If $$ \phi(x) =\int \left\{ \phi (x) \right\}^{-2} $$dx and $$ \phi ( 1) =0 $$ then $$ \phi (x) = $$

$$ \left\{ 3(x -1) \right\}^{1/3} $$

$$ \left\{ 2(x -1) \right\}^{1/4} $$

$$ \left\{ 5(x -2) \right\}^{1/5} $$

MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 7

The general solution of the differential equation

$e^xdy+(ye^x+2x)dx=0$ is

0%
$xe^x+x^2=C$
0%
$xe^x-y^2=C$
0%
$ye^x+x^2=C$
0%
$ye^x-x^2=C$

Explanation

$e^x dy +(ye^x+2x)dx=0$

$e^x dy=-(ye^x+2x)dx$

$\dfrac{dy}{dx}=\dfrac{-(ye^x+2x)}{e^x}$

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

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

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

This is of the form

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

where

$P=1$ and

$Q=-\dfrac{2x}{e^x}$

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

$=e^{\int 1 dx}=e^x$

Solution is :

$ye^x=\int \dfrac{-2x}{e^x} e^x \ dx+C$

$ye^x=-\int 2x \ dx +C$

$ye^x=-x^2+C$

$ye^x+x^2=C$

The solution of differential equation

$\cos x.\sin y dx+\sin x. \cos ydy=0$ is

0%
$\dfrac{\sin x}{\sin y}=c$
0%
$\sin x. \sin y=c$
0%
$\sin x+\sin y=c$
0%
$\cos x.\cos y=c$

Explanation

Given,

$\left( \cos { x } .\sin { y } \right) dx+\left( \sin { x } .\cos { y } \right) dy=0$

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

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

( All we have done is separated the variables )

$\Rightarrow \cot x\;dx=-\cot y\;dy$

( Now integrating both the sides we get )

$\Rightarrow \int \cot x\;dx =-\int \cot y\;dy$

$\Rightarrow \int \cot y\;dy =-\int \cot x\;dx$

$ln\;\sin y=-ln \sin x +ln C$ (

$ln$ C is interpretation const. )

$ln\sin y =ln \dfrac{1}{\sin x}+ln C$

$ln \sin y=ln \dfrac{C}{\sin x}$

$\left( \because ln\;a+ln\;b=ln \;ab\right)$

$\sin y=\dfrac{C}{\sin x}$

$\Rightarrow \sin x.\sin y=C$

Hence, the answer is

$\sin x.\sin y=C.$

Solution of the differential equation:

$\left( 2xcosy+{ y }^{ 2 }cosx \right) dx+\left( 2ysinx-{ x }^{ 2 }siny \right) dy=0$ is :

0%
${ x }^{ 2 }sinx+y^{ 2 }cosx=c$
0%
${ x }^{ 2 }siny+y^{ 2 }cosx=c$
0%
${ x }^{ 2 }cosy+y^{ 2 }sinx=c$
0%
None of these

For the given differential equation find the general solution:

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

0%
$\dfrac 1 5 [2sin x -cosx ]+C e ^{-2x}$
0%
$\dfrac 1 5 [2sin x -cosx ]+C$
0%
$\dfrac 1 2 [5sin x -cosx ]+C e ^{-2x}$
0%
None of these

The differential equation of the system of circles touching the

$x$ -axis at origin is

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

The solution of

$xdx+ydy=\frac { xdy-ydx }{ { x }^{ 2 }+{ y }^{ 2 } }$ is ________________________.

0%
${ x }^{ 2 }+{ y }^{ 2 }=2\quad { tan }^{ -1 }\frac { y }{ x } +C$
0%
${ x }^{ 2 }-{ y }^{ 2 }=2\quad { tan }^{ -1 }\frac { y }{ x } +C$
0%
${ x }^{ 2 }+{ y }^{ 2 }=2\quad { tan }^{ -1 }\frac { x }{ y } +C$
0%
${ x }^{ 2 }-{ y }^{ 2 }=2\quad { tan }^{ -1 }\frac { x }{ y } +C$

Solution of differential equation

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

$sin y \quad cos y \quad$ is

0%
$tany=(x-1)+ce-x$
0%
$coty=(x-1)+ce-x$
0%
$tany=(x-1)ex+c$
0%
$coty=(x-1)ex+c$

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

$sinx\dfrac { dy }{ dx } +ycosx=4x,x\in (0,\pi )$ . If

$y=\left( \dfrac { \pi }{ 2 } \right) =0,then\quad y\left( \dfrac { \pi }{ 6 } \right)$ is equal to :

0%
$\dfrac { 4 }{ 9\sqrt { 3 } } { \pi }^{ 2 }$
0%
$\dfrac { 8 }{ 9\sqrt { 3 } } { \pi }^{ 2 }$
0%
$-\dfrac { 8 }{ 9 } { \pi }^{ 2 }$
0%
$-\dfrac { 4 }{ 9 } { \pi }^{ 2 }$

Let

$f\left( x \right) ={ cos }^{ -1 }\left( cosx \right)$ then

0%
f\left( x \right) is differentable for all $\quad x\epsilon R$
0%
f\left( x \right) is not differentable at $\quad x\epsilon 0$
0%
f\left( x \right) is not differentable if $x=n\pi .n\epsilon I$
0%
f\left( x \right) is not differentable if $x=\frac { n\pi }{ 2 } n\epsilon I$

differential equation of all parabolas whose axis s y - axis.......

