MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 5

The solution of

$\frac{dy}{dx}=|x|$ is :

0%
$y=\frac{x|x|}{2}+c$
0%
$y=\frac{|x|}{2}+c$
0%
$y=\frac{x^2}{2}+c$
0%
$y=\frac{x^3}{2}+c$

Explanation

$\cfrac{dy}{dx}=x$ for

$x>0$

$\cfrac{dy}{dx}=-x$ for

$x\le 0$

For

$x>0,$

$\cfrac{dy}{dx}=x$

$\Rightarrow y=\cfrac{{x}^{2}}{2}+c$ ......... [By integrating]

For

$x\le 0$

$\cfrac{dy}{dx}=-x$

$y=\cfrac{{-x}^{2}}{2}+c$ ........ [By integrating]

$\Rightarrow$

$y=\cfrac { \left| x \right| x }{ 2 } +c$

The solution of the differential equation

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

0%
${ e }^{ y }=\left( { e }^{ x }+1 \right) +C{ e }^{ -{ e }^{ x } }$
0%
${ e }^{ y }=\left( { e }^{ x }-1 \right) +C$
0%
${ e }^{ y }=\left( { e }^{ x }-1 \right) +C{ e }^{ -{ e }^{ x } }$
0%
None of the above

Explanation

Given equation can be rewritten as

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

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

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

${ e }^{ y }=t\Rightarrow { e }^{ y }\dfrac { dy }{ dx } =\dfrac { dt }{ dx }$

$\therefore \dfrac { dt }{ dx } +{ e }^{ x }t={ e }^{ 2x }$
On comparing with

$\dfrac { dt }{ dx } +Pt=Q$ , we get

$P={ e }^{ x }$ and

$Q={ e }^{ 2x }$

$\therefore IF={ e }^{ \int { Pdx } }={ e }^{ \int { { e }^{ x }dx } }={ e }^{ { e }^{ x } }$
Required solution is

$t\cdot { e }^{ { e }^{ x } }=\displaystyle\int { { e }^{ 2x }{ e }^{ { e }^{ x } }dx } +C$

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

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

The solution of the differential equation

$\dfrac {dy}{dx} = \tan \left (\dfrac {y}{x}\right ) + \dfrac {y}{x}$ is:

0%
$\cos \left (\dfrac {y}{x}\right ) = cx$
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$\sin \left (\dfrac {y}{x}\right ) = cx$
0%
$\cos \left (\dfrac {y}{x}\right ) = cy$
0%
$\sin \left (\dfrac {y}{x}\right ) = cy$

Explanation

$\dfrac {dy}{dx} = \tan \left (\dfrac {y}{x}\right ) + \left (\dfrac {y}{x}\right )$ .....

$(i)$
Take,

$\dfrac {y}{x} = v$

$\implies y = vx$

$\implies \dfrac {dy}{dx} = v + x\dfrac {dv}{dx}$

$\therefore$ The given equation

$(i)$ becomes

$v + x\dfrac {dv}{dx} = \tan v + v$

$\implies \dfrac {1}{\tan v}dv = \dfrac {1}{x}dx$

$\implies \displaystyle \int \cot v\ dv = \int \dfrac {1}{x}dx$

$\implies \log |\sin v| = \log x + \log c=\log|xc|$

$\implies \sin v = xc$

$\therefore \sin \left (\dfrac {y}{x}\right ) = xc$

The solution of the differential equation

$\displaystyle (x^2-yx^2)\frac{dy}{dx}+y^2+xy^2=0$ is?

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$\displaystyle \log\left(\frac{x}{y}\right)=\frac{1}{x}+\frac{1}{y}+C$
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$\displaystyle \log\left(\frac{y}{x}\right)=\frac{1}{x}+\frac{1}{y}+C$
0%
$\displaystyle \log(xy)=\frac{1}{x}+\frac{1}{y}+C$
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$\displaystyle \log(xy)+\frac{1}{x}+\frac{1}{y}=C$

Explanation

The given differential equation is

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

$\Rightarrow x^2(1-y)dy+y^2(1+x)dx=0$

$\Rightarrow \displaystyle\frac{1-y}{y^2}dy+\frac{1+x}{x^2}dx=0$
On integrating, we get

$-\displaystyle\frac{1}{y}-\log y-\frac{1}{x}+\log x=C$

$\Rightarrow \log\left(\displaystyle\frac{x}{y}\right)=\dfrac{1}{x}+\dfrac{1}{y}+C$ .

The solution of

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

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$\tan^{-1}\left(\dfrac{2y+1}{\sqrt{3}}\right) = x+x^2+C$
0%
$4\tan^{-1}\left(\dfrac{2y+1}{\sqrt{3}}\right) = \sqrt{3}(2x+x^2)+C$
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$\sqrt{3}\tan^{-1}\left(\dfrac{3y+1}{3}\right)=4(1+x+x^2)+C$
0%
$\tan^{-1}\left(\dfrac{2y+1}{3}\right)=4(2x+x^2)+C$

Explanation

Given,

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

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

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

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

$4\tan^{-1}\left(\dfrac{2y+1}{\sqrt{3}}\right)=\sqrt{3}(2x+x^2)+C$

The solution of the differential equation

$\sec^{2} x \cdot \tan y \,dx + \sec^{2} y \cdot \tan x\ dy = 0$ is

0%
$\tan x \cdot \cot y = C$
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$\cot x \cdot \tan y = C$
0%
$\tan x \cdot \tan y = C$
0%
$\sin x \cdot \cos y = C$

Explanation

Given,

$\sec^{2} x \cdot \tan y dx+\sec^{2} y \cdot \tan x dy =0$

On separating the varaibales, we get

$\Rightarrow \sec^{2} x\cdot \tan y dx = - \sec^{2} y\cdot \tan \, x dy$

$\Rightarrow \dfrac{\sec^2x}{\tan \, x}dx=- \dfrac{\sec^{2}y}{\tan y}dy$

On integration both the sides, we get

$\displaystyle \int \frac{\sec^{2}x}{\tan x}dx= -\int \frac{\sec^{2}y}{\tan y}dy$

Let

$\tan x=u \Rightarrow \sec^{2}x=\dfrac{du}{dx}$

$\Rightarrow dx =\dfrac{du}{\sec^{2}x}$ and

$\tan \, y=v$

$\Rightarrow \sec^{2}y=\dfrac{dv}{dy}$

$\Rightarrow dy =\dfrac{dv}{\sec^2 y}$

$\displaystyle \therefore \int \frac{\sec^{2} x}{u}\cdot \frac{du}{\sec^{2} x}=-\int \frac{\sec^2y}{v}\cdot \frac{dv}{\sec^2y}$

