2}} +y^{2}3x^{2}dx+x^{3}2ydy=0$

$d(\dfrac{x^{2}}{y})+d(x^{3}y^{2})=0$

Integrating, we get the solution

$\dfrac{x^{2}}{y}+x^{3}y^{2}=c$

The solution of the differential equation

$x^{2}\dfrac{dy}{dx}\cos \dfrac{1}{x} - y\sin\dfrac{1}{x} = -1,$ where

$y \rightarrow -1$ as

$x \rightarrow \infty$ is

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$y=\sin\dfrac{1}{x}-\cos \dfrac{1}{x}$
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$y= \dfrac{x+1}{x\sin \dfrac{1}{x}}$
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$y= \cos \dfrac{1}{x} + \sin\dfrac{1}{x}$
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$y= \dfrac{x+1}{x\cos 1/x}$

Explanation

$x^{2}\dfrac{dy}{dx}\cos\dfrac{1}{x}-y\sin\dfrac{1}{x}=-1$

$\implies \dfrac{dy}{dx} - \dfrac{y}{x^{2}}\tan\dfrac{1}{x} =-\sec\dfrac{1}{x}\dfrac{1}{x^{2}}$ (linear)
I.F.

$= e^{\int\dfrac{1}{x^{2}}\tan\dfrac{1}{x}dx}=\sec\dfrac{1}{x}$

$\implies$ solutions is

$y \sec\dfrac{1}{x} = -\int\sec^{2}(\dfrac{1}{x})\dfrac{1}{x^{2}}dx=\tan\dfrac{1}{x}+c$
Given

$y \rightarrow -1, x \rightarrow \infty \implies x=-1$
Hence equation of curve is

$y=\sin\dfrac{1}{x} - \cos\dfrac{1}{x}$

The solution of the differential equation

$y(2x^{2}+y)\dfrac{dy}{dx}=(1-4xy^{2})x^{2}$ is given by

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$3(x^{2}y)^{2}+y^{3}-x^{3}=c$
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$xy^{2}+\dfrac{y^{3}}{3}-\dfrac{x^{3}}{3}= 0$
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$\dfrac{2}{5}yx^{5}+\dfrac{y^{3}}{3}=\dfrac{x^{3}}{3}-\dfrac{4xy^{3}}{3}+c$
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None of these

Explanation

$y(2x^{4}+y)\dfrac{dy}{dx}=91-4xy^{2})x^{2}$

$\implies 2x^{4}ydy+y^{2}dy+4x^{3}y^{2}dx-x^{2}dx=0$

$\implies 2x^{2}y(x^{2}dy+2xydx)+y^{2}dy-x^{2}dx=0$

$\implies 2x^{2}yd(x^{2}y)+y^{2}dy-x^{2}dx=0$
Integrating, we get

$(x^{2}y^{2})^{2}+\dfrac{y^{3}}{3}-\dfrac{x^{3}}{3}=c$
or

$3(x^{2}y)^{2}+y^{3}-x^{3}=c$

Solution of the differential equation

$(y+x\sqrt{xy}(x+y))dx + (y\sqrt{xy}(x+y)-x)dy = 0$ is

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$\dfrac{x^{2}+y^{2}}{2}+\tan^{-1}\sqrt{\dfrac{y}{x}} =c$
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$\dfrac{x^{2}+y^{2}}{2}+\tan^{-1}\sqrt{\dfrac{x}{y}} =c$
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$\dfrac{x^{2}+y^{2}}{2}+2\cot^{-1}\sqrt{\dfrac{x}{y}} =c$
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None of these

Explanation

The given equation is written as

$ydx-xdy+x\sqrt{xy}(x+y)dx+y\sqrt{xy}(x+y)dy=0$

$\implies ydx-xdy+(x+y)\sqrt{xy}(xdx+ydy)=0$

$\implies \dfrac{ydx-xdy}{y^{2}}+(\dfrac{x}{y}+1)\sqrt{x}{y}(d(\dfrac{x^{2}+y^{2}}{2}))=0$

$\implies d\dfrac{x^{2}+y^{2}}{2}+\dfrac{d(\dfrac{x}{y})}{(\dfrac{x}{y} +1)\sqrt{\dfrac{x}{y}}} =0$

$\implies \dfrac{x^{2}+y^{2}}{2}+2tan^{-1}\sqrt{\dfrac{x}{y}} = c$

Number of values of

$m\in N$ for which

$y=e^{mx}$ is a solution of the differential equation

$\dfrac{d^3y}{dx^3}-3\dfrac{d^2y}{dx^2}-4\dfrac{dy}{dx}+12y=0$

0%
$0$
0%
$1$
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$2$
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More than $2$

Explanation

$y=e^{mx}$ satisfies

$\dfrac{d^3y}{dx^3}-3\dfrac{d^2y}{dx^2}-4\dfrac{dy}{dx}+12y=0$
then

$e^{mx}(m^3-3m^2-4m+12)=0$

$\implies m=\pm 2,3$

$m\in N$ hence

$m\in (2,3)$

The solution of,

$ydx-xdy+(1+x^2)dx+x^2 \sin y dy=0$ ,

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$x+1-y^2+ \cos y+C=0$
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$y+1-x^2+ \cos y+C=0$
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$\dfrac{x}{y}+\dfrac{1}{y}- y+ \cos y +C=0$
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$\dfrac {y} {x} +\dfrac{1}{x}- x+ \cos y +C=0$

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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Both Assertion and Reason are incorrect

Explanation

$y=c_1 \cos 2x+c_2 \sin^2 x+c_3 \cos^2 x+c_4 e^{2x}+c_5e^{2x}$

$y=c_1 \cos 2x+c_2 \left[\dfrac {1-\cos 2x}{2}\right]+c_3 \left[\dfrac {\cos 2x+1}{2}\right]+c_4e^{2x}+c_5e^{2x}$

$=\left(c_1-\dfrac {c_2}{2}+\dfrac {c_3}{2}\right)\cos 2x +\left(\dfrac {c_2}{2}+\dfrac {c_3}{2}\right)+(c_4 +c_5)e^{2x}$

$=\lambda_1 \cos 2x+\lambda_2 e^{2x}+\lambda_3$

$\Rightarrow \$ Total number of independent parameters in the given general solution is

$3$ .

