MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 10

Solution of the differential equation

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

0%
$\ln|\dfrac{x}{y}|+\dfrac{xy}{x-y}=c$
0%
$\dfrac{xy}{x-y}=ce^{x/y}$
0%
$\ln|xy|=c+\dfrac{xy}{x-y}$
0%
$none\ of\ these$

The solution of the differential equation

$x^2\dfrac{dy}{dx}\cos\left(\dfrac{1}{x}\right) - y \sin\left(\dfrac{1}{x}\right) = -1 ; where \ y \rightarrow -1 \ as \ x \rightarrow \infty$ is

0%
$y = \sin\left(\dfrac{1}{x}\right) - \cos\left(\dfrac{1}{x}\right)$
0%
$y = \dfrac{x+1}{x\sin\left(\dfrac{1}{x}\right)}$
0%
$y = \cos\left(\dfrac{1}{x}\right) + \sin\left(\dfrac{1}{x}\right)$
0%
$y = \dfrac{x+1}{x\cos\left(\dfrac{1}{x}\right)}$

Explanation

$\displaystyle x^{2}\frac{dy}{dx}cos\left ( \frac{1}{x} \right )- ysin\left ( \frac{1}{x} \right )=-1$

$\displaystyle \frac{dy}{dx}-\frac{y}{x^{2}}tan\left ( \frac{1}{x} \right )= -\frac{1}{x^{2}}cos \left ( \frac{1}{x} \right )$

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

$I.F = e^{\displaystyle\int -\frac{1}{x^{2}}tan\left ( \dfrac{1}{x} \right )dx}$

$\displaystyle \frac{1}{x}=t$

$\displaystyle -\frac{dx}{x^{2}}=dt$

$=e^{\displaystyle\int tan(t)dt}$

$\displaystyle =e^{ln(sect)}$

$\displaystyle =sec\left ( \dfrac{1}{x} \right )$

$\displaystyle sec\left ( \frac{1}{x} \right ).y=\int \frac{1}{x^{2}cos\left ( \dfrac{1}{x} \right )}.sec\left ( \frac{1}{x} \right )$

$\displaystyle =-\int \frac{sec^{2}\left ( \dfrac{1}{x} \right )}{x^{2}}$

Let

$\displaystyle \frac{1}{x}=t$

$\displaystyle \frac{-dx}{x^{2}}=dt$

$\displaystyle =\int sec^{2}(t)dt$

$\displaystyle \boxed{sec\left ( \frac{1}{x} \right ).y=tan\left ( \frac{1}{x} \right )+C}$

$\displaystyle 1.-1= 0+C$

$C=-1$

$\displaystyle sec\left ( \frac{1}{x} \right ).y= tan\left ( \frac{1}{x} \right )-1,$

$\displaystyle \boxed{y=sin\left ( \frac{1}{x} \right )-cos\left ( \frac{1}{x} \right )}$

The solution of the differential equation

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

$y (1) = 1$ , is

0%
a hyperbola
0%
a circle
0%
$y^2 = x (1 + x) - 10$
0%
$(x - 2)^2 + (y - 3)^2 = 5$

Explanation

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

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

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

$y^2=u=2y\dfrac{dy}{dx}=\dfrac{du}{dx}$

$\dfrac{du}{dx}-\dfrac{u}{x}=x+\dfrac{1}{x}$

$\dfrac{du}{dx}+Pu=Q$

$\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=1+\dfrac{1}{x^2}$

$\int d\left ( \dfrac{u}{x} \right )=\int \left ( 1+ \dfrac{1}{x^2}\right )dx\Rightarrow \dfrac{u}{x} x-x^{-1}+C$

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

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

The general solution of differential equation

$\dfrac{dy}{dx}=\sin^{3}{x}\cos^{2}{x}+xe^{x}$

0%
$y=\dfrac{1}{5}\cos^{5}{x}+\dfrac{1}{3}csc^{3}{x}+(x+1)e^{x}+c$
0%
$y=\dfrac{1}{5}\cos^{5}{x}-\dfrac{1}{3}csc^{3}{x}+(x-1)e^{x}+c$
0%
$y=-\dfrac{1}{5}\cos^{5}{x}-\dfrac{1}{3}csc^{3}{x}-(x-1)e^{x}-c$
0%
None of these

The value of

$\displaystyle \lim_{x \rightarrow \infty}$ y(x) obtained from the differential equation