$\Rightarrow \displaystyle \int \dfrac{du}{u}=\int \dfrac{dv}{v}$

$\Rightarrow \log|u|=-\log |v|+\log|C|$

$\Rightarrow \log|\tan \, x|=-\log |\tan y| + \log |C|$

$\Rightarrow \log|\tan x\cdot \tan y| = \log |C|$

$(\because \log m + \log n=\log mn)$

$(\because \log m =\log n\Rightarrow m =n)$

$\Rightarrow \tan x\cdot \tan y = C$

The solution of the differential equation

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

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$\dfrac {1}{x + y} = c$
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$\sin^{-1} (x + y) = x + c$
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$\tan^{-1} (x + y) = c$
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$\tan^{-1} (x + y) = x + c$

Explanation

Given DE is

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

Let's substitute

$t=x+y$

$\Rightarrow 1+\cfrac{dy}{dx}=\cfrac{dt}{dx}$

$\Rightarrow \cfrac{dt}{dx}-1=\dfrac{dy}{dx}$

$\Rightarrow \cfrac{dt}{dx}-1=t^2$

$\Rightarrow \cfrac{dt}{dx}=t^2+1$

$\Rightarrow \cfrac{dt}{t^2+1}=dx$

On integrating both sides we get

$\displaystyle \int \cfrac{dt}{t^2+1}=\int dx$

$\Rightarrow \tan^{-1}{t}=x+c$

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

The solution of the differential equation

$(1+{ x }^{ 2 }{ y }^{ 2 })ydx+({ x }^{ 2 }{ y }^{ 2 }-1)xdy=0$ is

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$xy=\log { \dfrac { x }{ y } } +C$
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$xy=2\log { \dfrac { y }{ x } } +C$
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${x}^{2}{y}^{2}=\log { \dfrac { y }{ x } } +C$
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${x}^{2}{y}^{2}=2\log { \dfrac { y }{ x } } +C$

Explanation

Given differential equation is

$(1+{ x }^{ 2 }{ y }^{ 2 })ydx+({ x }^{ 2 }{ y }^{ 2 }-1)xdy=0$ ....

$(i)$

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

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

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

Integrating both sides

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

$\Rightarrow \displaystyle \int\left({\dfrac{dx}x-\dfrac{dy}y}\right)+\displaystyle \int\left({xy^2dx+x^2ydy}\right)=0$

$\Rightarrow \displaystyle \int d\left(\log{\dfrac xy}\right)+\displaystyle\int\dfrac12d(x^2y^2)+C=0$

$\therefore \log\dfrac xy+\dfrac12 x^2y^2+C=0$

What is the curve which passes through the point (1, 1) and whose slope is

$\dfrac{2y}{x}$ ?

0%
Circle
0%
Parabola
0%
Ellipse
0%
Hyperbola

Explanation

Let the curve be

$y=f\left( x \right)$ ,

$\therefore$ slope of the tangent drawn at any point on the curve is

$\dfrac { df\left( x \right) }{ dx } =f^{ ' }\left( x \right)$ ,

Given that slope at any point on the curve is

$\dfrac { 2y }{ x }$ ,

$\Longrightarrow \dfrac { dy }{ dx } =\dfrac { 2y }{ x } \\ \Longrightarrow \int { \dfrac { 1 }{ y } } dy=\int { \dfrac { 2 }{ x } } dx$

$\Longrightarrow \ln { y } =2\ln { x } +c$

Where

$c$ is the integration constant,

Given that the curve passes through the point

$(1,1)$ ,

$\Longrightarrow c=0$ ,

$\therefore y={ x }^{ 2 }$ is the equation of the curve which is a parabola.

$x dy = y dx + y^2 dy, y > 0$ and y(1) = 1, then what is y(-3) equal to?

0%
3 only
0%
-1 only
0%
Both -1 and 3
0%
Neither -1 nor 3

Explanation

Given Differential equation is

$xdy=ydx+{ y }^{ 2 }dy$ ,

$\Longrightarrow \dfrac { dx }{ dy } =\dfrac { x }{ y } -y$ ,

$\Longrightarrow \dfrac { dx }{ dy } -\dfrac { x }{ y } =-y$ this is a linear differential equation in

$y$ ,

The integrating factor is

$\dfrac { 1 }{ y }$ ,

Now the differential equation becomes,

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

$\Longrightarrow \dfrac { x }{ y } =\int { -1dy } \\ \Longrightarrow \dfrac { x }{ y } =-y+c\\$ ,

Given that

$y(1)=1$ ,

$c=2$ ,

to find

$y(-3)$ ,

$\Longrightarrow { y }^{ 2 }-2y-3=0$ ,

Solving the quadratic equation gives,

$y=-1\ and\ y=3$ ,

Given that

$y>0$ ,

$\therefore y=3\ only$

Solution of the differential equation

$\dfrac { dx }{ x } +\dfrac { dy }{ y } =0$ is

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

Explanation

$\dfrac { dx }{ x } +\dfrac { dy }{ y } =0$

Integrating , we get

$\Rightarrow \int { \dfrac { dx }{ x } } +\int { \dfrac { dy }{ y } } =0$

$\Rightarrow \log(x)+\log(y)=k$

$\Rightarrow xy={ e }^{ k }=c$

$\Rightarrow xy=c$

Hence, option C is correct.

A solution of the differential equation

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

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

Explanation

Let

$p = \dfrac{dy}{dx}$ .

Therefore, the given equation becomes

$p^2-xp+y=0\implies y = px-p^2$

On differentiating wrt x, we get

$\dfrac{dy}{dx} = p + x\dfrac{dp}{dx}-2p\dfrac{dp}{dx}$

$\Rightarrow p = p + x\dfrac{dp}{dx}-2p\dfrac{dp}{dx}$

$\Rightarrow \dfrac{dp}{dx}(x-2p) = 0$

$\Rightarrow \dfrac{dp}{dx} = 0$ or

$x-2p = 0$

Now,

$x-2p = 0$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{2}$

$\Rightarrow y=\dfrac{x^2}{4}+c$

Also from the other equality,

$\dfrac{dp}{dx} = 0 \Rightarrow p = k$ (constant)

Substituting

$p=k$ in the given equation, we get

$y = kx-k^2$

On comparing the given options with the obtained solutions

$y = kx-k^2$ and

$y = \dfrac{x^2}{4}+c$

We get that the only possible solution is

$y=2x-4$ .

The solution of the differential equation

${ y }^{ ' }\left( { y }^{ 2 }-x \right) =y$ is

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${ y }^{ 3 }-3xy=C$
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${ y }^{ 3 }+3xy=C$
0%
${ x }^{ 3 }-3xy=C$
0%
${ y }^{ 3 }-xy=C$
0%
${ x }^{ 3 }-xy=C$

Explanation

Given,

${ y }^{ ' }\left( { y }^{ 2 }-x \right) =y$

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

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

$\Rightarrow \dfrac { dx }{ dy } =y-\dfrac { x }{ y }$

$\Rightarrow \dfrac { dx }{ dy } +\dfrac { 1 }{ y } x=y$
Above differential equation is a linear differential equation in

$x$ .