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

Explanation

Parabola with axis on y- axis is given by,

$x^{2}= 4a(y-c),$ Differentiating,

$\left \{ center = (0,c) \right \}$

$2x = \frac{4 a dy}{dx} \Rightarrow a = \frac{x}{2y'}$

$\Rightarrow x^{2} = 4(\frac{x}{2y'})y$

$\Rightarrow x^{2}y' = 2xy$

$\Rightarrow 2x = 4ay'$

$\Rightarrow a = \frac{x}{2y'}$

$\Rightarrow x^{2} = \frac{4x}{2y'}(y-c)$

$\Rightarrow 2y'x^{2}= 2xy-2cx$

$\Rightarrow {y}''x^{2}+2xy'= 2xy'+2y-2c$

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

$\Rightarrow x^{2} = \frac{2x}{y'}(y-y+{y}''\frac{x^{2}}{2})$

$\Rightarrow x^{2}y'= x({y}''x^{2})$

$\Rightarrow x \frac{dy}{dx} -x^{2}\frac{d^{2}y}{dx^{2}} = 0$

1179198_1277079_ans_6ecc0afcca68439abe4e69b802284546.jpg

$Let_y-y(x)$ be the solution of the differential
equation

$sinx\dfrac{dy}{dx} -ycosx -4x,_x\epsilon (0,\pi ). if y\left(\frac{\pi}{2}\right)$ = 0,
then y

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

0%
$-\dfrac{4}{9}\pi^2$
0%
$-\dfrac{4}{9\sqrt3}\pi^2$
0%
$-\dfrac{-8}{9\sqrt3}\pi^2$
0%
None of these

Explanation

$\Rightarrow \dfrac{dy}{dx}-\cot x\cdot y=\dfrac{4x}{\sin x}$

This is a linear D.E

Integrating factor(I.F)

$=e^{\displaystyle\int -\cot xdx}=e^{-ln |\sin x|}$

$\Rightarrow I.F=\dfrac{1}{\sin x}$

$\Rightarrow \dfrac{y}{\sin x}=\displaystyle\int\dfrac{4x}{\sin^2x}dx=\displaystyle\int \underset{I}{4x}\underset{II}{cosec^2}xdx$

Applying integration by parts,

$\dfrac{y}{\sin x}=4\left[x(-\cot x)-\displaystyle\int (-\cot x)dx\right]$

$=4\left[-x\cot x+ln \sin x\right]+c$

$\Rightarrow y=-4x\sin x+4\sin x\cdot ln \sin x+c\sin x$

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

$\Rightarrow 0$

$=-2\pi+0+c$

$\Rightarrow c=2\pi |ln|=0$

$\Rightarrow y=-4x\sin x+4\sin x\cdot ln \sin x+2\pi \sin x$

$y\left(\dfrac{\pi}{6}\right)\Rightarrow y=\dfrac{-2\pi}{3}\left(\dfrac{1}{2}\right)+4\left(\dfrac{1}{2}\right)(-ln2)+\pi$

$y=\dfrac{2\pi}{3}-2ln2$ .

1167349_1264967_ans_e57974dcb1b249c4818f719a9299d7ea.jpg

Integrating factors of the differential equation

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

0%
$x/{e^x}$
0%
${e^x}/x$
0%
$x{e^x}$
0%
${e^x}$

Solution of differential equation

$\dfrac{dy}{dx} + \dfrac{y}{x} = \dfrac{1}{(1 + \ell nx + \ell ny)}$ is (where C is an integration constant)

0%
$2(1 + \ell n (xy)) = x^2 + C$
0%
$xy \ell n (xy) = x^2 + C$
0%
$(1 + \ell n (xy)) = x^2 + C$
0%
$2xy \ell n (xy) = x^2 + C$

The solution of the differential equation

$(x^2-yx^2)\dfrac{y^3}{x}=k+y^2+xy^2=0$ 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%
$log(xy)=\dfrac{1}{x}+\dfrac{1}{y}+c$
0%
$log(xy)+\dfrac{1}{x}+\dfrac{1}{y}=c$

An integrating factor for the

$DE: (1+y^{2})dx-(\tan^{-1}y-x)dy=0$ is

0%
$\tan^{-1} y$
0%
$e^{\tan^{-1}y}$
0%
$\dfrac{1}{1+y^{2}}$
0%
$\dfrac{1}{x(1+y^{2})}$