The solution of

$\dfrac{dy}{dx}+x\sin 2y=x^{3}\cos^{2}y$ , is

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$e^{x^{2}}=(x^{2}-1)e^{x^{2}}\tan y+C$
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$e^{x^{2}}\tan y=\dfrac{1}{2}(x^{2}-1)e^{x^{2}}+C$
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$e^{x^{2}}\tan y=(x^{2}-1)\tan y+C$
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$None\ of\ these$

Differential equation

$\dfrac{dy}{dx}=f(x)g(x)$ can be solved by

separating variable

$\dfrac{dy}{g(y)}=f(x)dx$

The equation of the curve to the point

$(1,0)$ which
stratifies the differential equation

$(1+y^2)dx-xydy=0$

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$x^2+y^2=1$
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$x^2-y^2=1$
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$x^2+y^2=2$
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$x^2-y^2=2$

Explanation

$\dfrac{dx}{x}=\dfrac{ydy}{1+y^2}$

$\Rightarrow \ln$

$x=\dfrac{1}{2}.\ln\ (1+y^2)+\dfrac C2$

$\Rightarrow 2\ln$

$x=\ln\ (1+y^2)+ C$

$\Rightarrow \ln$

$x^2=\ln\ (1+y^2)+ C$

From the given condition,

$(x,y)=(1,0),$

$C=0$

$\Rightarrow \ln$

$x^2=\ln\ (1+y^2)$

$\Rightarrow x^2=1+y^2$

$\therefore x^2-y^2=1$

The solution of the differential equation

$(e^{x^2}+e^{y^2})y\dfrac{dy}{dx}+e^{x^{2}}(xy^2-x)=0$ is

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$e^{x^2}(y^2-1)+e^{y^2}=C$
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$e^{y^2}(x^2-1)+e^{x^2}=C$
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$e^{y^2}(y^2-1)+e^{x^2}=C$
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$e^{x^2}(y-1)+e^{y^2}=C$

Explanation

$y^2=t; 2y\dfrac{dy}{dx};$ hence the differential equation becomes

$\left(e^{x^2}+e^t\right) +2e^{x62}(xt-x)=0$

$e^{x^2}+e^t+2e^{x^2}x(t-1)\dfrac{dx}{dt}=0$
put

$e^{x^2}=z; e^{x^2}2x\dfrac{dx}{dt}=\dfrac{dz}{dt}$

$\implies \dfrac{dz}{dt}+\dfrac{z}{9t-1)}=\dfrac{e^t}{(t-1)};$ I.F.

$=e^{\int \dfrac{dt}{t-1}}=e^{\ln (t-1)} =t-1$

$\implies z(t-1)=-\int(e^t)dt$

$\implies z(t-1)=e^t+C$

$\implies e^{x^2}(y^2-1)=-e^{y^2}+C$

$\implies e^{x^2}(y^2-1) + e^{y^2}=C$

$y$ satisfies the differential equation

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$\dfrac{dy}{dx}+y=e^x (\cos x- \sin x) - e^{-x} (\cos x- \sin x)$
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$\dfrac{dy}{dx}-y=e^x (\cos x- \sin x) + e^{-x} (\cos x+ \sin x)$
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$\dfrac{dy}{dx}+y=e^x (\cos x+ \sin x) - e^{-x} (\cos x- \sin x)$
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$\dfrac{dy}{dx}-y=e^x (\cos x- \sin x) + e^{-x} (\cos x- \sin x)$

Explanation

$f(0)=2$

$f(x)=(e^x+e^{-x}) \cos x - 2x- \left [x \displaystyle \int _{ 0 }^{ x }{ f'(t) dt - \int _{ x}^{ 0 }{ \dfrac{t}{1} }\dfrac{f'(t)}{0} }dt \right]$

$f(x)=(e^x+e^{-x}) \cos x - 2x -[x(f(x)-f(0))- t.f(t)_0^x - \int _{ 0 }^{ x }{ f(t) dt }]$

$f(x)=(e^x+e^{-x}) \cos x - 2x -xf(x) +2x+\left[x f(x) - \int _{ 0 }^{ x }{ f(t) }dt\right]$

$\displaystyle f(x)=(e^x+e^{-x}) \cos x - \int _{ 0 }^{ x }{ f(t) }dt$ ...

$(i)$

On differentiating Eq.

$(i)$

$f'(x)+f(x)= \cos x (e^x - e^{-x}) - (e^x - e^{-x}) \sin x$ ...

$(ii)$

Hence,

$\dfrac{dy}{dx}+y=e^x (\cos x- \sin x) - e^{-x} (\cos x+ \sin x)$

Solution of the diffrential equation

$\dfrac{dy}{dx}+ \dfrac {1+y^2}{\sqrt {1-x^2}}=0$ is

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$\tan^{-1} y + \sin ^{-1} x=C$
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$\tan^{-1} x + \sin ^{-1} y=C$
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$\tan^{-1} x . \sin ^{-1} y=C$
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$\tan^{-1} x - \sin ^{-1} y=C$

Explanation

$\dfrac{dy}{dx}+ \dfrac {1+y^2}{\sqrt {1-x^2}}=0$

$\displaystyle \Rightarrow \dfrac{dy}{1+y^2}+ \dfrac {dx}{\sqrt {1-x^2}}=0$

$\Rightarrow \tan^{-1} y + \sin ^{-1} x=C$

Differential equation

$\dfrac{dy}{dx}=f(x)g(x)$ can be solved by

separating variable

$\dfrac{dy}{g(y)}=f(x)dx$

$\dfrac{dy}{dx}=1+x+y+xy$ and

$y(-1)=0$ , then

$y$ is equal to

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$e^{\dfrac{(1-x)^1}{2}}$
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$e^{\dfrac{(1+x)^2}{2}} -1$
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In $(1+x)-1$
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$1+x$

Explanation

$\dfrac{dy}{dx}=1+x+y+xy$

$\dfrac{dy}{dx}=(1+x).(1+y)$

$\dfrac {dy}{(1+y)}=(1+x)dx$

$\ln (1+y)=x+\dfrac {x^2}{2}+c$

$x=-1, y=0$

$\ln (1+0)=-1+\dfrac {1}{2}+c$

$0=\dfrac {-1}{2}+c \Rightarrow c=\dfrac12$

$\ln (1+y)=x+\dfrac {x^2}{2}+\dfrac 12$

$\ln (1+y)=\dfrac {(x+1)^2}{2}$

$y=e^{\dfrac {(x+1)^2}{2}}-1$

The solution of the differential equation

$\dfrac{x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+ \dots}{1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dots}=\dfrac{dx-dy}{dx+dy}$ is

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$2ye^{2x}=Ce^{2x}+1$
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$2ye^{2x}=Ce^{2x}-1$
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$ye^{2x}=Ce^{2x}+2$
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None of these

Explanation

Applying componendo and divedendo we get

$\dfrac{dy}{dx}=\dfrac{e^{-x}}{e^x}=e^{-2x}$

$\implies 2y=-e^{-2x}+C$

$\implies 2ye^{2x}=Ce^{2x}-1$

Integrating factor of the differential equation

$(1-x^2)\dfrac {dy}{dx}-xy=1$ is

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$-x$
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$\dfrac {x}{1+x^2}$
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$\sqrt {1-x^2}$
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$\dfrac 12 \log (1-x^2)$

Explanation

Given that,

$(1-x^2) \dfrac{dy}{dx}-xy=1$
Dividing both sides by

$(1-x^2)$ , we get

$\Rightarrow \dfrac{dy}{dx}-\dfrac{x}{1-x^2}y=\dfrac{1}{1-x^2}$ Which is a linear differential equation.