$\text{IF}={ e }^{ \int { \frac { 1 }{ y } dy } }={ e }^{ \log { y } }=y$
Hence, solution will be

$x\cdot y=\displaystyle\int { y\cdot ydy } +{ C }^{ ' }$

$\Rightarrow xy=\dfrac { { y }^{ 3 } }{ 3 } +{ C }^{ ' }$

$\Rightarrow 3xy={ y }^{ 3 }+3{ C }^{ ' }$

$\Rightarrow { y }^{ 3 }-3xy=-3{ C }^{ ' }$

$\Rightarrow { y }^{ 3 }-3xy=C$ .... [put

$C=-3{ C }^{ ' }$ , constant]

The solution of differential equation

$4xy\cfrac { dy }{ dx } =\cfrac { 3{ \left( 1+x \right) }^{ 2 }\left( 1+{ y }^{ 2 } \right) }{ \left( 1+{ x }^{ 2 } \right) }$ is

0%
$\log { (1+y) } =\log { x } +2\tan { x } +constant$
0%
$\log { \left( 1+{ y }^{ 2 } \right) } =3\log { \left( \cfrac { 1 }{ x } \right) } +6\tan ^{ -1 }{ x } +constant$
0%
$2\log { \left( 1+{ y }^{ 2 } \right) } =3\log { x } +6\tan ^{ -1 }{ x } +constant$
0%
None of the above

Explanation

Given,

$4xy\cfrac { dy }{ dx } =\cfrac { 3{ \left( 1+x \right) }^{ 2 }\left( 1+{ y }^{ 2 } \right) }{ \left( 1+{ x }^{ 2 } \right) }$

$\Rightarrow \cfrac { 4ydy }{ 1+{ y }^{ 2 } } =\cfrac { 3{ \left( 1+x \right) }^{ 2 } }{ x({ 1+x }^{ 2 }) } dx\quad$

$\Rightarrow \cfrac { 4ydy }{ 1+{ y }^{ 2 } } =\left( \cfrac { 3 }{ x } +\cfrac { 6 }{ { 1+x }^{ 2 } } \right) dx$

Integrating both sides

$\displaystyle \Rightarrow \int\cfrac { 4ydy }{ 1+{ y }^{ 2 } } =\int \left( \cfrac { 3 }{ x } +\cfrac { 6 }{ { 1+x }^{ 2 } } \right) dx$

$\Rightarrow 2\log { \left( 1+{ y }^{ 2 } \right) } =3\log { x } +6\tan ^{ -1 }{ x } +C\quad$

The solution of the differential equation

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

0%
$\log { \tan { \left( \dfrac { y }{ 2 } \right) } } =C-2\sin { x }$
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$\log { \tan { \left( \dfrac { y }{ 4 } \right) } } =C-2\sin { \left( \dfrac { x }{ 2 } \right) }$
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$\log { \tan { \left( \dfrac { y }{ 2 } +\dfrac { \pi }{ 4 } \right) } } =C-2\sin { x }$
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$\log { \tan { \left( \dfrac { y }{ 2 } +\dfrac { \pi }{ 4 } \right) } } =C-2\sin { \left( \dfrac { x }{ 2 } \right) }$

Explanation

Given differential equation can be rewritten as

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

$=2\cos { \left( \dfrac { x }{ 2 } \right) } \sin { \left( \dfrac { -y }{ 2 } \right) }$

$=-2\cos { \left( \dfrac { x }{ 2 } \right) } \sin { \left( \dfrac { y }{ 2 } \right) }$

$\Rightarrow \displaystyle\int { \dfrac { dy }{ 2\sin { \left( \dfrac { y }{ 2 } \right) } } } =-\displaystyle\int { \cos { \left( \dfrac { x }{ 2 } \right) } dx }$

$\Rightarrow \dfrac { 1 }{ 2 } \displaystyle\int { \csc { \left( \dfrac { y }{ 2 } \right) } dy } =-\dfrac { \sin { \left( \dfrac { x }{ 2 } \right) } }{ \left( \dfrac { 1 }{ 2 } \right) } +C$

$\Rightarrow \dfrac { 1 }{ 2 } \dfrac { \log { \left( \csc { \left( \dfrac { y }{ 2 } \right) } -\cot { \left( \dfrac { y }{ 2 } \right) } \right) } }{ \left( \dfrac { 1 }{ 2 } \right) } =-\dfrac { \sin { \left( \dfrac { x }{ 2 } \right) } }{ \left( \dfrac { 1 }{ 2 } \right) } +C$

$\Rightarrow \log { \tan { \left( \dfrac { y }{ 4 } \right) } } =-2\sin { \left( \dfrac { x }{ 2 } \right) } +C$

The solution of

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

0%
$\tan^{-1} x + \cos^{-1} x = C$
0%
$\sin^{-1} x + \sin^{-1} y = C$
0%
$\sec^{-1} x + \text{cosec} ^{-1} x = C$
0%
None of the above.

Explanation

Given, equation is

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

Dividing the above equation by

$\sqrt{1-y^{2}}$ on both sides

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

On integrating, we get

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

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

An integrating factor of the differential equation

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

0%
$\sin^2 x$
0%
$\dfrac{2}{\sin x}$
0%
$\log |\sin x|$
0%
$\dfrac{1}{\sin^2 x}$
0%
$2 \sin x$

Explanation

$\sin x\cfrac{dy}{dx}+2y\cos x=1$

$\Rightarrow \cfrac{dy}{dx}+2y\cot x=\csc x$

Here,

$P=2\cot x, Q=\csc x$

Integrating factor

$I=e^{\int P dx}$

$=e^{\int 2\cot x dx}=e^{2\int\cfrac{\cos x}{\sin x}dx}$

$=e^{2\ln \sin x}$ Integration of

$\int \cot x=\ln|\sin x|+C$

$=\sin^2x$

The general solution of the differential equation

$\left( x+y \right) dx+xdy=0$ is

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

Explanation

Given differential equation is

$\left( x+y \right) dx+xdy=0$

$\Rightarrow xdx=-\left( x+y \right) dy$

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

It is a homogeneous differential equation.