$\dfrac{dy}{dx} = y - y^2$ , where y(0) = 2 is

0%
1
0%
-1
0%
0
0%
$\dfrac{2}{2-e}$

Explanation

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

$y(0)=2$

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

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

$\dfrac{-dy}{y^{2}\left(\dfrac{1}{2}\right)^{2}-2\times \dfrac{1}{2}y-\dfrac{1}{4}}=dx$

$\dfrac{dy}{\left(y-\dfrac{1}{2}\right)^{2}-\left(\dfrac{1}{2}\right)^{2}}=-dx$

$\dfrac{1}{2\times \dfrac{1}{2}}\ln\left|\dfrac{y-\dfrac{1}{2}-\dfrac{1}{2}}{y-\dfrac{1}{2}+\dfrac{1}{2}}\right|=-x+c$

$\ln\left(\dfrac{y-1}{y}\right)=x+c$

Given

$y(0)=2$

$\ln\left(\dfrac{2-1}{2}\right)=C$

$\boxed{C=\ln\left(\dfrac{1}{2}\right)}$

$\ln\left(\dfrac{y-1}{y}\right)=-x+\ln\left(\dfrac{1}{2}\right)$

$\ln\left(\dfrac{y}{y-1}\right)=x+\ln(2)$

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

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

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

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

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

$\displaystyle\lim_{x\rightarrow \infty}\dfrac{1}{1-e^{-\infty}}=\dfrac{1}{1-0}=1$

The solution of the differential equation

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

0%
$\dfrac{1}{x}\cos y = \dfrac{1}{{2x}^{2}}+K$
0%
$\dfrac{1}{x}\cot y = \dfrac{1}{{2x}^{2}}+K$
0%
$\dfrac{1}{x}\csc y = \dfrac{1}{{2x}^{2}}+K$
0%
$\dfrac{1}{x}\cos x = \dfrac{1}{{2x}^{2}}+K$

The solution of the differential equation

$(3xy+y^{2})dx+(x^{2}+xy)dy=0$ is

0%
$x^2(2x+y)=3$
0%
$y^2(2x+y)=3$
0%
$x^2y(2x+y)=3$
0%
$x^2y(2x+3)=9$

The solution of the differential equation

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

$\left[ ify\left( 0 \right) =1 \right]$

0%
$y=2+{x}^{2}$
0%
$y=\dfrac { \left( 1+x \right) }{ \left( 1-x \right) }$
0%
$y=x\left (x-1\right)$
0%
$y=\dfrac { \left( 1-x \right) }{ \left( 1+x \right) }$

The solution of differential equation

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

0%
$y\sin{y}=x\ln{x}+c$
0%
$y\sin{y}=x^{2}\ln{x}+c$
0%
$\sin{y}=x^{2}\ln{x}+c$
0%
$y\cos{y}=x^{2}\ln{x}+c$

Explanation

Equation of family of circle is

$(n+a)^2+(y-a)^2=a^2$ or

$x^2+y^2+2ax-2ay +a^2$

$=0$ ------(1)

Diffentiating , we get

$2n+2y \dfrac{dy}{dx}+2a-2a \dfrac{dy}{dx}=0$

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

$\Rightarrow a=\dfrac{x+yy^1}{y^1-1}$

where

$y^1=\dfrac{dy}{dx}$

Substituting the value of a in eq

$(1)$ and Simplifying

$\Rightarrow (xy^1-x+x+yy^1)+(yy^1-y-x-yy^1)^2$

$=(x+(yy^1))^2$

$\Rightarrow (x+y)^2[(y^1)^2+1]=(x+yy^1)^2$

The given differential equation is :

$\dfrac{dy}{dx}=\dfrac{x(2\log x+1)}{\sim y +y \cos y}$

$\Rightarrow (\sin y+y \cos y)dy =x(2\log x+1)dx$

$\Rightarrow \int (\sin y +y \cos y )dy=\int x(2\log x+1)dx$ integrating both sides

$\Rightarrow -\cos y +y \sin y - \int \sin y \, dy$

$=2\log ^x \dfrac{x^2}{2}-2 \int \dfrac{1}{x} \dfrac{x^2}{2}dx +\dfrac{x^2}{2}$

$\Rightarrow -\cos y + y \sin y + \cos y =x^2 \log x- \dfrac{x^2}{2}+\dfrac{x^2}{2}+c$