$\therefore IF=e^{-\displaystyle \int{\dfrac{x}{1-x^2}}dx}$
Put

$1-x^2=t \Rightarrow -2xdx=dt \Rightarrow xdx =-\dfrac{dt}{2}$
Now,

$IF =e^{\dfrac 12 \displaystyle \int{\dfrac {dt}{t}}}=e^{\dfrac 12 \log x}=e^{\dfrac 12 \log (1-x^2)}=\sqrt{1-x^2}$

Which of the following is a second order differential equation?

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$(y')^2+x=y^2$
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$y'y''+y=\sin x$
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$y''' +(y'')^2+y=0$
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$y'=y^2$

Explanation

The second order differential equation is

$y'y"+y=\sin x$ .

$y=x$ is a particular solution of the differential equation

$\dfrac {d^2y}{dx^2}-x^2 \dfrac {dy}{dx}+xy=x$ .

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True
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False

Explanation

False,

$y=x$ because does not satisfy the given differential equation.

$x+y=\tan^{-1}y$ is a solution of differential equation

$y^2 \dfrac {dy}{dx}+y^2+1=0$ .

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True
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False

Explanation

True,

$x+y=\tan^{-1}y\Rightarrow 1+\dfrac {dy}{dx}=\dfrac {1}{1+y^2}\dfrac {dy}{dx}$

$\Rightarrow \ \dfrac {dy}{dx}\left(\dfrac {1}{1+y^2}-1\right)=1$ , i.e.,

$\dfrac {dy}{dx}=\dfrac {-(1+y^2)}{y^2}$ which satisfies the given equation.

State true or false.

The solution of

$\dfrac{dy}{dx}=\left(\frac{y}{x}\right)^{\tfrac{1}{3}}$ is

$y^{\tfrac{2}{3}}-x^{\tfrac{2}{3}}=C$ .

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True
0%
False

Explanation

Given differential equation

$\dfrac{dy}{dx}=\left(\dfrac{y}{x}\right)^{{1}/{3}}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{y^{{1} / {3}}}{x^{{1} / {3}}}$

$\Rightarrow y^{{-1} / {3}} dy = x^{{1} / {3}} dx$

On integrating both sides, we get

$\displaystyle \int y^{{-1} / {3}} dy =\int x^{{1} / {3}} dx$

$\Rightarrow \dfrac{y^{{-1} / {3+1}}}{\tfrac{-1}{3}+1}=\dfrac{x^{{1} / {3+1}}}{\tfrac{-1}{3}+1} + C^{'}$

$\Rightarrow \dfrac{2}{3}y^{2/3}=\dfrac{2}{3}x^{2/3} +C^{'}$

$\Rightarrow y^{2/3}-x^{2/3} =C^{'}$

$[where,\ \dfrac{2}{3}C^{'}=C]$

The order of the differential equation

$(2 x^{2} \cfrac{d^{2} y}{d x^{2}}-3 \cfrac{d y}{d x}+y=0)$

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$2$
0%
$1$
0%
$0$
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Not defined

Explanation

Order: Order of a differential equation is the order of the highest order derivative (also known as differential coefficient) present in the equation

Here the highest order derivative is of order

$2$ .

hence the order is

$2$ .

hence the correct answer is (A)

The solution of the differential equation

$\dfrac {dy}{dx}=\dfrac {1+y^2}{1+x^2}$ is :

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$y=\tan^{-1}x$
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$y-x=k(1+xy)$
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$x=\tan^{-1}y$
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$\tan ( xy)=k$

Explanation

Given that,

$\dfrac{1+y^2}{1+x^2}$

$\Rightarrow \dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}$

On integrating both sides, we get

$\displaystyle \int{\dfrac{dy}{1+y^2}}=\displaystyle \int{\dfrac{dx}{1+x^2}}$

$\tan^{-1}y=\tan^{-1}x+C$

$\Rightarrow \tan^{-1}y-\tan^{-1}x=C$

$\Rightarrow \tan^{-1}\left( \dfrac{y-x}{1+xy}\right)=C$

$\Rightarrow \dfrac{y-x}{1+xy}=\tan C$

$\Rightarrow y-x=\tan C (1+xy)$

$\Rightarrow y-x=K(1+xy)$

Where,

$k=\tan C$

Which of the following is the general solution of

$\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0$ ?

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$y=(Ax+B) e^x$
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$y=(Ax+B) e^{-x}$
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$y=Ae^x +Be^{-x}$
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$y=A\cos x+B\sin x$

Explanation

Given that,

$\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0$

$D^2y-2Dy+y=0$ ,

Where,

$D=\dfrac{d}{dx}$

$(D^2-2D+1)y=0$

The auxiliary equation is

$m^2-2m+1=0$

$(m-1)^2=0 \Rightarrow m=1, 1$

Since, the roots are real and equal.

$\therefore CF=(Ax+B)e^x\\ \Rightarrow y=(Ax+B)e^x$

[ since, if roots of Auxiliary equation are real and equal say (m), then

$CF=(C_1x+C_2)e^{mx}]$

Solution of

$\displaystyle \sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\frac{dy}{dx}=0$ , is:

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$\log\left(\displaystyle \frac{x}{1+\sqrt{1+x^{2}}}\right)+\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$
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$\displaystyle \log\left(\frac{x}{\sqrt{1+x^{2}}}\right)+\sqrt{1-x^{2}}+\sqrt{1+y^{2}}=c$
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$\displaystyle \log\left(\frac{x}{\sqrt{1+x^{2}}}\right)=c$
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$\displaystyle \log\left(\sqrt{1+x^{2}}-\sqrt{1+y^{2}}\right)+\log\left(\frac{x}{\sqrt{1+x^{2}}}\right)=c$

Explanation

Given,

$\displaystyle \sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\dfrac{dy}{dx}=0$

$\Rightarrow \sqrt{1+y^{2}}\sqrt{1+x^{2}} = -xy\dfrac{dy}{dx}$

Integrating both sides, we get

$\Rightarrow \displaystyle\int {\dfrac{\sqrt{1+x^{2}}}{x}dx} = -\displaystyle\int{\dfrac{y}{\sqrt{1+y^{2}}}}dy$
Substitute,

$1+ y^{2}=t^{2}$

$\Rightarrow$

$2y dy = 2t dt$

$\Rightarrow$

$\displaystyle\int {\dfrac{\sqrt{1+x^{2}}}{x}dx} = - \displaystyle\int dt=- t+c = -\sqrt{1+y^{2}} +c$
Now substitute,