Explanation

We have,

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

$\\ Hence,\, option\, B\, is\, the\, correct\, answer.$

Solution of differential equation

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

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

Solution of the differential equation

$\left( { x }^{ 2 }+1 \right) y'+2xy=4{ x }^{ 2 }$ is

0%
$y\left( 1+{ x }^{ 2 } \right) =\dfrac { 4{ x }^{ 3 } }{ 3 } +C$
0%
$y\left( 1-{ x }^{ 2 } \right) ={ x }^{ 3 }+C$
0%
$y\left( 1-{ x }^{ 2 } \right) =\dfrac { { x }^{ 3 } }{ 2 } +C$
0%
None of these

Explanation

We have,

$\left( {{x}^{2}}+1 \right)y'+2xy=4{{x}^{2}}$

$\Rightarrow \left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}+2xy=4{{x}^{2}}$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{2x}{\left( {{x}^{2}}+1 \right)}y=\dfrac{4{{x}^{2}}}{\left( {{x}^{2}}+1 \right)}$

On comparing that,

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

Now,

$P=\dfrac{2x}{{{x}^{2}}+1},\,\,\,\,\,Q=\dfrac{4{{x}^{2}}}{{{x}^{2}}+1}$

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

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

$={{e}^{\log \left( {{x}^{2}}+1 \right)}}\,\,\,\,\,\,\therefore {{e}^{\log x}}=x$

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

Now,

$y\times I.F.=\int{Q.IFdx+C}$

$y\times \left( {{x}^{2}}+1 \right)=\int{\dfrac{4{{x}^{2}}}{\left( {{x}^{2}}+1 \right)}\times }\left( {{x}^{2}}+1 \right)dx+C$

$y\left( {{x}^{2}}+1 \right)=\int{4{{x}^{2}}}dx+C$

$y\left( {{x}^{2}}+1 \right)=4\int{{{x}^{2}}}dx+C$

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

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

Hence, this is the answer.

Consider the differential equation,

$ydx-(x+y^{2})dy=0$ . If for

$y=1$ , x takes value

$1$ , then value of

$x$ when

$y=4$ is:

0%
$16$
0%
$36$
0%
$64$
0%
$9$

General solution of the differential equation

$\frac{{dy}}{{dx}} = 1 + xy$ is

0%
$y = c \cdot \,{e^{ - x2/2}}$
0%
$y = c \cdot \,{e^{ x2/2}}$
0%
$y = (x + c) \cdot {e^{ - x2/2}}$
0%
None

Explanation

$\frac{dy}{dx}= 1+xy \Rightarrow \frac{dy}{dx}-ky=1$

$IF = e^{\int -xdx}= e^{-\frac{x^{2}}{2}}$

$\frac{dy}{dx}=1+xy\Rightarrow e^{-\frac{x^{2}}{2}}\frac{dy}{dx}-2yxe^{-\frac{x^{2}}{2}}=1.e^{-\frac{x^{2}}{2}}$

$\therefore \frac{d}{dx}(ye^{-\frac{x^{2}}{2}})=e^{-\frac{x^{2}}{2}}$

$ye^{-\frac{x^{2}}{2}}=e^{-\frac{x^{2}}{2}}+c$

1210672_1324668_ans_c44685bda53c4c5c9efb6232f65994ef.jpg

The solution of the differentiable 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

0%
$y=\sin \dfrac {1}{x}-\cos \dfrac {1}{x}$
0%
$y=\dfrac {x+1}{x\sin \dfrac {1}{x}}$
0%
$y=\cos \dfrac {1}{x}-\sin \dfrac {1}{x}$
0%
$y=\dfrac {x+1}{x\cos \dfrac {1}{x}}$

The population

$p(t)$ at time

$t$ of a certain mouse species satisfies the differential equation

$\dfrac { d p ( t ) } { d t } = 0.5 p ( t ) - 450.$ If

$p ( 0 ) = 850 ,$ then the time at which the population becomes zero is

0%
$\dfrac { 1 } { 2 } \ln 18$
0%
$\ln 18$
0%
$2 \ln 18$
0%
$\ln 9$

Explanation

We have,

$\begin{array}{l} \dfrac { { d\left( { p\left( t \right) } \right) } }{ { dt } } =\dfrac { 1 }{ 2 } p\left( t \right) -450 \\ \dfrac { { d\left( { p\left( t \right) } \right) } }{ { dt } } =\dfrac { { p\left( t \right) -900 } }{ 2 } \\ 2\int { \dfrac { { d\left( { p\left( t \right) } \right) } }{ { p\left( t \right) -900 } } } =\int { dt } \\ 2\, In\left| { \, p\left( t \right) -900 } \right| =t+c \\ t=0 \\ \Rightarrow 2\, In\, 50=0+c \\ \therefore \, 2\, In\, 900=t+2\, In\, 50 \\ t=2\, \, \left( { In\, 900-\, In\, 50 } \right) \\ =2\, In\, \left( { \dfrac { { 900 } }{ { 50 } } } \right) \\ =2\, In\, 18 \end{array}$

Hence, option

$C$ is correct answer.