$\Rightarrow y \sin y =x^2 \log x +c$ .......(2)

It is given that when

$y=\dfrac{\pi}{2},x=1$

So, putting

$y=\dfrac{\pi}{2}, x=1$ in eq (2),

$\Rightarrow \dfrac{\pi}{2} \sin \dfrac{\pi }{2}=1.\log 1+c$

$\Rightarrow c=\dfrac{\pi}{2}$

Putting

$c=\dfrac{\pi}{2}$ in eq (2) we oftain

$\boxed{y \sin y =x^2\log x+\dfrac{\pi}{2}}$

2124354_1164597_ans_5d8596f6f45c417ab4b5a70575aadbf9.png

The equation of a curve passing through the point (0, 0) and whose differential equation is

${y'} = {e^x}\sin x$ is

0%
$2y - 1 = {e^x}\left( {\sin x - \cos x} \right)$
0%
$2y - 1 = {e^{ - x}}\left( {\cos x - \sin x} \right)$
0%
$2y + 1 = {e^{ - x}}\left( {\cos x - \sin x} \right)$
0%
$None\,of\,these$

Explanation

given,

$\dfrac {dy}{dx}=e^x\sin x$

$dy=e^x\sin x.dx$

$\displaystyle \int dy=\displaystyle \int e^x\sin x.dx\ \rightarrow$ integrating both sides.

$det,\ \displaystyle \int e^x\sin x\ dx=I----(1)$

$y=\displaystyle \int \underset{f(x)}{\underset{\downarrow}{e^x}}\underset{g(x)}{\underset{\downarrow }{\sin x}}\ dx=I$

Since,

$\displaystyle \int f(x)\,g(x)dx =f(x).\displaystyle \int g(x)dx-\displaystyle \int \left [f'(x) \displaystyle \int g(x)dx\right]dx$

$\therefore \ \Rightarrow \ e^x.\displaystyle \int \sin x\ dx-\displaystyle \int \left [e^x.\displaystyle \int \sin x\ dx\right]$

$I=e^x.(-\cos x)-\displaystyle \int e^x.(-\cos x).dx$

$I=-e^x\cos x+\displaystyle \int \underset{f(x)}{\underset{\downarrow}{e^x}}.\underset{g(x)}{\underset{\downarrow}{\cos x}}\ dx$

Now apply same formula as above

$I=-e^x\cos x.+\left [e^x.\displaystyle \int \cos x.dx-\displaystyle \int (ex.\displaystyle \int \cos x.dx)dx\right]$

$I=-{ e }^{ x }.\cos { x } +\left[ { e }^{ x }.\sin { x } -\underbrace { \displaystyle \int { { e }^{ x }.\sin { x } .dx } }_{ \left( I \right) } \right]$

$I=-e^x\cos x+e^x\sin x-I+c$

$2y=e^x(\sin x-\cos x)+c$

Since this curve passes through

$(0,0)$ therefore

$2\times 0=e^0(\sin 0\cos 0)+c$

$c=1$

$\therefore \$ equation of curve

$2y=e^x(\sin x-\cos x)+1$

$(2y-1)=e^x(\sin x-\cos x)$

Solution of the differential equation

$ye^{x/y}dx=(xe^{\frac {x}{y}}+y^{2})dy(y \neq 0)$ is ?

0%
$e^{x/y}=x+C$
0%
$e^{x/y}+x=C$
0%
$e^{y/x}=x+C$
0%
$e^{x/y}=y+C$

Solution of differential equation

$xdy=(y-{x}^{2}-{y}^{2})dx$ is (where

$c$ is arbitrary constant)

0%
$y=x\cos{(c+x)}$
0%
$y=x\tan{(x-c)}$
0%
$y=x\csc{(c+x)}$
0%
none of these

Let

$y=y(x)$ be the solution of the differential equation

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

$y(0)=0$ , then

$y\left(\dfrac{1}{2}\right)$ is equal to

0%
$\dfrac{\pi}{3\sqrt3}$
0%
$\dfrac{\pi}{\sqrt3}$
0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{3}$

If the general solution of the differential equation if

$y'=\frac { y }{ x } +\Phi \left( \frac { x }{ y } \right)$ , for some function

$\Phi$ , is given by

$yin|cx|=x$ , where c is an arbitrary constant, then

$\Phi$ (2) is equal to :