So, putting

$y=vx\Rightarrow \dfrac { dy }{ dx } =v+x\dfrac { dv }{ dx }$ , we get

$v+x\dfrac { dv }{ dx } =-\dfrac { x+vx }{ x } =-\dfrac { 1+v }{ 1 }$

$\Rightarrow x\dfrac { dv }{ dx } =-1-2v$

$\Rightarrow \displaystyle\int { \dfrac { dv }{ 1+2v } } =-\displaystyle\int { \dfrac { dx }{ x } }$

$\Rightarrow \log { \left( 1+2v \right) } =-\log { x } +\log { { C }_{ 1 } }$

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

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

$\Rightarrow { x }^{ 2 }+2xy=C$ (where,

$C={ C }_{ 1 }^{ 2 }$ )

The solution of the differential equation

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

0%
$\log { \left( \dfrac { x }{ y } \right) } =\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +C$
0%
$\log { \left( \dfrac { y }{ x } \right) } =\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +C$
0%
$\log { \left( xy \right) } =\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +C$
0%
$\log { \left( xy \right) } +\dfrac { 1 }{ x } +\dfrac { 1 }{ y } =C$

Explanation

Given differential equation is

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

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

$\Rightarrow \left( \dfrac { 1 }{ { y }^{ 2 } } -\dfrac { 1 }{ y } \right) dy+\left( \dfrac { 1 }{ { x }^{ 2 } } +\dfrac { 1 }{ x } \right) dx=0$
On integrating both sides, we get the required solution

$\dfrac { -2 }{ y } -\log { y } -\dfrac { 2 }{ x } +\log { x } =C$

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

The general solution of the differential equation

$xdy-ydx={y}^{2}dx$ is

0%
$y=\cfrac { x }{ C-x }$
0%
$x=\cfrac { 2y }{ C+x }$
0%
$(C+x)(2x)$
0%
$y=\cfrac { 2x }{ C+x }$
0%
$x=\cfrac { y }{ C-x }$

Explanation

We have

$xdy-ydx={ y }^{ 2 }dx$

$\Rightarrow x\cfrac { dy }{ dx } -y={ y }^{ 2 }$

$\Rightarrow \cfrac { x\cfrac { dy }{ dx } -y }{ { y }^{ 2 } } =1\quad$

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

$\Rightarrow -\cfrac { d }{ dx } \cfrac { x }{ y } =1$

$\Rightarrow \cfrac { x }{ y } =-x+C$

$\Rightarrow y=\cfrac { x }{ C-x }$

The degree of the differential equation

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

0%
$3$
0%
$2$
0%
$1$
0%
None of the above

Explanation

The given differential equation is not a polynomial equation in

$\dfrac {dy}{dx}$ . Therefore, its degree is not defined.

Solution of

$\cfrac { dx }{ dy } +mx=0$ , where

$m< 0$ is:

0%
$x=C{ e }^{ my }$
0%
$x=C{ e }^{ -my }$
0%
$x=my+C$
0%
$x=C$

Explanation

Given

$\dfrac{dx}{dy}+mx=0, m<0$

$\Rightarrow \dfrac{dx}{dy}=-mx$

$\Rightarrow \dfrac{dx}{x}=-m\:dy$

Integrating both sides we get

$\Rightarrow ln\: x=-my+c$

Raising to power

$e$ on both sides we get,

$x=e^{-my+c}$

$\Rightarrow x=e^{-my} \cdot e^c$

$\Rightarrow x=Ce^{-my}$

The solution of the differential equation

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

0%
$e^{y} = (e^{x} + 1) + Ce^{-x}$
0%
$e^{y} = (e^{x} - 1) + C$
0%
$e^{y} = (e^{x} - 1) + Ce^{-x}$
0%
None of the above

Explanation

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

$\Rightarrow e^{y}\dfrac {dy}{dx} + e^{x}\cdot e^{y} = e^{x} \cdot e^{x}$
Let

$e^{y} = t$

$\Rightarrow e^{y} \dfrac {dy}{dx}\cdot \dfrac {y}{x} = \dfrac {dt}{dx}$
Then, given equation reduces to

$\dfrac {dt}{dx} + e^{x}t = e^{2x}$
Here,

$P = e^{x}$
and

$Q = e^{2x}$

$\therefore IF = e^{\int Pdx} = e^{\int e^{x}dx} = e^{e^{x}}$
Required solution is

$t\cdot e^{e^{x}} = \int e^{2x} \cdot e^{e^{x}} dx + C$

$\Rightarrow e^{y} = (e^{x} - 1) + C \cdot e^{-e^{x}}$ .

The solution of

$\left( y-3{ x }^{ 2 } \right) dx+xdy=0$ is

0%
$y(x)=\sin { x } +\cfrac { 1 }{ { x }^{ 2 } } +C$
0%
$y(x)=\cos { x } -\cfrac { 1 }{ { x }^{ 2 } } +C\quad$
0%
$y(x)={ x }^{ 2 }+\cfrac { C }{ x }$
0%
$y(x)=\sqrt { x } +\cfrac { C }{ x }$

Explanation

Given,

$\left( y-3{ x }^{ 2 } \right) dx+xdy=0\quad$

$\left( y-3{ x }^{ 2 } \right) =-x\cfrac { dy }{ dx }$

$\cfrac { dy }{ dx } =\cfrac { -y }{ x } +3x$

$\cfrac { dy }{ dx } +\cfrac { 1 }{ x } y=3x$

$\text{I.F}={ e }^{ \displaystyle \int { \cfrac { 1 }{ x } dx } }={ e }^{ \log { x } }=x$
Thus

$y.x=\int { 3{ x }^{ 2 } } dx=3.\cfrac { { x }^{ 3 } }{ 3 } +c$

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

Let

$f(x)$ be differentiable on the interval

$(0,\infty)$ and

$\lim _{ t\rightarrow x }{ \cfrac { { t }^{ 3 }f(x)+{ x }^{ 3 }f(t) }{ t-x } } =2$ gives a linear differential equation whose integrating factor is

0%
${ x }^{ 3 }$
0%
$1/{ x }^{ 3 }$
0%
${ x }^{ 2 }$
0%
$1/{ x }^{ 2 }$

Explanation

$\lim _{ t\rightarrow x }{ \cfrac { { t }^{ 3 }f(x)+{ x }^{ 3 }f(t) }{ t-x } } =2$

$\Rightarrow \lim _{ t\rightarrow x }{ \cfrac { 3{ t }^{ 2 }f(x)-{ x }^{ 3 }f'(t) }{ 1 } } =2$
Using L-Hospital rule

$\Rightarrow 3{ x }^{ 2 }f(x)-{ x }^{ 3 }f'(x)=2\Rightarrow { x }^{ 3 }\cfrac { dy }{ dx } -3{ x }^{ 2 }y=2$

$\Rightarrow \cfrac { dy }{ dx } -\cfrac { 3 }{ x } y=\cfrac { 2 }{ { x }^{ 3 } }$

$\Rightarrow IF={ e }^{ \int { \cfrac { -3 }{ x } dx } }=\cfrac { 1 }{ { x }^{ 3 } }$

The solution of

$x \log x \displaystyle\frac{dy}{dx}+y=1$ is?