$1+x^{2}= k^{2}$

$\Rightarrow$

$2x dx = 2k dk$

$\Rightarrow \displaystyle\int\dfrac{k^{2}}{x^{2}}dk = -\sqrt{1+y^{2}} +c$

$\Rightarrow \displaystyle\int\dfrac{k^{2}}{k^{2}-1}dk = -\sqrt{1+y^{2}} +c$

$\Rightarrow \displaystyle\int\left[\dfrac{k^{2}-1}{k^{2}-1}+\dfrac{1}{k^{2}-1}\right]dk = -\sqrt{1+y^{2}} + c$

$\Rightarrow k +\dfrac{1}{2}\log{\dfrac{k-1}{k+1}} = -\sqrt{1+y^{2}} +C$

Since ,

$\displaystyle\int\dfrac{1}{x^{2}-d^{2}}dx=\dfrac{1}{2d}\log{\dfrac{x-d}{x+d}}+\mbox{constant}$

$\therefore \sqrt{1+x^{2}} +\dfrac{1}{2}\log{\dfrac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}}+\sqrt{1+y^{2}}= C$

$\Rightarrow \sqrt{1+x^{2}} +\log\left(\dfrac{x}{\sqrt{1+x^{2}}+1}\right)+\sqrt{1+y^{2}}= C$

$f (x)$ and

$g (x)$ are two solutions of the differential equation

$a\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$ , then

$f (x) - g (x)$ is the solution of

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$a^{2}\displaystyle \frac{d^{2}y}{dx^{2}}+\displaystyle \frac{dy}{dx}+y=e^{\displaystyle x}$
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$\displaystyle {a}^{2}\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$
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$a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=0$
0%
$a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$

Explanation

Given, Differential equation as

$a\dfrac{\mathrm{d^2} y}{\mathrm{d} x^2}+x^2\dfrac{\mathrm{d} y}{\mathrm{d} x}+y=e^x$ has

$f(x),g(x)$ as two solutions.

On substituting

$f(x)$ in the given Differential equation,

$\Rightarrow af''(x)+x^2f'(x)+f(x)=e^x ................ A$

On substituting

$g(x)$ in the given differential equation,

$\Rightarrow ag''(x)+x^2g'(x)+g(x)=e^x ................B$

Subtracting B from A,

$\Rightarrow a\times (f''(x)-g''(x))+x^2\times(f'(x)-g'(x))+(f(x)-g(x))=0$

It is in the form of

$a\dfrac{\mathrm{d^2} y}{\mathrm{d} x^2}+x^2\dfrac{\mathrm{d} y}{\mathrm{d} x}+y=0$ , which have a solution of

$y=f(x)-g(x)$

Solution of

$\dfrac{d^{2}y}{dx^{2}}$ =

$\log x$ is:

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$y=\dfrac{1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$
0%
$y=\dfrac{1}{2}x^{2}\log x+\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$
0%
$y=\dfrac{-1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$
0%
$y=3x^{2}+4x+c_{1}$

Explanation

$\dfrac{d^{2}y}{dx^{2}}=\log x$

$\Rightarrow \dfrac{d}{dx}y_{1}=\log x$

$\Rightarrow \int dy_{1}=\int \log x dx+c_{1}$

$\int \log dx=x\log x-x+{ c }_{ 1 }$

$\Rightarrow y_{1}=x\log x-x+c_{1}$

$\int dy = \int (x \log x-x+c_{1})dx+c_{2}$

$y=\int x\log xdx-\dfrac{x^{2}}{2}+c_{1}x+c_{2}$

$\int x\log xdx=x(x\log x-x)-\int(x\log x-x)dx$

$\rightarrow 2\int x\log xdx=x^{2}\log x-x^{2}+x^{2}/_2=x^{2}\log x^-x^{2}/_2$

$\Rightarrow \int x\log xdx=\dfrac{x^{2}\log x}{2}-x^{2}/_4$

$\Rightarrow y=\dfrac{x^{2}\log x}{2}-\dfrac {3x^{2}}{4}+c_{1}x+c_{2}$

lf the solution of the differential equation

$\displaystyle \frac{1}{\sin^{-1}x}(\frac{dy}{dx})=1$ is

$y=x \sin^{-1}x+f(x)+c$ then f (x) is

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$1-x^{2}$
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$\sqrt{1-x^{2}}$
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$1+x^{2}$
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$\sqrt{1+x^{2}}$

The solution of

$\displaystyle \frac{dy}{dx}+\frac{x(1+y^{3})}{y^{2}(1+x^{2})}=0$ is:

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$(1-x^{2})+(1+y^{3})=c$
0%
$(1+y^{2})+(1+x^{3})=c$
0%
$(1+x^{2})^{3}(1+y^{3})^{2}=c$
0%
$(1+x^{2})+(1+y^{3})=cxy^{2}$

Explanation

$\dfrac{dy}{dx}+\dfrac{x\left ( 1+y^{3} \right )}{y^{2}\left ( 1+x^{2} \right )}=0$

$\Rightarrow \dfrac{1}{3}\left ( \dfrac{3y^{2}}{1+y^{3}}dy \right )+\dfrac{1}{2}\left ( \dfrac{2x}{1+x^{2}}dx \right )=0$

put

$1+x^{2}=t;$

$1+y^{3}=k$

$\Rightarrow 2xdx=dt$

$\Rightarrow 3y^{2}dy=dk$

$\displaystyle\Rightarrow \dfrac{1}{3}\int \dfrac{dk}{k}+\dfrac{1}{2}\int \dfrac{dt}{t}=c$

$\Rightarrow \log k^{\frac{1}{3}}t^{\frac{1}{2}}=c$

$\Rightarrow \log k^{2}+3=log c$

$\left ( 1+y^{3} \right )^{2}\left ( 1+x^{2} \right )^{3}=c$

A tangent drawn to the curve

$\mathrm{y}=\mathrm{f}(\mathrm{x})$ at

$\mathrm{P}(\mathrm{x}, \mathrm{y})$ cuts the

$\mathrm{x}$ -axis and

$\mathrm{y}$ -axis at A and

$\mathrm{B}$ respectively such that BP: AP

$=3$ : 1, given that

$\mathrm{f}(1)=1$ , then

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equation of curve is $\displaystyle \mathrm{x}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-3\mathrm{y}=0$
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normal at (1, 1) is $\mathrm{x}+3\mathrm{y}=4$
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curve passes through (2, 1/8)
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equation of curve is $\displaystyle \mathrm{x}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+3\mathrm{y}=0$