0%
4
0%
$\frac { 1 }{ 4 }$
0%
-4
0%
$-\frac { 1 }{ 4 }$

$(2y+xy^{3})dx+(x+x^{2}y^{2})dy=0$ solution of differential equation is

0%
$3x^{2}y+x^{3}y^{2}=c$
0%
$3x^{2}y^{2}+xy^{2}=c$
0%
$yx^{2}+\dfrac{xy}{3}=c$
0%
$x^{2}y+\dfrac{x^{3}y^{3}}{3}=c$

Solution of differential equation

${ x }^{ 6 }dy+3{ x }^{ 5 }ydx=xdy-2y\ dx$ is

0%
${ x }^{ 3 }y=\dfrac { y }{ { x }^{ 2 } } +C$
0%
${ x }^{ 3 }y=\dfrac { 2y }{ { x }^{ 2 } } +C$
0%
${ x }^{ 3 }{ y }^{ 2 }=\dfrac { y }{ { x }^{ 2 } } +C$
0%
${ x }^{ 3 }=\dfrac { y }{ { x }^{ 2 } } +C$

The solution of equation

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

0%
$x={ ce }^{ y }-y-2$
0%
$y=x+{ ce }^{ y }-2$
0%
$x+{ ce }^{ y }-2-2=0$
0%
$y=x$

Let

$y=y(x)$ be the solution of the differential equation

$\sin x\dfrac {dy}{dx}+y\cos x=4x,x\ \in\ (0,\pi)$ , if

$y\left(\dfrac {\pi}{2}\right)=0$ , then

$y\left(\dfrac {\pi}{6}\right)$ is equal to

0%
$-\dfrac {4}{9}\pi^{2}$
0%
$\dfrac {4}{9\sqrt {2}}\pi^{2}$
0%
$\dfrac {-8}{9\sqrt {2}}\pi^{2}$
0%
$-\dfrac {8}{9}\pi^{2}$

Solution of the differential equation

$dr=a\left( r\sin\theta d\theta-\cos\theta dr \right )$ is

0%
$r\left( 1+a\cos { \theta } \right) =c$
0%
$r\left( 1+a\cos { \theta } \right) =ac$
0%
$r\left( 1-a\cos { \theta } \right) =c$
0%
$r\left( 1-a\cos { \theta } \right) =ac$

The solutions of the differential equation

$\frac{dy}{dx}= \frac{siny+x}{sin2y-x cos y }$ is

0%
$sin^{2}y = xsin y +\frac{x^{2}}{2}+c$
0%
$sin^{2}y = xsin y -\frac{x^{2}}{2}+c$
0%
$sin^{2}y = x+sin y +\frac{x^{2}}{2}+c$
0%
$sin^{2}y = x-sin y +\frac{x^{2}}{2}+c$

The general solution of the differential equation

$\dfrac{dy}{dx} + \sin \dfrac{x+y}{2} = \sin \dfrac{x-y}{2}$ is

0%
$\log \left(\tan \dfrac{y}{2}\right) = c-2\sin x$
0%
$\log\left( \tan \dfrac{y}{4}\right) = c-2\sin \left(\dfrac{x}{2}\right)$
0%
$\log \left(\tan \left(\dfrac{y}{2} +\dfrac{\pi}{4}\right) \right)= c-2 \sin x$
0%
None of the above

$y\left(x\right)$ is the solution of the differential equation

$\left(x+2\right)\dfrac{dy}{dx}=x^{2}+4x-9,x\neq -2$ and

$y\left(0\right)=0$ , then

$y\left(-4\right)$ is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
$2$

The solution of differential equation,

$\left(x+\tan y\right)dy=\sin 2y\,dx$ is

0%
$xcoty=\dfrac{1}{2}\log |cosec2y-cot2y|+c$
0%
$x=\tan y+c\sqrt{\tan y}$
0%
$x=\cot y+c\sqrt{\cot y}$
0%
$None\ of\ these$