The solution of differential equation

$\cos ^ { 2 } x \dfrac { d y } { d x } - ( \tan 2 x ) y = \cos ^ { 4 } x , | x | < \dfrac { \pi } { 4 } ,$ where

$y \left( \dfrac { \pi } { 6 } \right) = \dfrac { 3 \sqrt { 3 } } { 8 }$

0%
$y = \tan 2 x \cos ^ { 2 } x$
0%
$y = \cot 2 x \cos ^ { 2 } x$
0%
$2 y = \tan 2 x \cos ^ { 2 } x$
0%
$2 y = \cot 2 x \cos ^ { 2 } x$

Explanation

Given that:-

$\cos ^ { 2 } x \dfrac { d y } { d x } - ( \tan 2 x ) y = \cos ^ { 4 } x , | x | < \dfrac { \pi } { 4 } ,$ where

$y \left( \dfrac { \pi } { 6 } \right) = \dfrac { 3 \sqrt { 3 } } { 8 }$

To find:- Solution of the given equation.

Solution:-

$\because \cos^2x(\dfrac{dy}{dx})-(\tan 2x)y=\cos^4x$

$\Rightarrow \cos^2 x(\dfrac{dy}{dx})-\tan 2 xy =\cos^4x$

$\Rightarrow \dfrac{dy}{dx}-\dfrac{\tan 2 x}{\cos^2x}.y=\cos^4x$

Now, we will compat the eqn i to

$\dfrac{dy}{dx}+Py=B$

$\therefore P=f(x)$

$B=f(x)$

So, Integration factor (I.F)

$=e^{\int Pdx}$

Here we will find the value of

$-\dfrac{\tan 2x}{\cos^2x}$

$\displaystyle -\int \dfrac{\tan 2x}{\cos^2x}dx$

$\displaystyle =-\int \dfrac{2\sin 2x}{\cos 2x(2\cos^2x)}dx$

$\displaystyle =-\int \dfrac{2\sin 2x\,dx}{\cos 2x(1+\cos 2x)}$

Let,

$t=\cos 2x$

$\Rightarrow dt=-2\sin x\,d\,x$

$\displaystyle \Rightarrow =\int \dfrac{dt}{t(1+t)}$

$\displaystyle =\int \dfrac{1}{t}-\dfrac{1}{1+t}dt$

$=ln\left|\dfrac{t}{1+t}\right|$

$=ln \left|\dfrac{\cos 2x}{1+\cos 2x}\right|$

Now, putting

$ln\left|\dfrac{\cos 2x}{1+\cos 2x}\right|$ in I.F

$\therefore e^{Pdx}$

$=e^{ln}\left|\dfrac{\cos 2x}{1+\cos 2x}\right|$

$=\dfrac{\cos 2x}{1+\cos 2x}$

$=\dfrac{\cos 2x}{2\cos^2x}$

Since, multiplying eqn i by I.F both side, we get.

$\dfrac{\cos 2x}{2\cos^2x}.\dfrac{dy}{dx}-\dfrac{\sin 2x}{\cos 4 x}y=\dfrac{\cos 2x}{2}$

$\dfrac{\cos 2x}{2\cos^2x}.dy-(\dfrac{\sin 2x}{\cos 4 x})dx.y=\dfrac{\cos 2x}{2}dx$

$\displaystyle \Rightarrow \int d(y\dfrac{\cos 2x}{2\cos^2x})=\int \dfrac{\cos 2x}{2}dx$

$\Rightarrow y.\dfrac{\cos 2x}{2\cos^2x}=\dfrac{\sin 2x}{4}+C$ ...........(II)

$\Rightarrow \dfrac{3\sqrt{3}}{8}(\dfrac{1/2}{2.\dfrac{3}{4}})=\dfrac{\sqrt{3}}{8}+c$

$\Rightarrow x=\dfrac{\pi}{6}\,\,y=\dfrac{3\sqrt{3}}{8}$

$\Rightarrow \dfrac{\sqrt{3}}{8}=\dfrac{\sqrt{3}}{8}+c$

$\therefore c=0$

putting the value of c in eq ii.

$\therefore y.\dfrac{\cos 2x}{2\cos^2x}=\dfrac{\sin 2x}{4}+0$

$\Rightarrow y=\dfrac{\sin 2x.2\cos^2x}{\cos 2x.4}$

$\therefore y=\dfrac{\tan 2x.\cos^2x}{2}$

$\therefore 2y=\tan 2x.\cos^2x$

hence, the required solution is

$2y=\tan 2x.\cos^2x$

The order of the differential equation

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

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

The general solution of the differential equation

$\dfrac{ydx-xdy}{y}=0$ is

0%
$xy=C$
0%
$x=Cy^2$
0%
$y=Cx$
0%
$y=Cx^2$

Explanation

$\dfrac{ydx-xdy}{y}=0$

$\dfrac{ydx-xdy}{xy}=0$

$\dfrac{1}{x}dx-\dfrac{1}{y}dy=0$

Integrating both sides, we get,

$log|x|-log|y|=log|k|$

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

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

$y=Cx$ where,

$C=\dfrac{1}{k}$

Hence, C is the correct option.

