CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 3 - MCQExams.com

The solution of $$\dfrac{dy}{dx}=\dfrac{x^{2}+4x-9}{x+2}$$ is:
  • $$y=(x+2)^{2}-13log\left | x+2 \right |+c$$
  • $$y=(x+2)^{2}-5log\left | x+2 \right |+c$$
  • $$y=\frac{x^{2}}{2}+2x+13log\left | x+2 \right |+c$$
  • $$y=\frac{x^{2}}{2}+2x-13log\left | x+2 \right |+c$$
The solution of $$(x^{2}-y^{2}x^{2})\dfrac{dy}{dx}+(y^{2}+x^{2}y^{2})=0$$ is:
  • $$x+y+\dfrac{1}{x}+\dfrac{1}{y}=c$$
  • $$2(x+y)-\left ( \dfrac{1}{x}+\dfrac{1}{y} \right )=c$$
  • $$2(x-y)+\left ( \dfrac{1}{x}-\dfrac{1}{y} \right )=c$$
  • $$(x-y)-\left ( \dfrac{1}{x}+\dfrac{1}{y} \right )=c$$
The solution of $$\frac{dy}{dx}=e^{3x-2y}+x^{2}e^{-2y}$$
  • $$y=log(e^{3x}+x^{2})+c$$
  • $$y=2log(e^{3x}+x^{2})+c$$
  • $$2e^{2x}=3(e^{3y}+y^{3})+c$$
  • $$3e^{2y}=2(e^{3x}+x^{3})+c$$
The solution of $$x\dfrac{dy}{dx}=2\sqrt{y-1}$$ is:
  • $$e^{\sqrt{y-1}}=cx$$
  • $$\sqrt{y-x}=ex$$
  • $$\dfrac{1}{\sqrt{y-1}}=cx$$
  • $$x(\sqrt{y-x})=c$$
The solution of $$x^{2}\dfrac{dy}{dx}=\sqrt{4-y^{2}}$$ is:
  • $$sin^{-1}\left ( \dfrac{y}{2} \right )+\dfrac{1}{x}=c$$
  • $$\dfrac{x^{3}}{3}+sin^{-1}\left ( \dfrac{y}{4} \right )=c$$
  • $$\sin^{-1}\left ( \dfrac{y}{2} \right )+2x=c$$
  • $$\sqrt{4-y^{2}}+2x=c$$
The solution of 

$$3e^{x}\cos^{2}y\ dx+(1-e^{x}) \cot y\ dy=0$$
  • $$\tan y=c(e^{x}-1)$$
  • $$\tan ^{2}y=(e^{x}-1)c$$
  • $$\cot y=c(e^{x}-1)^{2}$$
  • $$\tan y=c(e^{x}-1)^{3}$$
The solution of $$xcos^{2}ydx=ycos^{2}xdy$$ is:
  • $$\tan x \tan y =c$$
  • $$y \tan y  =x \tan x + c$$
  • $$\tan x \cos y = \tan y \cos x +c$$
  • $$y \tan y - x \tan x + log \left( \dfrac{cosy}{cosx} \right )=c$$
The solution of $$y (dx) +(1+x^{2})tan^{-1} x (dy)=0$$
  • $$ytan^{-1}x=c$$
  • $$xtan^{-1}x=0$$
  • $$y(1+x^{2})=c$$
  • $$y^{2}(1+x^{2})=c$$
The solution of $$e^{y}(1+x^{2})\dfrac{dy}{dx}=2x(1+e^{y})$$ is:
  • $$\dfrac{1+e^{y}}{1+x^{2}}=c$$
  • $$e^{y}(1+x^{2})=c$$
  • $$(1+e^{y})+(1+x^{2})=c$$
  • $$(e^{y}+1)x^{2}=c$$
Solution of $$(e^{x}+1)ydy+(y+1)dx=0$$ is:
  • $$(y+1)(1+e^{-x})=ce^{y}$$
  • $$(e^{x}+1)y=c$$
  • $$(1+e^{x})(y+1)=c$$
  • $$(e^{x}+1)x=c$$
The solution of $$x\sqrt{1+y^{2}}dx+y\sqrt{1+x^{2}}dy=0$$ is:
  • $$sinh^{-1}+sinh^{-1}y=c$$
  • $$\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$$
  • $$(1+x^{2})(1+y^{2})=c$$
  • $$\sqrt{\frac{1+x^{2}}{1+y^{2}}}=c$$
Equation of the curve whose polar sub tangent $$r^{2}\displaystyle \frac{d\theta}{dr}$$ is constant
  • $$r(\theta+c)+k=0$$
  • $$r^{2}(\theta+c)=2k$$
  • $$r(\theta-c)=k^{2}$$
  • $$r\theta=c$$
The solution of $$ydx+xdy=dx+dy$$ is:
  • $$xy=x+y+c$$
  • $$x-y\dfrac{x}{y}+c=0$$
  • $$xy-x+y=c$$
  • $$x+y\dfrac{x}{y}+c=0$$
The solution of $$\displaystyle \frac{dy}{dx}=\sqrt{1+x+y+xy}$$ is:
  • $$\sqrt{1+y}=\sqrt{1+x}+c$$
  • $$\frac{2}{3}\sqrt{1+y}=\left ( 1+x \right )^{\frac{1}{2}}+c$$
  • $$3\sqrt{1+y}=\left ( 1+x \right )^{\frac{3}{2}}+c]$$
  • $$\sqrt{1+x}=\sqrt{1+y}=c$$
The solution of:  $$e^{x}\displaystyle \sqrt{1-y^{2}}dx+\frac{y}{x}dy=0$$
  • $$(x -1)^{2}e^{x}=(1-y^{2})+c$$
  • $$(x +1)e^{x}=\sqrt{1-y^{2}}+c$$
  • $$x.\ e^{x}=\sqrt{1-y^{2}}+c$$
  • $$(x -1)e^{x}=\sqrt{1-y^{2}}+c$$
The solution of $$\dfrac{dy}{dx}=\dfrac{ax+h}{by+k}$$ represents a parabola when :
  • $$a\ne0,b=0$$
  • $$a=1,b=2$$
  • $$a=0,b\neq 0$$
  • $$a=2,b=1$$
The solution of $$cosec^{2}x\dfrac{dy}{dx}=\dfrac{1}{y}$$ is:
  • $$y^{2}= x - \sin x \cos x + c$$
  • $$y^{2}=\dfrac{x^{2}}{2}- \sin 2x + c$$
  • $$2y^{2}= x + \sin 2x + c$$
  • $$y^{2}=\dfrac{x^{2}}{2} + \sin x + c$$
The solution of $$\displaystyle \frac{dy}{dx}=xy+2x-3y-6$$ is:
  • $$(y+2)^{2}=c.e^{(x-3)^{2}}$$
  • $$\log(y+2)=x^{2}-3x+c$$
  • $$y+2=2(x-3)+c$$
  • $$(y+2)(x-3)=c$$
Solution of $$y-x\displaystyle \frac{dy}{dx}=3[1-x^{2}\frac{dy}{dx}]$$ is:
  • $$(y+3)(1+3x)=cx$$
  • $$(y-3)(1-3x)=cx$$
  • $$(y-3)(1+3x)=cx$$
  • $$(y+3)(1-3x)=cx$$

Solution of $$\displaystyle \frac{dy}{dx}+\frac{y^{2}+y+1}{x^{2}+x+1}=0$$ is:
  • $$\tan ^{ -1 }{ \left( \frac { 2x+1 }{ \sqrt { 3 } } \right) } +\tan ^{ -1 } (\frac { 2y+1 }{ \sqrt { 3 } } )=c$$
  • $$\displaystyle \sin^{-1}(\frac{2x+1}{\sqrt{3}})+\sin^{-1}(\frac{2y+1}{\sqrt{3}})=c$$
  • $$\displaystyle \sinh^{-1}(\frac{2x+1}{\sqrt{3}})+\sinh^{-1}(\frac{2y+1}{\sqrt{3}})=c$$
  • $$\displaystyle \sin^{-1}(\frac{2x+1}{\sqrt{3}})=c$$
The solution of $$e^{x-y}dx+e^{y-x}dy=0$$ is:
  • $$e^{x}+e^{y}=c$$
  • $$e^{2x}+e^{2y}=c$$
  • $$e^{x+y}+e^{x-y}=c$$
  • $$e^{x}-e^{y}=c$$
General solution of $$\displaystyle \frac{dy}{dx}=\frac{1}{\log_{x}e}$$ is given as $$y = $$
  • $$\displaystyle \frac{1}{x}+c$$
  • $$\displaystyle \frac{x^{2}}{2}+c$$
  • $$x\log_{e}x-x+c$$
  • $$\displaystyle \frac{x}{2}+c$$
Solution of $$xe^{x^{2}+y}.dx=y.dy$$ is:
  • $$x.e^{x^{2}}+2y.e^{y}=c$$
  • $$e^{x^{2}}+2(y+1)e^{-y}=c$$
  • $$x.e^{x^{2}}+y=2y+c$$
  • $$x.e^{x^{2}}y=c$$
If $$\frac{{dy}}{{dx}} = {e^{ - 2y}}$$ and $$y = 0$$ when $$x = 5$$, then value of x for $$y = 3$$ is
  • $$e^5$$
  • $$e^6+1$$
  • $$\frac{{{e^6} + 9}}{2}$$
  • $$log_e6$$
The solution of $$\cos \mathrm{x}\cos \mathrm{y}\mathrm{d}\mathrm{x}+ \sin \mathrm{x} \sin \mathrm{y} d\mathrm{y} =0$$ is 
  • $$cosx=c\sin y$$
  • $$\sin x=c\cos y$$
  • $$\sec x-\sec y=c$$
  • $$\tan x=c$$
Solution of $$\displaystyle \frac{dy}{dx}=4+4x-3y-3xy$$ is:
  • $$2\log(4-3y)+3x^{2}+6x=c$$
  • $$\log(3-4y)+3y^{2}+6y=c$$
  • $$(4-3y)(1+x)=c$$
  • $$\log(4-3y)+x^{2}+3x=c$$

The solution of $$\displaystyle \frac{dy}{dx}=xy+x+y+1$$
  • $$y=\displaystyle \frac{x^{2}}{2}+xc$$