If a curve

$y=f(x)$ passes through the point

$(1,-1)$ and satisfies the differential equation,

$y(1+xy)dx=xdy$ , then

$f\left(-\dfrac {1}{2}\right)$ is equal to

0%
$-\dfrac {4}{5}$
0%
$\dfrac {2}{5}$
0%
$\dfrac {4}{5}$
0%
$-\dfrac {2}{5}$

General solution of differential equation

$x^{2}(x+y\frac{dy}{dx})+(x\frac{dy}{dx}-y)\sqrt{x^{2}+y^{2}}=0$ is

0%
$\frac{1}{\sqrt{x^{2}+y^{2}}}+\frac{y}{x}=c$
0%
$\sqrt{x^{2}+y^{2}}-\frac{y}{x}=c$
0%
$\sqrt{x^{2}+y^{2}}+\frac{y}{x}=c$
0%
$2\sqrt{x^{2}+y^{2}}+\frac{y}{x}=c$
0%
$\frac{1}{\sqrt{x^{2}+y^{2}}}-\frac{y}{x}=c$

Solution of the differential equation

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

0%
$\log|xy|+x-y=C$
0%
$\log|xy|-x+y=C$
0%
$\log|xy|+x+y=C$
0%
$\log|xy|-x-y=C$

Solution of

$(x+y-1)dx+(2x+2y-3)dy=0$ is

0%
$y+x+log(x+y-2)=c$
0%
$y+2x+log(x+y-2)=c$
0%
$2y+x+log(x+y-2)=c$
0%
$2y+2x+log(x+y-2)=c$

The solution of

$\dfrac{xdy}{x^{2}+y^{2}}=\left(\dfrac{y}{x^{2}+y^{2}}-1\right)dx$ is

0%
$y=x\cot(c-x)$
0%
$\cos^{-1}u/x=-x+C$
0%
$y=x\tan (c-x)$
0%
$y^{2}/x^{2}=x\tan (c-x)$

General solution of differential equation

$x^2dy + y(x + y)dx = 0$ is

0%
$2x + y = Ax^2Y$
0%
$2x - y = \dfrac{Ax^2}{y}$
0%
$2x + y = \dfrac{Ax^2}{y}$
0%
$x - 2y = Ax^2y$

$y(t)$ is solution of

$(t + 1) \dfrac{dy}{dt} - ty = 1, y(0) = -1$ , then

$y(1) =$

0%
$\dfrac{1}{4}$
0%
$-2$
0%
$-\dfrac{1}{2}$
0%
$\dfrac{1}{2}$

Solution of the differential equation

$\cos{\,}x {\,}dy=y(\sin{\,}x-y)dx,0<x<\dfrac{\pi}{2}$ is

0%
y sec x $=$ tan x $+$ c
0%
y sec x $=$ sec x $+$ c
0%
y sec x $=$ (sec x $+$ c) y
0%
sec x $=$ y (tan x $+$ c)

The solution of the differential equation

$ydx-xdy+3x^2y^2e^{x^3} dx=0$ is

0%
$\dfrac{x}{y}+e^{x^3}=c$
0%
$\dfrac{x}{y}-e^x=c$
0%
$\dfrac{x}{y}+ex^{x^3}-x^2y=c$
0%
$\dfrac{x}{y}-e^{x^3}=c$

Solution of the equation

$(x+y)^2\dfrac{dy}{dx}=4,y(0)=0$ is

0%
$y=2tan^{-1}\left(\dfrac{x+y}{2}\right)$
0%
$y=4tan^{-1}\left(\dfrac{x+y}{4}\right)$
0%
$y=4tan^{-1}\left(\dfrac{x+y}{2}\right)$
0%
$y=2tan^{-1}\left(\dfrac{x+y}{4}\right)$

The solution of the differential equation

$\log\left(\dfrac {dy}{dx}\right)=4x-2y-2,y=1$ when

$x=1$ is

0%
$2e^{2y+2}=e^{4x}+e^{2}$
0%
$2e^{2y-2}=e^{4x}+e^{2}$
0%
$2e^{2y+2}=e^{4x}+e^{4}$
0%
$3e^{2y+2}=e^{3x}+e^{2}$

the solution of

$ydx-xdy+3{ x }^{ 2 }{ y }^{ 2 }{ e }^{ { x }^{ 3 } }dx=0$ is

0%
$\frac { x }{ y } +{ e }^{ { x }^{ 3 } }$ =C
0%
$\frac { x }{ y } -{ e }^{ { x }^{ 3 } }$ =0
0%
$-\frac { x }{ y } +{ e }^{ { x }^{ 3 } }$ =C
0%
none of these

The solution of the differential equation

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

0%
$\dfrac{x}{y^{2}}=y+C$
0%
$\dfrac{x}{y}=y^{2}+C$
0%
$\dfrac{x^{2}}{y}=y^{2}+C$
0%
$\dfrac{x}{y}=x^{2}+C$

General solution of the differential equation (x+y-2) dy=(x+y) dx is

0%
y+x=log(y-x+1)+c
0%
y-x log (x+y-1)+c
0%
y-2x=log (x+y-1)+c
0%
y+2x=log (x+y+1)+c

The solution of

$x^2dy-y^2dx+xy^2(x-y)dy=0$ is?