A differential equation associated with the primitive

$y=a+b\ e^{5x}+c\ e^{7x}$ is

0%
$y_{3}+2y_{2}-y_{1}=0$
0%
$y_{3}+2y_{2}-35y_{1}=0$
0%
$4y_{3}+5y_{2}-20y_{1}=0$
0%
$none\ of\ these$

$\sqrt { \dfrac { \upsilon }{ \mu } } +\sqrt { \dfrac { \mu }{ \upsilon } } =6$ , then

$\dfrac { d\upsilon }{ d\mu } =$

0%
$\dfrac { 17\mu -\upsilon }{ \mu -17\upsilon }$
0%
$\dfrac { \mu -17\upsilon }{ 17\mu -\upsilon }$
0%
$\dfrac { 17\mu +\upsilon }{ \mu -17\upsilon }$
0%
$\dfrac { \mu +17\upsilon }{ 17\mu -\upsilon }$

Explanation

$\sqrt{\dfrac{v}{\mu}}+\sqrt{\dfrac{\mu}{v}}=6$

$\Rightarrow \dfrac{v+\mu}{\sqrt{v\mu}}=6$

$\Rightarrow {\left(v+\mu\right)}^{2}=36v\mu$

$\Rightarrow {v}^{2}+{\mu}^{2}+2v\mu-36v\mu=0$

$\Rightarrow {v}^{2}+{\mu}^{2}-34v\mu=0$

$\Rightarrow 2v\dfrac{dv}{d\mu}+2\mu-34\left(v+\mu\dfrac{dv}{d\mu}\right)=0$

$\Rightarrow \left(2v-34\mu\right)\dfrac{dv}{d\mu}=34v-2\mu$

$\Rightarrow -2\left(17\mu-v\right)\dfrac{dv}{d\mu}=-2\left(\mu-17v\right)$

$\therefore \dfrac{dv}{d\mu}=\dfrac{\mu-17v}{17\mu-v}$

The number of arbitrary constant in the particular solution of a differential equation is

0%
$3$
0%
$4$
0%
infinite
0%
zero

Explanation

According to the definition of particular solution of a differential equation, it does not contain any arbitrary constants.

So number of arbitrary constants in a particular solution of a differential equation is

$0$ .

Solution of differential equation

$\sin y.\dfrac {dy}{dx}+\dfrac {1}{x}\cos y=x^{4}\cos^{2}y$ is

0%
$x\sec y=x^{6}+C$
0%
$6x\sec y=x+C$
0%
$6x\sec y=x^{6}+C$
0%
$6x\sec y=6x^{6}+C$

The general solution of the differential equation

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

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

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

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

$\dfrac{dy}{e^y}=e^xdx$

On integrating both side we get

$\displaystyle\int \dfrac{dy}{e^y}=\displaystyle\int e^xdx$

$\Rightarrow -e^{-y}=e^x+c$

$e^{-y}+e^x=c$ .

The solution of the differential equation

${ 2x }^{ 2 }y\dfrac { dy }{ dx } =tan\left( { x }^{ 2 }{ y }^{ 2 } \right) -{ 2xy }^{ 2 }$ given

$y(1)=\sqrt { \dfrac { \pi }{ 2 } }$ is

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

The solution of the equation

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

0%
$log x=(x-y)+\frac{y}{x}+c$
0%
$logx=2log(x-y)+\frac{y}{x}+c$
0%
$logx=log(x-y)+\frac{x}{y}+c$
0%
none of these above

Explanation

Given:

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

To find: Solution of the eq

Now,

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

divide

$x^2$ bo both side.