Explanation

Given

$BP:AP=3:1.$ Then equation of tangent is

$\ Y-y=f'\left( x \right) \left( X-x \right)$

The intercept on the coordinate axes are

$\displaystyle\ A\left( x-\frac { y }{ f'\left( x \right) } =0 \right)$ and

$\ B\left[ 0,y-xf'\left( x \right) \right]$

Since,

$P$ is internally intercepts a line

$AB,$

$\left [x=\left (\dfrac{mx_1+nx_2} {m+n} \right )\right ]$ by using this formula

$\displaystyle\ \therefore x=\frac { 3\left( x-\dfrac { y }{ f'\left( x \right) } \right) +1\times 0 }{ 3+1 }$

$\displaystyle\ \Rightarrow \frac { dy }{ dx } =\frac { y }{ -3x }$

$\displaystyle\ \Rightarrow \frac { dy }{ y } =-\frac { 1 }{ 3x } dx$

On integrating both sides , we get

$\ x{ y }^{ 3 }=C$

Since, curve passes through

$\\ \left( 1,1 \right) , then\quad c=1$

$\ \therefore x{ y }^{ 3 }=1$

$\displaystyle\ \therefore$ At

$\displaystyle x=\frac { 1 }{ 8 } \Rightarrow y=2$

The solution of

$(1-x^{2})\displaystyle \frac{dy}{dx}+xy=xy^{2}$ is:

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$y(c+\sqrt{1-x^{2}})=1-x^{2}$
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$y(c-\sqrt{1-x^{2}})=1-x^{2}$
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$y(c+\sqrt{1-x^{2}})=\sqrt{1-x^{2}}$
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$y(c-\sqrt{1-x^{2}})=\sqrt{1-x^{2}}$

Explanation

$\dfrac{dy}{dx}+\dfrac{xy}{1-x^{2}}=\dfrac{xy^{2}}{1-x^{2}}$

$\dfrac{dy}{dx}=\dfrac{xy(y-1)}{1-x^{2}}$

$\Rightarrow \dfrac{dy}{y(y-1)}=\dfrac{x}{1-x^{2}}dx$

$\Rightarrow\displaystyle \int \left(\frac{1}{y-1}-\dfrac{1}{y}\right)dy=-\dfrac{1}{2}\int \dfrac{-2x}{1-x^{2}}dx+c_{1}$

$\Rightarrow \log\left(\dfrac{y-1}{y}\right)=-\dfrac{1}{2}\log(1-x^{2})+\log{c_{1}}$

$\Rightarrow \dfrac{y-1}{y}=\dfrac{c_{1}}{\sqrt{1-x^{2}}}$

$\Rightarrow y\sqrt{1-x^{2}}-\sqrt{1-x^{2}}=c_{1}y$

$\Rightarrow y(\sqrt{1-x^{2}}-c_{1})=\sqrt{1-x^{2}}$

Put

$c_{1}=-c$

$\Rightarrow y(c+\sqrt{1-x^{2}})=\sqrt{1-x^{2}}$

Solution of

$\>y-x\displaystyle \frac{dy}{dx}=5\left(y^{2}+\frac{dy}{dx}\right)$ , is:

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$(5+x)(1-5y)=cy$
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$\dfrac { 5+x }{ 1-5y } =cy$
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$(5-x)(1+5y)=cy$
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$(5+x)(5-x)=c$

Explanation

$5y^{2} - y = -\dfrac{dy}{dx} (5+x)$

$\Rightarrow -\dfrac{dx}{5+x} = \dfrac{dy}{y(5y-1)}$

$\Rightarrow -\dfrac{dx}{5+x} = \dfrac{5y - (5y-1)}{y(5y-1)}dy$

Integrating both sides, we get

$\Rightarrow\displaystyle\int\dfrac{dx}{5+x} = \int\left(\dfrac{5}{1-5y} + \frac{1}{y}\right)dy$

$\Rightarrow \log(5+x) = -\log(1-5y)+\log y + \log c; (\log c$ is a constant

$)$

$\Rightarrow yc = (5+x)(1-5y)$

Solve :

$\dfrac{dy}{dx}=\dfrac{x\left ( 2\ln x+1 \right )}{\sin y+y\cos y}$

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$y\sin y=x^{2}\ln x+c$
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$y\sin x=y^{2}\ln x+c$
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$y\sin y=x^{2}\ln x-c$
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$y\sin x=y^{2}\ln x-c$

Explanation

$\displaystyle \frac { dy }{ dx } =\frac { x\left( 2\log { x } +1 \right) }{ \sin { y } +y\cos { y } } \Rightarrow \frac { dy }{ dx } \left( \sin { y } +y\cos { y } \right) =x+2x\log { x }$
Integrating both sides

$\displaystyle \int { \frac { dy }{ dx } \left( \sin { y } +y\cos { y } \right) dx } =\int { \left( x+2x\log { x } \right) dx } \\ \Rightarrow y\sin { y } ={ x }^{ 2 }\log { x } +c$

$\dfrac{dy}{dx}=xy+2x+3y+6$ , then find the value of

$y(-1)-e^2y(-3)$ .

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$e^2-1$
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$e^2+1$
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$2(e^2-1)$
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$-2(e^2-1)$

Explanation

$\dfrac{dy}{dx}=(x+3)(y+2)$ , If

$X=x+3, Y=y+2$

$\dfrac{dY}{Y}=XdX$

$\Rightarrow Y=A. e^{X^2/2}, y=-2 +Ae^{(x+3)^2/2}$

$y(-1)=-2+Ae^2, y(-3)=-2+A$

$y(-1)-e^2 y(-3)=-2 +2e^2 =2(e^2-1)$

Solve:

$y-x\dfrac{dy}{dx}=a\left ( y^{2}+\dfrac{dy}{dx} \right )$

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$y=c(1-ay)(x+a)$
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$2y=c(1-ay)(x+a)$
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$y=c(1-ay)(x+a)/2$
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$2y=c(1+ay)(x+a)$

Explanation

Given,

$y-\dfrac{xdy}{dx}=ay^{2}+\dfrac{a.dy}{dx}$

$y-ay^{2}=(x+a).\dfrac{dy}{dx}$

$\dfrac{dx}{x+a}=\dfrac{dy}{y(1-ay)}$

$\int \dfrac{dx}{x+a}=\int \dfrac{1-ay+ay}{y(1-ay)}.dy$

$ln(x+a)=\int \dfrac{1}{y}+\dfrac{a}{1-ay}.dy$

$ln(x+a)=ln(y)-ln(1-ay)+lnc$

$ln(x+a)=ln(\dfrac{cy}{1-ay})$

$(x+a)=\dfrac{cy}{1-ay}$ .