0%
$\log x=\displaystyle\frac{c}{(y-1)}$
0%
$y\log x\displaystyle\frac{dy}{dx}+y=1$
0%
$xy=\log (\log x)+c$
0%
$\displaystyle\frac{x}{y}\log y=c$

Explanation

$\left( x\log { x } \right) dy=\left( 1-y \right) dx$

$\cfrac { dy }{ 1-y } =\cfrac { dx }{ x\log { x } }$
Integrating both sides

$\int { \cfrac { dy }{ 1-y } } =\int { \cfrac { dx }{ x\log { x } } }$
Let

$\log { x } =t,dt=\cfrac { 1 }{ x } dx$

$-\ln { \left( 1-y \right) } =\int { \cfrac { dt }{ t } } =\ln { t } +\ln { c }(constant)$

$\ln { \left( \cfrac { 1 }{ 1-y } \right) } =\ln { \left( tc \right) }$

$tc=\cfrac { 1 }{ 1-y }$

$\log { x } =\cfrac { 1 }{ c\left( 1-y \right) } =\cfrac { k }{ y-1 }$
where

$k=-\cfrac { 1 }{ c }$

Solution of the differential equation

$(x^{2} + y^{3}) (2x^{2}dx + 3ydy) = 12x\ dx + 18y^{2}dy$ is

0%
$\dfrac {2}{3}x^{3} + \dfrac {3}{2}y^{2} = 6ln (x^{2} + y^{3}) + c$
0%
$x^{2} + y^{3} = 9ln (x^{2} + y^{3}) + c$
0%
$\dfrac {2}{3}x^{3} + \dfrac {3}{2}y^{2} = 6ln (x^{3} + y^{2}) + c$
0%
$x^{3} + y^{2} = 6ln (x^{2} + y^{3}) + c$

Explanation

$LHS\longrightarrow (x^{ 2 }+y^{ 3 })(2{ x }^{ 2 }dx+3ydy)\\ =2{ x }^{ 4 }dx+3{ x }^{ 2 }ydy+2{ x }^{ 2 }y^{ 3 }dx+3{ y }^{ 4 }dy\\ \Rightarrow 2{ x }^{ 4 }dx+3{ x }^{ 2 }ydy+2{ x }^{ 2 }y^{ 3 }dx+3{ y }^{ 4 }dy=12xdx+18{ y }^{ 2 }dy\\ (or)\; the\; given\; question\; can\; be\; written\; as\\ (2{ x }^{ 2 }dx+3ydy)=\cfrac { 6(2{ x }^{ }dx+3y^2dy) }{ x^{ 2 }+y^{ 3 } } \\ integrating\; \Rightarrow \cfrac { 2 }{ 3 } { x }^{ 3 }+\cfrac { 3 }{ 2 } { y }^{ 2 }=6ln(x^{ 2}+y^{ 3 })+C\\$

The solution of

$y' = e^{x - y} + x^{2} e^{-y}$ is

0%
$3(e^{y} - e^{x}) - x^{3} = c$
0%
$e^{y} - e^{x} - x^{3} = c$
0%
$e^{y} - e^{x} + x^{3} = c$
0%
$3(e^{y} - e^{x}) + x^{3} = c$

Explanation

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

$e^y dy =(e^x +x^2)dx$

integrating both side,

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

$3(e^y -e^x) -x^3 = c$

$y=\sqrt{(a-x)(x-b)}-(a-b)\tan^{-1}\sqrt{\displaystyle\frac{a-x}{x-b}}(a > b)$ then

$\displaystyle\frac{dy}{dx}=$ .

0%
$\sqrt{\displaystyle \frac{a-x}{x-b}}$
0%
$\sqrt{(a-x)(x-b)}$
0%
$0$
0%
$1$

Explanation

Given,

$y=\sqrt {(a-x) (x-b)}-(a-b) \tan^{-1} \sqrt {\dfrac {a-x}{x-b}}$ Let

$u=(a-x) (x-b), \surd =\sqrt {\dfrac {a-x}{x-b}}$

$\Rightarrow \ y=\sqrt u - (a-b)\tan^{-1}v$ Now, differentiating wrt

$x$

$\Rightarrow \ \dfrac {dy}{dx} =\dfrac {d}{dx}(\sqrt u)- (a-b)\dfrac {d}{dx} \tan^{-1}v$

$=\dfrac {d(\sqrt u)}{du}.\dfrac {du}{dx}-(a-b) \dfrac {d\tan^{-1}v}{dv}.\dfrac {dv}{dx}$ [chain rule]

$=\dfrac {1}{2\sqrt u}. \dfrac {du}{dx} -(a-b) \dfrac {1}{1+v^2}.\dfrac {dv}{dx}......(1)$

$\dfrac {du}{dx} =\dfrac {d}{dx} (a-x) (x-b) =\dfrac {d}{dx} (ax-av-x^2 +bx)=a-2x+b$

$\dfrac {dv}{dx}=\dfrac {d}{dx} \left (\sqrt {\dfrac {a-x}{x-b}}\right)=\dfrac {1}{2\sqrt {\dfrac {a-x}{x-b}}}\times \dfrac {(x-b)(-1)-(a-x)(1)}{(x-b)^2}$ [quotent rule]

$=\dfrac {\sqrt {x-b}}{2\sqrt {a-x}}\times \dfrac {(b-a)}{(x-b)^2}$ substituting in

$(1)$

$\dfrac {dy}{dx}=\dfrac {1}{2\sqrt {(a-x)(x-b)}}\times (a-2x+b)-(a-b)\dfrac {1}{1+\dfrac {a-x}{x-b}}\times \dfrac {\sqrt {x-b}}{\sqrt {b-x}}\times \dfrac {(b-a)}{(x-b)^2}$

$=\dfrac {a-2x+b}{2\sqrt {(a-x)(x-b)}}-\dfrac {(ab)(xb)}{(a-b)}\times \dfrac {\sqrt {x-b}}{2\sqrt {a-x}}\dfrac {(b-a)}{(x-b)^2}$

$=\dfrac {a-2x+b+a-b}{2\sqrt {(a-x)(x-b)}}$

$=\dfrac {2(a-x)}{2\sqrt {(a-x)(x-b)}}$

$\Rightarrow \boxed {\dfrac {dy}{dx}=\sqrt {\dfrac {a-x}{x-b}}}$ option

$A$ is corect

The solution of the differential equation

$3xy'-3y+{ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 1/2 }=0$ , satisfying the condition

$y(1)=1$ is

0%
$3\cos ^{ -1 }{ \left( \cfrac { y }{ x } \right) } =\ln { \left| x \right| }$
0%
$3\cos { \left( \cfrac { y }{ x } \right) } =\ln { \left| x \right| }$
0%
$3\cos ^{ -1 }{ \left( \cfrac { y }{ x } \right) } =2\ln { \left| x \right| }$
0%
$3\sin ^{ -1 }{ \left( \cfrac { y }{ x } \right) } =\ln { \left| x \right| }$

Explanation

The differential equation is

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

$3xdy-3ydx+(x^{2}-(y)^2)^{\frac{1}{2}}dx=0$

Let

$y=vx$

$\implies dy=vdx+xdv$

Substituting these values in the differential equation, we get

$3x(vdx+xdv)-3vxdx+(x^{2}-(vx)^2)^{\frac{1}{2}}dx=0$

$3x(vdx+xdv)-3vxdx+x(1-v^2)^{\frac{1}{2}}dx=0$

$3x^{2}dv+x(1-v^2)^{\frac{1}{2}}dx=0$

$3xdv+(1-v^2)^{\frac{1}{2}}dx=0$

$-\dfrac{3dv}{(1-v^2)^{\frac{1}{2}}}=\dfrac{dx}{x}$

Integrating the above equation, we get

$3\int -\dfrac{dv}{(1-v^2)^{\frac{1}{2}}}=\int \dfrac{dx}{x}$

$3cos^{-1}v=ln|x| +c$

$3cos^{-1}(\dfrac{y}{x})=ln|x| +c$

Using the goundary condition y(1)=1, we get

$3cos^{-1}1=ln|1|+c$

$c=0$

$\implies3cos^{-1}\dfrac{y}{x}=ln|x|$ is the solution of the differential equation.