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

or ,

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

Let,

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

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

$\therefore (1+v)(v+\dfrac{xdv}{dx})=(1+v^2)$

or,

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

or,

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

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

Integrating both side

$\therefore \int (\dfrac{1+v}{1-v})dv=\int \dfrac{dx}{x}$

or,

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

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

or,

$-2 \log (1-v)-v=\log +c$

$2\log (1-v)+\log x+v=c$

or,

$\log (1-v)^2+\log x+v=c$

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

$\log \{x(\dfrac{x-y}{x})^2\}=c-\dfrac{y}{x}$

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

$(x-y)^2=xe^{-y/x}.c$

Solve the given differential equation

$\dfrac{dy}{dx}=(cosx-sinx),$

0%
$y=sinx+cosx+c$
0%
$y=sinx-cosx+c$
0%
$y=tanx+secx$
0%
$\text{None of these}$

Explanation

Given

$\dfrac {dy}{dx}=\cos x-\sin x$

$dy=(\cos x-\sin x) dx$

Integrating on Both sides

$\displaystyle \int dy=\int (\cos x-\sin x) dx$

$y=\displaystyle \int \cos xdx-\int \sin x dx$

$y=\sin x+\cos x +c$

The solution of the differential equation

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

0%
$y=xe^{cx}$
0%
$y+xe^{cx}=0$
0%
$y+e^{x}$
0%
none of these

Explanation

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

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

Put

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

$\Rightarrow\,v+x\dfrac{dv}{dx}=v\left(\log{v}+1\right)$

$\Rightarrow\,v+x\dfrac{dv}{dx}=v\log{v}+v$

$\Rightarrow\,x\dfrac{dv}{dx}=v\log{v}$

$\Rightarrow\,\dfrac{dv}{v\log{v}}=\dfrac{dx}{x}$

Integrating both sides,

$\Rightarrow\,\log{\left(\log{v}\right)}=\log{x}+\log{c}=\log{cx}$

$\Rightarrow\,\log{v}=cx$

$\Rightarrow\,\log{\dfrac{y}{x}}=cx$

$\Rightarrow\,y=x{e}^{cx}$

Which of the following functions is differentiable at x=0?

0%
${ e }^{ -\left| x \right| }-\left| x \right|$
0%
${ e }^{ \left| x \right| }+\left| x \right|$
0%
$\left| x \right| -{ e }^{ \left| x \right| }$
0%
$\left| x \right| -{ e }^{ -\left| x \right| }$

Solution of the differential equation of

${ (y }^{ 2 }-{ x }^{ 3 })dx-xydy=0\quad$ is

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

Explanation

${ (y }^{ 2 }-{ x }^{ 3 })dx-xydy=0$

$\quad \Rightarrow y(ydx-xdy)={ x }^{ 3 }dx\Rightarrow \cfrac { y }{ x } \left( \cfrac { ydx-xdy }{ { x }^{ 2 } } \right) =dx\Rightarrow \cfrac { y }{ x } d\left( \cfrac { y }{ x } \right) =dx\Rightarrow -\cfrac { 1 }{ 2 } { \left( \cfrac { y }{ x } \right) }^{ 2 }=x+k$

$\Rightarrow -{ y }^{ 2 }=2{ x }^{ 3 }+2{ x }^{ 2 }k\Rightarrow { y }^{ 2 }+2{ x }^{ 3 }+c{ x }^{ 2 }=0\quad$

The integrating factor (I.F) of differential equation

$\cfrac { dy }{ dx } \left( 1+x \right) -xy=1-x$ is _____

0%
$\left( 1+x \right) { e }^{ x }$
0%
$\left( x-1 \right) { e }^{ x }\quad$
0%
$\left( 1+x \right) { e }^{ -x }$
0%
$\left( 1-x \right) { e }^{ -x }\quad$

Explanation

$\cfrac { dy }{ dx } -$

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

$\dfrac{1-x}{1+x}$

here

$p =-\dfrac{x}{x+1}$

integration factor is

$e^{\int(pdx)}$

${ e }^{ \int { -\cfrac { x }{ 1+x } } dx }$

$=\quad { e }^{ -\int { \left( 1-\cfrac { x }{ 1+x } \right) } dx }$

$={ e }^{ -\left( x-\ln { 1+x } \right) }$

$={ e }^{ -x }.{ e }^{ \ln { 1+x } }$

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

$\cos { x } \cfrac { dy }{ dx } -y\sin { x } =6x,(0<x<\cfrac { \pi }{ 2 } )$ and

$\quad y\left( \cfrac { \pi }{ 3 } \right) =0\quad$ then

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

0%
$-\cfrac { { \pi }^{ 2 } }{ 4\sqrt { 3 } }$
0%
$-\cfrac { { \pi }^{ 2 } }{ 2 }$
0%
$-\cfrac { { \pi }^{ 2 } }{ 2\sqrt { 3 } }$
0%
$\cfrac { { \pi }^{ 2 } }{ 2\sqrt { 3 } }$

Explanation

$\cfrac { dy }{ dx } -y\tan { x } =6x\sec { x }$

$\quad y\left( \cfrac { \pi }{ 3 } \right) =0\quad ;y\left( \cfrac { \pi }{ 6 } \right) =7$

${ e }^{ \int { pdx } }={ e }^{ -\int { \tan { x } dx } }={ e }^{ \ln { \cos { x } } }=\cos { x }$

$y.\cos { x } =\int { 6x\sec { x } \cos { x } } dx$

$y.\cos { x } =\cfrac { 6{ x }^{ 2 } }{ 2 } +C\quad \quad$

$y=3{ x }^{ 2 }\sec { x } +C\sec { x }$

$0=3.\cfrac { { \pi }^{ 2 } }{ 9 } .(2)+C(2)$

$2C=\cfrac { -2{ \pi }^{ 2 } }{ 3 } \Rightarrow C=-\cfrac { 2{ \pi }^{ 2 } }{ 3 }$