0%
$ln\left|\dfrac{x-y}{xy}\right|=\dfrac{y^2}{2}+c$
0%
$ln\left|\dfrac{xy}{x-y}\right|=\dfrac{x^2}{2}+c$
0%
$ln\left|\dfrac{x-y}{xy}\right|=\dfrac{x^2}{2}+c$
0%
$ln\left|\dfrac{x-y}{xy}\right|=x+c$

Solution of

$\left(1+xy\right)ydx+\left(1-xy\right)xdy=0$ is

0%
$\log\dfrac{x}{y}+\dfrac{1}{xy}=c$
0%
$\log\dfrac{x}{y}=c$
0%
$\log\dfrac{x}{y}-\dfrac{1}{xy}=c$
0%
$\log\dfrac{y}{x}-\dfrac{1}{xy}=c$

The general solution of the differential equation

$y-x\dfrac{dy}{dx}=y^{2}\cos x\left(1-\sin x\right)$ is

0%
$y\left(\sin x+c\right)=\tan x+\sec x$
0%
$y\left(\sin x+c\right)=\tan x-\sec x$
0%
$\dfrac{x}{y}=\sin x+\dfrac{1}{4}\cos 2x+c$
0%
$\dfrac{x}{y}=\sin x+\dfrac{1}{8}\cos 2x+c$

Solution of

$\dfrac { d y } { d x } + 2 x y = y$ is

0%
$y = c e ^ { x - x ^ { 2 } }$
0%
$y = c e ^ { x ^ { 2 } } - x$
0%
$y = c e ^ { x }$
0%
$y = c e ^ { - x ^ { 2 } }$

Let =y(x) be the solution of the differential equation, x

$\frac { dy }{ dx } +y=x{ log }_{ e }x,(x>1).$ If

$2y(2)={ log }_{ e }4-1,$ then y(e) is equal to :

0%
$-\dfrac { { e }^{ 2 } }{ 2 }$
0%
$\dfrac { e } { 4 }$
0%
$\dfrac { { e }^{ 2 } }{ 4 }$
0%
$-\dfrac { { e }^{ 2 } }{ 4}$

The solution of the Differential Equation

$(x+y)(dx-dy)=dx+dy$ is

0%
$l\ n(x-y)=x+y+C$
0%
$l\ n(x+y)=x-y+C$
0%
$l\ n(x-y)=x-y-C$
0%
$none\ of\ these$

The integrating factor of the equation

$y^{2}dx+(3xy-1)dy=0$ is

0%
$y^{2}$
0%
$y^{3}$
0%
$\dfrac {1}{y^{2}}$
0%
$\dfrac {1}{y^{3}}$

The solution of the differential equation

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

0%
$y-\tan\dfrac{x+y}{2}=c$
0%
$y+\tan\dfrac{x+y}{2}=c$
0%
$y+2\tan\dfrac{x+y}{2}=c$
0%
$None\ of\ these$

The solution of differential equation

$dy/dx -y/x+\tan y/x$ is:

0%
$x=C'\cos (y/x)$
0%
$x=C'\sin (y/x)$
0%
$x=C'\csc (y/x)$
0%
$none\ of\ these$

The solution of the equation

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

0%
$\log xy+x+y=c$
0%
$\log (\dfrac{x}{y})+x-y=c$
0%
$y-x-\log xy=c$
0%
$None\ of\ these$

Solution of

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

0%
$y=3x^{2}+9$
0%
$y=3x+9$
0%
$y=\dfrac {4}{3}x^{2}$
0%
$y=9x+3$

Number of values of m

$\in$ N, for which

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

$D^3y-3D^2y-4Dy+12y=0$ , is?

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

The solution the differential equation

$\left(\dfrac{dy}{dx}\right)^{2}-\dfrac{dt}{dx}(e^{x}+e^{-x})+1=0$ is/are

0%
$y+e^{-x}=c$
0%
$y-e^{-x}=c$
0%
$y+e^{x}=c$
0%
$y-e^{x}=c$

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 10 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 10 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.