Hence, the answer is option (A).

The solution of equation

$\dfrac{dy}{dx}=\dfrac{ax+b}{cy+d}$ represents :

0%
A straight line if $a=c=0$ and $b,d \neq 0$
0%
A parabola if $a=2, c=0, d \neq 0$
0%
A huberbola if $b=d=0, a$ and $c\neq 0$
0%
A rectangular hyberbola of $b=d=0, a=c=2$

Explanation

Given Differential equation is

$\cfrac { dy }{ dx } =\cfrac { ax+b }{ cy+d }$ ,

Separating the variables and integrating gives,

$\int { (cy+d)dy } =\int { (ax+b)dx }$ ,

$\Longrightarrow \cfrac { c{ y }^{ 2 } }{ 2 } +dy=\cfrac { a{ x }^{ 2 } }{ 2 } +bx+K$ ,

Where

$K$ is the integration constant,

$a=c=0$ and

$b,d\neq 0$ , the equation becomes,

$dy-bx=K$ which representss a straight line.

The solution of differential equation

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

0%
$y=1+ cxy$
0%
$y=\ln(cxy)$
0%
$y+1=cxy$
0%
$y=c+xy$

Explanation

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

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

$\therefore \dfrac{dy}{y-1}-\dfrac{dy}{y}=\dfrac{dx}{x}$

$\therefore \ln(y-1)-\ln y=\ln x+C^{'}$

$\therefore \ln\left(\dfrac{y-1}{yx}\right)=C^{'}$

$\therefore \dfrac{y-1}{yx}=e^{C^{'}}$

$\therefore \dfrac{y-1}{yx}=c$ ........

$[e^{C^{'}}=c]$

$\therefore y=1+cxy$

The solution of the differential equation

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

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

Explanation

Given,

$(x + 3y^2) \dfrac{dy}{dx} = y , y > 0$

we can rewrite it as,

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

$\Rightarrow \dfrac{dx}{dy} = \dfrac{x}{y} + 3y$

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

As it is a linear equation of the form

$\dfrac{dx}{dy} + px = Q$

Here,

$P = \dfrac{-1}{y} , Q = 3y$

$I. F = e^{\displaystyle\int p dy} = e^{\displaystyle-\int \dfrac{1}{y} dy}$

$= e^{-\ln \ y} = e^{\ln \ y^{-1}}$

$= y^{-1} = \dfrac{1}{y}$

$\therefore \dfrac{1}{y} . x = \displaystyle \int 3y \times \dfrac{1}{y} dy + c$

$\Rightarrow \dfrac{x}{y} = 3 \displaystyle \int dy + c$

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

The solution of the differential equation

$y'=\cfrac { 1 }{ { e }^{ -y }-x }$ , is

0%
$x={ e }^{ -y }\left( y+c \right)$
0%
$y+{ e }^{ -y }=x+c$
0%
$x={ e }^{ y }(y+c)$
0%
$x+y={ e }^{ -y }+c$

Explanation

$y^{'}=\dfrac{1}{e^{-y}-x}$

$y^{'}e^{-y}-y^{'}x=1$

$y^{'}e^{-y}=y^{'}x+1$

$y^{'}=e^{y}y^{'}x+e^{y}$

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

$\dfrac{dy}{dx}=\dfrac{d(xe^{y})}{dx}$

Integrating the above equation, we get

$y=xe^{y}-c$

$xe^{y}=y+c$

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

Hence, the answer is option (A).

The general solution of the differential equation

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

0%
$\log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =-2\cos { \cfrac { x }{ 2 } } +C$
0%
$\log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =2\cos { \cfrac { x }{ 2 } } +C$
0%
$\log _{ e }\left|{ \tan { \cfrac { y }{ 2 } } }\right| =2 \sin { \cfrac { x }{ 2 } } +C$
0%
$\log _{ e }\left|{ \tan { \cfrac { y }{ 2 } } }\right| =-2\sin { \cfrac { x }{ 2 } } +C\quad$

Explanation

We have

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

$\Rightarrow \cfrac { dy }{ dx } =\sin { \cfrac { x-y }{ 2 } } -\sin { \cfrac { \\ x+y }{ 2 } }$

$\Rightarrow \cfrac { dy }{ dx } =2\cos { \cfrac { \left( \cfrac { x-y }{ 2 } +\cfrac { x+y }{ 2 } \right) }{ 2 } } \sin { \cfrac { \left( \cfrac { x-y }{ 2 } -\cfrac { x+y }{ 2 } \right) }{ 2 } }$ .....

$\left[ \because \sin { C } -\sin { D } =2\cos { \cfrac { C+D }{ 2 } } .\sin { \cfrac { C-D }{ 2 } } \right]$

$\Rightarrow \cfrac { dy }{ dx } =2\cos { \cfrac { x }{ 2 } } .\sin { \cfrac { -y }{ 2 } }$

$\Rightarrow \cfrac { dy }{ dx } =-2\cos { \left( \cfrac { x }{ 2 } \right) } .\sin { \left( \cfrac { y }{ 2 } \right) }$

$\Rightarrow \cfrac { dy }{ \sin { \left( \cfrac { y }{ 2 } \right) } } =-2\cos { \left( \cfrac { x }{ 2 } \right) }$
On integrating both sides, we get

$\Rightarrow \displaystyle \int \text{cosec} { \left( \cfrac { y }{ 2 } \right) } dy=-2\int _{ }^{ }{ \cos { \left( \cfrac { x }{ 2 } \right) } dx }$

$\Rightarrow 2\log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =-4\sin { \left( \cfrac { x }{ 2 } \right) } +C$

$\Rightarrow \log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =-2\sin { \left( \cfrac { x }{ 2 } \right) } +C$

The differential equation

$\dfrac{dy}{dx} = e^x.e^y$ has solution ____________

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

Explanation

$\dfrac{dy}{dx}=e^x.e^y$

$\implies e^{-y}dy=e^xdx$

Integrating both sides-

$\int e^{-y}dy=\int e^xdx$

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

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

The particular solution of differential equation

$\cfrac { dy }{ dx } =-4x{ y }^{ 2 },y(0)=1$ is ______

0%
$y=\left( 2{ x }^{ 2 }+1 \right) =1$
0%
${ x }^{ 2 }=\cfrac { 1 }{ { y }^{ 2 } }$
0%
$y={ x }^{ 2 }+\log { x }$
0%
$4{ e }^{ x }+\cfrac { 1 }{ { y }^{ 2 } } =8\quad$