$\dfrac{dy}{dx}+\dfrac{\sqrt{\left ( x^{2}-1 \right )\left ( y^{2}-1 \right )}}{xy}=0$
Solve the above equation:

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$\sqrt{x^{2}-1}-\sec ^{-1}x+\sqrt{y^{2}-1}=c$
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$\sqrt{x^{2}-1}-\sec ^{-1}x+\sqrt{y^{2}-1}=-c$
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$\sqrt{x^{2}-1}+\sec ^{-1}x+\sqrt{y^{2}-1}=c$
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$\sqrt{x^{2}-1}-\sec ^{-1}x-\sqrt{y^{2}-1}=c$

Explanation

$\displaystyle \frac { dy }{ dx } =\frac { \sqrt { { x }^{ 2 }-1 } \sqrt { { y }^{ 2 }-1 } }{ xy } \Rightarrow \frac { y }{ \sqrt { { y }^{ 2 }-1 } }dy =\frac { \sqrt { { x }^{ 2 }-1 } }{ x } dx$

$\displaystyle \Rightarrow \int { \frac { y }{ \sqrt { { y }^{ 2 }-1 } } dy } =\int { \frac { \sqrt { { x }^{ 2 }-1 } }{ x } dx }$

$\displaystyle \Rightarrow -\sqrt { { y }^{ 2 }-1 } =\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) } +\sqrt { { x }^{ 2 }-1 } +c\\ \Rightarrow \sqrt { { x }^{ 2 }-1 } +\sqrt { { y }^{ 2 }-1 } -\sec ^{ -1 }{ x } =c$

The general solution of differential equation

$x\left ( 1+y^{2} \right )\mathrm{d} x+y\left ( 1+x^{2} \right )\mathrm{d} y=0$ is/are:

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$\displaystyle \frac{1}{2} \ln\left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=k$
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$\displaystyle \left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=c$
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$\displaystyle \left ( 1+y^{4} \right )=c\left ( 1+x^{2} \right )$
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$\displaystyle \ln\left [\frac{(1+x^{2})}{ (1+y^{2}) } \right ]= k$

Explanation

Given equation is

$x\left ( 1+y^{2} \right )\mathrm{d} x+y\left ( 1+x^{2} \right )\mathrm{d} y=0$

$\Rightarrow x\left ( 1+y^{2} \right )\mathrm{d} x=-y\left ( 1+x^{2} \right )\mathrm{d} y$

integrating on both sides

$\Rightarrow\displaystyle \int \frac{x}{1+x^{2}}\mathrm{d} x=-\int \frac{y}{1+y^{2}}\mathrm{d} y$

Put

$1+{ x }^{ 2 }=t \Rightarrow 2xdx=dt$

and

$1+y^{ 2 }=u \Rightarrow 2ydy=du$

$\displaystyle \frac { 1 }{ 2 } \int { \frac { dt }{ t } } =-\frac { 1 }{ 2 } \int { \frac { du }{ u } }$

$\Rightarrow\displaystyle \frac{1}{2}ln\left ( 1+x^{2} \right )=-\frac{1}{2}ln\left ( 1+y^{2} \right )+k$

$\Rightarrow ln\left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=2k$

$\Rightarrow \left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=e^{2k}$

If the length of tangent at any point on the curve

$y=f(x)$ intercepted between the points and the x-axis is of length

$1$ . Find the equation of the curve.

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$\displaystyle \sqrt { 1-{ y }^{ 2 } } -\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =x+c$
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$\displaystyle \sqrt { 1-{ y }^{ 2 } } -\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =-x+c$
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$\displaystyle \sqrt { 1-{ y }^{ 2 } } +\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =x+c$
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$\displaystyle \sqrt { 1-{ y }^{ 2 } } +\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =-x+c$

Explanation

Since, the length of tangent

$\displaystyle =\left| y\sqrt { 1+{ \left( \frac { dx }{ dy } \right) }^{ 2 } } \right| =1$

$\displaystyle \Rightarrow { y }^{ 2 }\left( 1+{ \left( \frac { dx }{ dy } \right) }^{ 2 } \right) =1$

$\displaystyle\therefore \frac { dy }{ dx } =\pm \frac { y }{ \sqrt { 1-{ y }^{ 2 } } }$

$\displaystyle \Rightarrow \int { \frac { \sqrt { 1-{ y }^{ 2 } } }{ y } } dy=\pm \int { x } dx$

$\displaystyle\Rightarrow \int { \frac { \sqrt { 1-{ y }^{ 2 } } }{ y } } dy=\pm x+c$

Substitute

$y=\sin { \theta } \Rightarrow dy=\cos { \theta } d\theta$

$\displaystyle \therefore \int { \frac { \cos { \theta } }{ \sin { \theta } } } .\cos { \theta } d\theta =\pm x+c\Rightarrow \int { \frac { \cos ^{ 2 }{ \theta } }{ \sin ^{ 2 }{ \theta } } } .\sin { \theta } d\theta =\pm x+c$

Again substitute

$\cos { \theta } =t\Rightarrow -\sin { \theta } d\theta =dt$

$\displaystyle \therefore \int { \frac { { t }^{ 2 } }{ 1-{ t }^{ 2 } } } dt=\pm x+c,\Rightarrow \int { \left( 1-\frac { 1 }{ 1-{ t }^{ 2 } } \right) } dt=\pm x+c$

$\displaystyle \Rightarrow t-\log { \left| \frac { 1+t }{ 1-t } \right| = } \pm x+c\Rightarrow \sqrt { 1-{ y }^{ 2 } } -\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =\pm x+c$

The solution of differential equation

$(e^x + 1) y dy = (y + 1) e^x dx$ is :

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$(e^x + 1) (y + 1) = ce^y$
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$(e^x + 1) (y + 1) = ce^{-y}$
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$(e^x + 1) (y + 1) = ce^{2y}$
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none of the above

Explanation

The given differential equation is

$(e^x + 1) y dy = (y + 1) e^x dx$

$\Rightarrow \displaystyle \frac{ydy}{(y+1)}=\displaystyle \frac{e^{x}}{(e^{x}+1)}dx;$ Integrating both sides

$\Rightarrow y- log |y + 1| = log (e^x + 1) + log k$

$\Rightarrow y = log |(y + 1)(e^x + 1)| + log k$

$\Rightarrow (y + 1)(e^x + 1) = ce^y$

The solution of differential equation

$(e^x + 1)y dy = (y + 1) e^x dx$ is:

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$(e^x + 1) (y + 1) = c . e^y$
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$(e^x + 1) |(y + 1)| = ce^{-y}$
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$(e^x + 1) (y + 1) = c$
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none of these

Explanation

The given differential equation is

$(e^x + 1) y . dy = (y + 1) . e^x . dx$

$\Rightarrow \displaystyle \frac{y.dy}{(y+1)}=\displaystyle \frac{e^{x}}{(e^{x}+1)}.dx$

$\Rightarrow \int\left ( 1-\displaystyle \frac{1}{y+1} \right ).dy=\int \displaystyle \frac{e^{x}}{e^{x}+1}.dx$

$\Rightarrow y - log |y +1| = log (e^x + 1) + log k$

$\Rightarrow (y+1) (e^x+1)=e^y.c$

Solution of

$\displaystyle \frac { dy }{ dx } =\frac { y\left( x\log { y } -y \right) }{ x\left( y\log { x } -x \right) }$ is:

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${ x }^{ y }=c{ y }^{ x }$
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$xy=c$
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${ \left( xy \right) }^{ x }=c$
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None of these

Explanation

$\displaystyle \frac { dy }{ dx } =\frac { y\left( x\log { y-y } \right) }{ x\left( y\log { x-x } \right) }$

$\displaystyle \Rightarrow x\left( y\log { x-x } \right) \frac { dy }{ dx } =y\left( x\log { y-y } \right)$

$\displaystyle \Rightarrow \left( \log { x } -\frac { x }{ y } \right) \frac { dy }{ dx } =\log { y } -\frac { y }{ x }$

$\displaystyle \Rightarrow \frac { y }{ x } +\log { x } .\frac { dy }{ dx } =\log { y } +\frac { x }{ y } \frac { dy }{ dx }$

$\displaystyle \Rightarrow \frac { d }{ dx } \left( x\log { x } \right) =\frac { d }{ dx } \left( x\log { y } \right)$

$\Rightarrow y\log { y } =x\log { y } +\log { c } \Rightarrow \log { { x }^{ y } } =\log { { y }^{ x } } +\log { c } \Rightarrow { x }^{ y }=c{ y }^{ x }.$

$\int_{a}^{x} ty (t) dt = x^{2} + y(x)$ then y as a function of x is:

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$y = 2 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$
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$y = 1 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$
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$y = 2 - (1 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$
0%
none

Explanation

Given

$\int_{a}^{x} ty(t)dt = x^{2} + y(x)$ .....(1)

$\Rightarrow xy = 2x + \displaystyle \frac{dy}{dx}$

$\Rightarrow x(y - 2) = \displaystyle \frac{dy}{dx}$

$\Rightarrow \int xdx = \int \displaystyle \frac{dy}{y - 2}$

$\Rightarrow \displaystyle \frac{x^{2}}{2} = ln | y - 2 | + lnc$

$\Rightarrow e^{\displaystyle \frac{x^{2}}{2}} = c(y - 2)$ ....(2)

Now, put

$x = a$ in eqn (1)

$\Rightarrow y = -a^{2}$
So, by eqn (2)

$\therefore e^{ \frac { a^{ 2 } }{ 2 } }=c(-a^{ 2 }-2)$

$\Rightarrow c= -\dfrac { e^{ \frac { a^{ 2 } }{ 2 } } }{ (a^{ 2 }+2) }$
Put this value in (2)

$\therefore e^{\displaystyle \frac{x^{2}}{2}} = - \displaystyle \frac{e^{\frac{a^{2}}{2}}}{(a^{2} + 2)} (y - 2)$

$\Rightarrow -y + 2 = (a^{2} + 2) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$

$\Rightarrow y = 2 - (2 + a^{2}) e^{\dfrac{x^{2} -a^{2}}{2}}$

Through any point

$\left ( x,y \right )$ of a curve which passes through the origin, lines are drawn parallel to the coordiante axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of

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circles
0%
parabolas
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hyperbolas
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straight lines

Explanation

Let

$P\left ( x,y \right )$ be the point on the curve passing through the origin

$O\left ( 0,0 \right )$ ,
and let

$PN$ and

$PM$ be the lines parallel to the

$x$ - and

$y$ -axes, respectively. If the equation of the curve is

$y= y\left ( x \right )$ ,
the area

$POM$ equals

$\displaystyle \int_{0}^{x}ydx$ and the area

$PON$ equals

$xy-\displaystyle \int_{0}^{x}ydx$ .
Assuming that

$2\left ( POM \right )= PON$ , we therefore have

$2\displaystyle \int_{0}^{x}ydx= xy-\displaystyle \int_{0}^{x}ydx\Rightarrow 3\displaystyle \int_{0}^{x}ydx= xy$ .
Differentiating both sides of this gives

$3y= x\displaystyle \frac{dy}{dx}+y\Rightarrow 2y= x\displaystyle \frac{dy}{dx}\Rightarrow \displaystyle \frac{dy}{y}=2\displaystyle \frac{dx}{x}$

$\Rightarrow \log \left | y \right |= 2\log \left | x \right |+C\Rightarrow y= Cx^{2}$ with

$C$ being a constant.
This solution represents a parabola.

The solutions of

$v=u\displaystyle\frac{dv}{du}+\left (\displaystyle\frac{dv}{du} \right )^{2}$ where

$u = y$ and

$v=xy$ are:

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$y=0$
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$y=-4x$
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$xy=cy+c^{2}$
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$x^{2}y=cy+c^{2}$

Explanation

$\displaystyle v=u\frac { dv }{ du } +{ \left( \frac { dv }{ du } \right) }^{ 2 }$

Differentiating w.r.t

$u$ , we get

$\displaystyle \frac { dv }{ du } =\frac { dv }{ du } +u\frac { { d }^{ 2 }v }{ d{ u }^{ 2 } } +2\frac { dv }{ du } \frac { { d }^{ 2 }v }{ d{ u }^{ 2 } }$

$\displaystyle \Rightarrow \left( u+2\frac { dv }{ du } \right) \frac { { d }^{ 2 }v }{ d{ u }^{ 2 } } =0$

$\displaystyle \Rightarrow \frac { dv }{ du } =c$ or

$\displaystyle \frac { dv }{ du } =-\frac { u }{ 2 }$

Putting

$\displaystyle \frac { dv }{ du } =c$ , we get

$v=cu+{ c }^{ 2 }\Rightarrow xy=cy+{ c }^{ 2 }$

Again putting

$\displaystyle \frac { dv }{ du } =-\frac { u }{ 2 }$ , we get

$\displaystyle v=-\frac { { u }^{ 2 } }{ 4 } \Rightarrow { y }^{ 2 }=-4xy\Rightarrow y=0$ or

$y=-4x$

$\displaystyle y-x\frac{dy}{dx}=b\left ( 1+x^{2}\frac{dy}{dx} \right ).$
Solve the above differential equation.