$y(\pi /6)=3.\cfrac { { \pi }^{ 2 } }{ 36 } .\left( \cfrac { 2 }{ \sqrt { 3 } } \right) +\left( \cfrac { 2 }{ \sqrt { 3 } } \right) .\left( -\cfrac { { \pi }^{ 2 } }{ 3 } \right) \Rightarrow y=-\cfrac { { \pi }^{ 2 } }{ 2\sqrt { 3 } }$

Let

$f(x)$ be a function such that

$f(0)=f'(0)=0, f''(x)=\sec^{4}x+4$ , then the function is

0%
$\log (\sin x)+\dfrac{1}{3}\tan^{3}x+xb$
0%
$\dfrac{2}{3}\log (\sec^2 x)+\dfrac{1}{6}\tan^{2}x+2x^{2}$
0%
$\log (\cos x)+\dfrac{1}{6}\cos^{2}x+\dfrac{x^{2}}{5}$
0%
$None\ of\ these$

$y = y(x)$ and

$\dfrac{ 2 + sinx }{y + 1}(\dfrac{dy}{dx}) = -cosx, y(0) = 1, then y(\pi/2)$ equals

0%
1/3
0%
2/3
0%
-1/3
0%
1

Explanation

$\dfrac{1}{y + 1}dy = -\dfrac{cosx}{2 + sinx}dx$
Integrating, we get

$log (y + 1) + log k + log(2+sinx) = 0$

$\therefore k(y + 1)(2 + sinx) = 1$ when

$x=0, y=1$ where

$k$ is constant.

$\therefore 4k = 1$ or

$k =1/4$

$\therefore (y+ 1)(2 + sinx) = 4$
Now put

$x = \pi/2 \therefore (y + 1)3 = 4$

$\therefore y = \dfrac{1}{3}$

$x\dfrac{dy}{dx} = y(\log y - \log x + 1)$ , then the solution of the equation is

0%
$\log \dfrac{x}{y} = cy$
0%
$\log \dfrac{y}{x} = cy$
0%
$\log \dfrac{x}{y} = cx$
0%
None of these

Explanation

$\dfrac{dy}{dx} = \dfrac{x}{y}[\log \dfrac{y}{x} +1]$
Put

$y = vx$

$v + x\dfrac{dy}{dx} = v\log v = v$

$\therefore \dfrac{dv}{v \log v} = \dfrac{dx}{x}$

$\therefore \log (\log v) = \log x + \log c = \log cx$

$\therefore log \dfrac{y}{x} = cx$

Solution of

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

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

Explanation

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

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

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

$\implies log y = x - x^{2} + c_{1}$

$\implies y = e^{x - x^{2}} e^{c_{1}} = ce^{x - x^{2}}$ where

$c = e^{c_{1}}$

$\implies y = ce^{x - x^{2}}$ is the required solution.

The solution of differential equation

$yy' = x\big(\dfrac{y^{2}}{x^{2}} + \dfrac{f (y^{2}/x^{2})}{f' (y^{2}/x^{2})}\big)$ is

0%
$f(y^{2}/x^{2}) = cx^{2}$
0%
$x^{2}f(y^{2}/x^{2}) = c^{2}y^{2}$
0%
$x^{2}f(y^{2}/x^{2}) = c$
0%
$f(y^{2}/x^{2}) = cy/x$

Explanation

The given equation can be written as

$\dfrac{dy}{dx} = \big[\dfrac{y^{2}}{x^{2}} + \dfrac{f(y^{2}/x^{2})}{f'(y^{2}/x^{2})}\big]$
Above equation is a homogeneous equation.
Putting

$y = vx$ , we get

$v [v + x\dfrac{dv}{dx}] = v^{2} + \dfrac{f(v^{2})}{f'(v^{2})}$

$\implies vx\dfrac{dv}{dx} = \dfrac{f(v^{2})}{f'(v^{2})}$ variable separable

$\implies \dfrac{2vf'(v^{2})}{f(v^{2})}dv = 2\dfrac{dv}{x}$
Now integrating both side, we get

$\log f(v^{2}) = \log x^{2} + \log c$ [

$\log c =$ constant]
or

$f(v^{2}) = \log cx^{2}$
or

$f(y^{2}/x^{2}) = cx^{2}$

The solution of

$\dfrac{dv}{dt} + \dfrac{k}{m}v = -g$ is

0%
$v = ce^{-\dfrac{k}{m}t} - \dfrac{mg}{k}$
0%
$v = c - \dfrac{mg}{k}e^{-\dfrac{k}{m}t}$
0%
$ve^{-\dfrac{k}{m}t} = c - \dfrac{mg}{k}$
0%
$ve^{-\dfrac{k}{m}t} = c - \dfrac{mg}{k}$