Explanation

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

$\cfrac { dy }{ dx } =-4x{ y }^{ 2 }$

$\therefore \cfrac { dy }{ { y }^{ 2 } } =-4xdx$

$\therefore \int { { y }^{ -2 } } dy=-4\int { x } dx+c\quad$

$\therefore \cfrac { { y }^{ -1 } }{ -1 } =-4\left( \cfrac { { x }^{ 2 } }{ 2 } \right) +c$

$\therefore \cfrac { -1 }{ y } =-2{ x }^{ 2 }+c$
Take

$x=0$ and

$y=1$

$\therefore -1=-2(0)+c$

$\therefore c=-1$

$\therefore \cfrac { -1 }{ y } =-2{ x }^{ 2 }-1$

$\therefore \cfrac { 1 }{ y } =1+2{ x }^{ 2 }$

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

The order of the differential equation whose general solution is

$y \, = \, c, \, cos \, 2x \, + \, c_2 \, cos^2x \, + \, c_3 \, sin^2x \, + \, c_4$

0%
$2$
0%
$4$
0%
$3$
0%
None of these

Explanation

$y=c_1\cos2x+c_2\cos^2x+c_3sin^2x+c_4$

$=c_1(2\cos^2x-1)+$

$=(2c_1+c_2)\cos^2x+c_3\sin^2x +(c_4-c_1)$

$=t_1\cos^2x+c_3\sin^2x+t_2$

So order will be

$3$ Because of three unknown present

The solution of the differential equation

$2x \dfrac{dy}{dx} = y; y(1) = 2$ represents

$=$ ____.

0%
parabola
0%
ellipse
0%
circle
0%
line

Explanation

$2x\dfrac{dy}{dx}=y$

$\implies 2\dfrac{dy}{y}=\dfrac{dx}{x}$

Integrating both sides-

$2\log y=\log x+c$

For

$y(1)=2-$

$2\log 2=\log 1+c\implies c=2\log 2$

Hence,

$2\log y=\log x+2\log 2$

$\implies \log y^2=\log {4x}$

$\implies y^2=4x$

This is the equation of parabola.

What are the order and degree, respectively, of the differential equation

${ \left( \cfrac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }={ y }^{ 4 }+{ \left( \cfrac { dy }{ dx } \right) }^{ 5 }\quad$ ?

0%
$4,5$
0%
$2,3$
0%
$3,2$
0%
$5,4$

Explanation

Order is the highest derivative of the dependent variable with respect to the independent variable and degree is the highest power to which the highest order derivative in the differential equation is raised.

so,

$Order=3$ and

$Degree=2$

Therefore, Answer is

$C$

What is the degree of the differential eqaution :

$\dfrac{d^3y}{dx^3}-6(\dfrac{dy}{dx})^2-4y=0$

0%
1
0%
2
0%
3
0%
None of these

Explanation

Given the differential equations is

$\dfrac{d^3y}{dx^3}-6(\dfrac{dy}{dx})^2-4y=0$ .

The highest order term involved in the equation is

$\dfrac{d^3y}{dx^3}$ .

And the degree of the highest ordered term is

$1$ , so degree of the differential equations is

$1$ .

the general solution of differential equation

$x^4 \, \frac{dy}{dx} \, + \, x^3 y \, + cosec \, xy \, =0$ , is

0%
2 cos ( x y ) + $\dfrac{1}{x^2}$ = C
0%
2 cos ( x y ) + $\dfrac{1}{y^2}$ = C
0%
2 sin y + $\dfrac{1}{x^2}$ = C
0%
2 sin ( x y ) + $\dfrac{1}{y^2}$ = C

Explanation

$x^4dy \, + \, x^3 \, ydx \, + \, cosec (xy) \, dx \, = \, 0$

$x^3(x \, dx \, + \,y \, dx )+ \, + \, cosec (xy) \, dx \, = \, 0$

$\therefore \int \dfrac{d (xy)}{cosec (xy)} \, + \,\int \dfrac{dx}{x^3} \, = \, 0$

$-cos \, xy \, + \, \dfrac{x^{-3+1}}{-3 + 1} \, = \, C$

$\therefore \,2\, cos\,(xy)\,+\, \dfrac{1}{x^2 }\,=\,C$

Solution of the differential equation

$\tan { y } .\sec ^{ 2 }{ x } dx+\tan { x } .\sec ^{ 2 }{ y } dy=0$ is

0%
$\tan { x }+\tan { y }=k$
0%
$\tan { x }-\tan { y }=k$
0%
$\cfrac { \tan { x } }{ \tan { y } } =k$
0%
$\tan { x }.\tan { y }=k$

Explanation

$\tan { y } .\sec ^{ 2 }{ x } dx+\tan { x } .\sec ^{ 2 }{ y } dy=0$

$\Rightarrow \tan { y } .\sec ^{ 2 }{ x } dx=-\tan { x } .\sec ^{ 2 }{ y } dy$

$\Rightarrow \cfrac { \sec ^{ 2 }{ x } dx }{ \tan { x } } =-\cfrac { \sec ^{ 2 }{ y } dx }{ \tan { y } }$
Integrating,

$\int { \cfrac { \sec ^{ 2 }{ x } dx }{ \tan { x } } } =-\int { \cfrac { \sec ^{ 2 }{ y } dx }{ \tan { y } } }$

$\Rightarrow \int { \cfrac { d\left( \tan { x } \right) }{ \tan { x } } } =-\int { \cfrac { d\left( \tan { y } \right) }{ \tan { y } } }$

$\Rightarrow \ln { \left| \tan { x } \right| } =-\ln { \left| \tan { x } \right| } +c$

$\Rightarrow \ln { \left| \tan { x } \right| } +\ln { \left| \tan { x } \right| } =c$

$\Rightarrow \ln { \left| \tan { x } .\tan { y } \right| } =c$

$\Rightarrow \tan { x } \tan { y } ={ e }^{ c }$

$\Rightarrow \tan { x } \tan { y } =k$ (where

${ e }^{ c }=k$ )

Let the function f satisfies

$f(x) .f'(-x) =f(-x) .f'(x)$ for all

$x$ and

$f(0) =3$
The value of

$f(x) .f(-x)$ for all

$x$ , is

0%
$4$
0%
$9$
0%
$12$
0%
$6$

Explanation

$f(x)\cdot\cfrac{d(f(-x))}{d(-x)}=f(-x)\cfrac{d(f(x))}{dx}$

$\cfrac{d(f(-x))}{f(x)}$ =

$\cfrac{-(d(f(x))}{f(x))}$

$\Rightarrow \log(f(-x))=-\log(f(x))+\log C$

$\log (f(x)f(-x))=\log C$

$f(x)\cdot f(-x)=C$

Put

$x=0$

and

$f(0)=3$

$3\times 3=C$

$C=9$

Therefore,

$f(x)\cdot f(-x)=9$

The order of differential equation of family of all concentric circles centered at