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$\displaystyle y=k\left ( y-b \right )\left ( 1+bx \right ).$
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$\displaystyle y=k\left ( y+b \right )\left ( 1-bx \right ).$
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$\displaystyle y=k\left ( y+b \right )\left ( 1+bx \right ).$
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$\displaystyle y=k\left ( y-b \right )\left ( 1-bx \right ).$

Explanation

After rearranging the terms, we get

$y-b=\dfrac{dy}{dx}(bx^{2}+x)$

$\dfrac{dx}{x(bx+1)}=\dfrac{dy}{y-b}$

$\int \dfrac{bx+1-bx}{x(bx+1)}.dx=ln(y-b)$

$\int \dfrac{1}{x}-\dfrac{b}{bx+1}.dx=ln(y-b)$

$ln(x)-ln(bx+1)+ln(c)=ln(y-b)$

$\dfrac{cx}{bx+1}=y-b$

The solution of

$\displaystyle\frac{dy}{dx}+x=xe^{(n-1)y}$ is:

0%
$\displaystyle\frac{1}{n-1}\log \left (\displaystyle\frac{e^{(n-1)y}-1}{e^{(n-1)y}} \right )=x^{2}/2+C$
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$e^{(n-1)y}=Ce^{(n-1)y+(n-1)x^{2/2}}+1$
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$\log \left (\displaystyle\frac{e^{(n-1)y}-1}{(n-1)e^{(n-1)y}} \right )=x^{2}+C$
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$e^{(n-1)}y=ce^{(n-1)y^{2/2+x}}+1$

Explanation

Given

$\displaystyle \frac { dy }{ dx } +x=x{ e }^{ y\left( n-1 \right) }$

$\Rightarrow \dfrac { dy }{ dx } =x\left( { e }^{ y\left( n-1 \right) }-1 \right)$

$\displaystyle \Rightarrow \frac { dy }{ \left( { e }^{ y\left( n-1 \right) }-1 \right) } =xdx$

integrating on both sides

$\displaystyle \Rightarrow \frac { 1 }{ n-1 } \int { \frac { \left( n-1 \right) }{ \left( { e }^{ y\left( n-1 \right) }-1 \right) { e }^{ y\left( n-1 \right) } } { e }^{ y\left( n-1 \right) } } dy=\frac { x^{ 2 } }{ 2 } +c$

$\displaystyle \Rightarrow { e }^{ y\left( n-1 \right) }=c{ e }^{ \left( n-1 \right) +\left( n-1 \right) \frac { x^{ 2 } }{ 2 } }+1$

The general solution of

$xy^{5}=y_{4}(y_{n}=\displaystyle\frac{d^{n}y}{dx^{n}})$ is given by:

0%
${ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{3}+C_{3}x^{2}+C_{4}x+C_{5}$
0%
${ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{4}+C_{3}x^{2}+C_{4}x+C_{5}$
0%
${ y }^{ -1 }=C_{1}+C_{2}x+C_{3}x^{2}+C_{4}x^{3}+C_{5}x^{4}$
0%
none of these

Explanation

$\displaystyle x{ y }^{ 5 }=\frac { { d }^{ 4 }y }{ d{ x }^{ 4 } } \Rightarrow \frac { { d }^{ 4 }y }{ d{ x }^{ 4 } } { y }^{ -5 }=x$

$\displaystyle \Rightarrow \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \frac { { y }^{ -4 } }{ -4 } =\frac { { x }^{ 2 } }{ 2 } +c$

$\displaystyle \Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \frac { { y }^{ -3 } }{ 12 } =\frac { { x }^{ 3 } }{ 6 } +cx+d$

$\displaystyle \Rightarrow \frac { dy }{ dx } \frac { { y }^{ -2 } }{ -24 } =\frac { { x }^{ 4 } }{ 24 } +\frac { c{ x }^{ 2 } }{ 2 } +dx+e$

$\displaystyle \Rightarrow \frac { { y }^{ -1 } }{ 24 } =\frac { { x }^{ 5 } }{ 120 } +\frac { c{ x }^{ 3 } }{ 6 } +\frac { d{ x }^{ 2 } }{ 2 } +ex+f$

$\Rightarrow { y }^{ -1 }={ C }_{ 1 }{ x }^{ 5 }+{ C }_{ 2 }{ x }^{ 3 }+{ C }_{ 3 }{ x }^{ 2 }+{ C }_{ 4 }x+{ C }_{ 5 }$

The normal to a curve at

$P\left ( x,y \right )$ meets the

$x$ -axis at

$G$ . If the distance of

$G$ from the origin is twice the abscicca of

$P$ , then the curve is a:

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circle
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hyperbola
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ellipse
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parabola

Explanation

Equation of normal at

$P(x,y)$ is given by,

$\displaystyle Y-y=-\frac{dx}{dy}(X-x)$

$\displaystyle \Rightarrow G=\left( x+y\frac{dy}{dx},0\right)$
Now given condition is,

$\displaystyle \left|x+y\frac{dy}{dx}\right|=\left|2x\right|$

$\displaystyle \Rightarrow y\frac{dy}{dx}=x$ or

$\displaystyle y\frac{dy}{dx}=-3x$

$\Rightarrow \displaystyle ydy=xdx$ or

$ydy=-3xdx$
After integrating,we get

$\displaystyle \frac{y^{2}}{2}=\frac{x^{2}}{2}+c$ or

$\displaystyle \frac{y^{2}}{2}=-\frac{3x^{2}}{2}+c$

$\Rightarrow x^{2}-y^{2} = -2c$ or

$\displaystyle 3x^{2}+y^{2}=2c$

$\displaystyle \Rightarrow x^{2}-y^{2}=c_{1}$ or

$\displaystyle 3x^{2}+y^{2}=c_{2}$

The differential equation

$\displaystyle \frac{dy}{dx}= \displaystyle \frac{\sqrt{1-y^{2}}}{y}$ determines a family of circles with:

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variable radii and a fixed centre $\left ( 0,1 \right )$
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variable radii and a fixed centre $\left ( 0,-1 \right )$
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fixed radius $1$ and variable centres along the $x$ -axis
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fixed radius $1$ and variable centres along the $y$ -axis

Explanation

$\displaystyle \frac{dy}{dx}= \displaystyle \frac{\sqrt{1-y^{2}}}{y}$

$\Rightarrow \displaystyle \frac{y}{\sqrt{1-y^{2}}}\: dy= dx$
Integrating we have

$-\sqrt{1-y^{2}}= x+C$

$\Rightarrow \left ( x+C \right )^{2}+y^{2}= 1$ .
This represents a family of circles with variable centre

$\left ( -C,0 \right )$ (which lies on

$x$ -axis) and radius

$1$

The general solution of the equation

$\displaystyle \frac { dy }{ dx } =\frac { { x }^{ 2 } }{ { y }^{ 2 } }$ is:

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$x^3-y^3=c$
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$x^3+y^3=c$
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$x^2+y^2=c$
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$x^2-y^2=c$

Explanation

$\displaystyle \frac { dy }{ dx } =\frac { { x }^{ 2 } }{ { y }^{ 2 } } \Rightarrow { y }^{ 2 }dy={ x }^{ 2 }dx$

Integrating both sides

$\displaystyle \int { { y }^{ 2 }dy } =\int { { x }^{ 2 }dx } \Rightarrow \frac { { y }^{ 3 } }{ 3 } =\frac { { x }^{ 3 } }{ 3 } +\frac { c }{ 3 }$

$\Rightarrow { x }^{ 3 }-{ y }^{ 3 }=c$

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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