Explanation

$\dfrac{dv}{dt} + \dfrac{k}{m} v = -g$

$\implies \dfrac{dv}{dt} = \dfrac{k}{m} (v + \dfrac{mg}{k})$

$\implies \dfrac{dv}{v+mg/k} = \dfrac{k}{m} dt$

$\implies log (v + \dfrac{mg}{k}) = -\dfrac{k}{m}t + logc$

$\implies v + \dfrac{mg}{k} = c^{-(k/m)t}$

$\implies v = ce^{\dfrac{k}{m}t} - \dfrac{mg}{k}$

The solution of the equation

$(x^{2}y + x^{2})dx + y^{2}(x-1)dy = 0$ is given by

0%
$x^{2} + y^{2} + 2(x-y) + 2 ln \dfrac{(x-1)(y+1)}{c} = 0$
0%
$x^{2} + y^{2} + 2(x-y) + ln \dfrac{(x-1)(y+1)}{c} = 0$
0%
$x^{2} + y^{2} + 2(x-y) - 2 ln \dfrac{(x-1)(y+1)}{c} = 0$
0%
None of these

Explanation

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

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

$\implies \int [x+1+\dfrac{1}{x-1}]dx = -\int[y-1+\dfrac{1}{y+1}]dy$

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

$\implies \dfrac{x^{2} + y^{2}}{2} + (x-y) + ln(\dfrac{(x-1)(y+1)}{c} = 0$

Solution of

$2y sin x \frac {dv}{dx}= 2 sin x cos x -y^2 cos x,$ for

$x = \frac { \pi}{2} , y = 1$ is

0%
$y^2 = sin x$
0%
$y = sin^2 x$
0%
$y^2 = 1 +cos x$
0%
None of these

$\phi(x) =\int \left\{ \phi (x) \right\}^{-2}$ dx and

$\phi ( 1) =0$ then

$\phi (x) =$

0%
$\left\{ 2(x -1) \right\}^{1/4}$
0%
$\left\{ 5(x -2) \right\}^{1/5}$
0%
$\left\{ 3(x -1) \right\}^{1/3}$
0%
None of these

Solution of differential equation

$dy - \sin x \sin y dx = 0$ is

0%
$e^{\cos x}\tan \dfrac{y}{2} = c$
0%
$e^{\cos x}\tan y = c$
0%
$\cos x \tan y = c$
0%
$\cos x \sin y = c$

Explanation

$dy - \sin x \sin y dx = 0$

$\implies\ dy = \sin x \sin y dx$

$\implies cosec\ y\ dy = \sin x dx$

integating both side

$\implies log\tan\dfrac{y}{2}+log k=-\cos x$

$\implies \dfrac{\tan\dfrac{y}{2}}{c}=e^{-\cos x}$

$\implies e^{\cos x}\tan\dfrac{y}{2}=c$

if integrating factor of

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

$e^{\int pdx}$ , then P is equal to

0%
$\dfrac{2x^{2} - ax^{3}}{x(1-x^{2})}$
0%
$2x^{3} - 1$
0%
$\dfrac{2x^{2} - a}{ax^{3}}$
0%
$\dfrac{2x^{2} - 1}{x(1-x^{2})}$

Explanation

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

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

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

$\implies \dfrac{dy}{dx} + \dfrac{2x^{2}-1}{x(1-x^{2})}y= \dfrac{ax^{3}}{x(1-x^{2})}$
which is of the form

$\dfrac{dy}{dx} +Py = Q$
Its integrating factor is

$e^{\int Pdx}$
Here

$P=\dfrac{2x^{2}}{x(1-x^{2})}$

The solution of differentiation equation

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

0%
$x^{2}y+\dfrac{x^{3}y^{3}}{3}= c$
0%
$xy^{2}+\dfrac{x^{3}y^{3}}{3}= c$
0%
$x^{2}y+\dfrac{x^{4}y^{4}}{4}= c$
0%
None of these

Explanation

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

$\implies (2ydx + xdy) + (xy^{3}dx+x^{2}y^{2}dy) + 0$
Multiplying by

$x$ , we get

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

$\implies d(x^{2}y)+\dfrac{1}{3}d(x^{3}y^{3}) = 0$
Integrating, we get

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

The solution of differential equation

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

0%
$\tan(x^{2}+y^{2})=\dfrac{x^{2}}{y^{2}}+c$
0%
$\cot(x^{2}+y^{2})=\dfrac{x^{2}}{y^{2}}+c$
0%
$\tan(x^{2}+y^{2})=\dfrac{y^{2}}{x^{2}}+c$
0%
$\tan(x^{2}+y^{2})=\dfrac{y^{2}}{x^{2}}+c$

Explanation

The given equation can be written as

$\dfrac{xdx+ydy}{(ydx-xdy)/y^{2}}=y^{2}\dfrac{x}{y^{3}}\cos^{2}(x^{2}+y^{2})$

$\implies \dfrac{xdx+ydy}{\cos^{2}(x^{2}+y^{2})}=\dfrac{x}{y}(\dfrac{ydx-xdy}{y^{2}})$

$\implies \dfrac{1}{2}\sec^{2}(x^{2}+y^{2})d(x^{2}+y^{2})=\dfrac{x}{y}d(\dfrac{x}{y})$
On integrating, we get

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

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

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 7 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 7 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.