$(h,k)$ is

0%
1
0%
2
0%
3
0%
4

Explanation

Equation of circle :

$(x-h) ^{2}+(y-k)^{2}=r^{2}$ where 'r' is radius of the circle

Here (h, k) are given constants and only radius 'r' is the variable

Hence given equation contains only 1 variable

Therefore order of DE =1

The solution of

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

0%
$\dfrac { 1 }{ { x }^{ 3 } } =\dfrac { 3 }{ 2 } \left( 1-{ y }^{ 2 } \right) +c{ e }^{ -{ y }^{ 2 } }$
0%
$\dfrac { 1 }{ { x }^{ 3 } } =\dfrac { 3 }{ 2 } \left( 1+{ y }^{ 2 } \right) +c{ e }^{ -{ y }^{ 2 } }$
0%
$\dfrac { 1 }{ { x }^{ 4 } } =\dfrac { 3 }{ 2 } \left( 1-{ y }^{ 2 } \right) +c{ e }^{ { y }^{ 2 } }$
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$\dfrac { 1 }{ { x }^{ 4 } } =\dfrac { 3 }{ 2 } \left( 1+{ y }^{ 2 } \right) +c{ e }^{ { y }^{ 2 } }$

The solution of the equation

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

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$\dfrac{e^{-2x}}{4} = y$
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$y = \dfrac{e^{2x}}{2} + c$
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$\dfrac{1}{4} e^{-2x} + cx^2 + d = y$
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$\dfrac{1}{4} e^{-2x} + cx + d = y$

Explanation

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

$\dfrac{dy}{dx}+0y=e^{2x}$

we know that,

$P=0, Q=e^{2x}$

integrating factor, I.F

$I.F=e^{\int pdx}$

$I.F=e^{\int 0dx}$

$I.F=e^{0x}=e^0=1$

$I.F=1$

solution of the equation,

$y\times I.F=\int Q\times I.F.dx+c$

$y\times 1=\int e^{2x}\times 1\times dx+c$

$y=\int e^{2x}dx+c$

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

option

$B$ will be the right answer

The solution of

$x\dfrac{dy}{dx}+y\log{y}=xy{e}^{x}$ is

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$x\log{y}=(x+1){e}^{x}+c$
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$\log{y}=(x-1){e}^{x}+c$
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$(x-1)\log{y}=x{e}^{x}+c$
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$x\log{y}=(x-1){e}^{x}+c$

Explanation

$x\dfrac{{dy}}{{dx}} + y\log y = xy{e^x}$

dividing both sides by

$y$

$\dfrac{x}{y}\dfrac{{dy}}{{dx}} + \log y = xy{e^x}$

putting

$\log y = t$

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

$= \dfrac{{xdt}}{{dx}} + t = x{e^x}$

$= \dfrac{{dt}}{{dx}} + \dfrac{t}{x} = {e^x}$

$p = \dfrac{1}{x}\,\,\,\,\,Q = {e^x}$

$I.F = {e^{\int_{}^{} {\dfrac{1}{x}dx} }} = {e^{\log x}} = x$

solution of given D.E. is

$t \times I.F = \int_{}^{} {Q.I.Fdx + c}$

$t \times I.F = \int_{}^{} {{e_{}}^x} xdx + c$

$x\log = x{e^x}\int_{}^{} {i.{e^x}dx + c}$

$x\log = x{e^x} - {e^x} + c = x\log y = {e^x}\left( {x - 1} \right) + c$

The solution of

$\sec ^{2}y\dfrac {dy}{dx}+2x\tan y=x^{3}$ is

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

Explanation

$\begin{array}{l} { \sec ^{ 2 } }y\frac { { dy } }{ { dx } } +2x\tan y={ x^{ 3 } }\to (i) \\ let\, \tan y=t\to (ii) \\ different\, \, equation(ii)\, \, w.\, r.\, t\, \ x \\ we\, \, get \\ { \sec ^{ 2 } }y\frac { { dy } }{ { dx } } =\frac { { dt } }{ { dx } } \\ now\, \, equation\, (i)\, \, becomes \\ \frac { { dt } }{ { dx } } +2xt+{ x^{ 3 } } \end{array}$

This is linear equation

$\begin{array}{l} I.F=\int _{ e }{ pdx=\int _{ e }{ 2xdx } } ={ e^{ { x^{ 2 } } } } \\ tx{ e^{ { x^{ 2 } } } }=\int { { x^{ 3 } }\cdot { e^{ { x^{ 2 } } } }dx+c } \\ t{ e^{ { x^{ 2 } } } }=\int { x\cdot { x^{ 2 } }\cdot { e^{ { x^{ 2 } } } }dx+c } \, \, \, \, \, \, \, \, \left[ \begin{array}{l} put\, { x^{ 2 } }=u \\ difference\, \, we\, \, get \\ 2x\, dx=du \end{array} \right] \\ t{ e^{ { x^{ 2 } } } }=\frac { 1 }{ 2 } \int { x\cdot { e^{ { x^{ 2 } } } }dx+c } \, \, \, \, \, \, \, \, \left[ { by\, parts\, \, { { int } }eger } \right] \\ t{ e^{ { x^{ 2 } } } }=\frac { 1 }{ 2 } \left[ { x\cdot { e^{ x } }-\int { 1\cdot { e^{ x } }dx } } \right] \\ t{ e^{ { x^{ 2 } } } }=\frac { 1 }{ 2 } \left[ { 4{ e^{ x } }-{ e^{ x } } } \right] +c \\ t{ e^{ { x^{ 2 } } } }=\frac { { { e^{ { x^{ 2 } } } } } }{ 2 } \left( { { x^{ 2 } }-1 } \right) +c \\ \tan y=\frac { 1 }{ 2 } \left[ { { x^{ 2 } }-1 } \right] +c\cdot { e^{ { x^{ 2 } } } } \end{array}$

The solution of

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

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$\dfrac{1}{y} = cx - xlogx$
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$\dfrac{1}{x} = cy - ylogy$
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$\dfrac{1}{x} = cx + xlogy$
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$\dfrac{1}{y} = cx - ylogx$

Explanation

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

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

Let

$\dfrac{1}{x} = 1$

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

$\dfrac{-dt}{dy}+\dfrac{t}{y} = 1$

$\dfrac{dt}{dy}+(\dfrac{-1}{y})t = -1$

I.f

$\displaystyle e\int \dfrac{1}{y}dy$

I.f

$= \dfrac{1}{y}$

$\displaystyle t.\dfrac{1}{y} = \int -1\times \dfrac{1}{y} = -\int \dfrac{1}{y}$

$\dfrac{1}{xy} = -lny+c$

$\dfrac{1}{x} = -y\,ln\,y+cy$

$\dfrac{1}{x} = cy-y\,ln\,y$

1066829_1071112_ans_1a009d83df3540bc99c42a0889fd2c39.jpg

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 5 - 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 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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