  • $$x=\displaystyle \frac{y^{2}}{2}+y+c$$
  • $$y+1=c.e(\displaystyle \frac{x^{2}+2x}{2})$$
  • $$y+1=x(x+1)$$
The solution of $$\sin^{-1}ydx+\displaystyle \dfrac{x}{\sqrt{1-y^{2}}}dy=0$$ is:
  • $$y\sin^{-1}x=c$$
  • $$y=c\sin^{-1}x$$
  • $$y=\displaystyle \sin(\frac{c}{x})$$
  • $$x=c\mathrm{s}in\mathrm{y}$$
lf the primitive of $$\displaystyle \frac{1}{f(x)}$$ is equal to $$\log\{f(x)\}^{2}+c$$, then $$f(x)$$ is:
  • $$ x+d$$
  • $$ \displaystyle \frac{x}{2}+d$$
  • $$\displaystyle \frac{x^{2}}{2}+d$$
  • $$ x^{2}+d$$
Find the solution of $$\displaystyle \frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}$$.
  • $$\displaystyle e^{y}=e^{x}-\frac{x^{3}}{3}+c$$
  • $$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{4}+c$$
  • $$\displaystyle e^{y}=e^{x}-\frac{x^{3}}{4}+c$$
  • $$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{3}+c$$
Find the solution of $$\displaystyle \left ( e^{y}+1 \right )\cos x dx+e^{y}\sin x dy=0$$
  • $$\displaystyle \sin x\left ( e^{y}+1 \right )=c.$$
  • $$\displaystyle \sin x\left ( e^{y}-1 \right )=c.$$
  • $$\displaystyle \sin x\left ( 2e^{y}+1 \right )=c.$$
  • $$\displaystyle \sin x\left ( 3e^{y}-1 \right )=c.$$
Let $$f$$ be the differentiable for all $$x$$, If $$f(1)=-2$$ and $${f}'\left ( x \right )\geq 2$$ for $$[1, 6]$$, then:
  • $$f(6)< 8$$
  • $$f(6)\geq 8$$
  • $$f(6)=5$$
  • $$f(6)< 5$$
$$\displaystyle x\cos ^{2}ydx=y\cos ^{2}x dy$$
  • $$\displaystyle x\tan x+\log \sec x= y\tan y+\log \sec y+c.$$
  • $$\displaystyle x\tan x+\log \sec x= y\tan y-\log \sec y+c.$$
  • $$\displaystyle x\tan x-\log \sec x= y\tan y+\log \sec y+c.$$
  • $$\displaystyle x\tan x-\log \sec x= y\tan y-\log \sec y+c.$$
$$\displaystyle e^{x-y}dx+e^{^{y-x}}dy=0$$
Solve the differential equations.
  • $$\displaystyle e^{2x}+e^{2y}=k$$
  • $$\displaystyle e^{-2x}+e^{-2y}=k$$
  • $$\displaystyle e^{-2x}+e^{-2y}=-k$$
  • $$\displaystyle e^{2x}+e^{2y}=-k$$
Find the solution of $$\displaystyle \left ( e^{x}+1 \right )y dy=\left ( y+1 \right )e^{x}dx$$.
  • $$\displaystyle k\left ( y+1 \right )\left ( e^{x}+1 \right )=e^{y}$$
  • $$\displaystyle k\left ( y-1 \right )\left ( e^{x}+1 \right )=e^{y}$$
  • $$\displaystyle k\left ( y+1 \right )\left ( e^{x}-1 \right )=e^{y}$$
  • $$\displaystyle k\left ( y-1 \right )\left ( e^{x}-1 \right )=e^{y}$$
$$e^{\frac{dy}{dx}}=x+1$$ given that when x = 0, y = 3, if x=1, then $$y=\left ( 2 \right ).\ln \left (2 \right )+k$$, what is k?
  • 1
  • 2
  • 4
  • 5
The differential equation $$\displaystyle \frac{dy}{dx}=\frac{\sqrt{1-y^{2}}}{y}$$determines a family of circles with:
  • variable radii and a fixed centre at $$\displaystyle (1,1)$$
  • variable radii and a fixed centre at $$\displaystyle (0, -1)$$
  • fixed radius 1 and variable centres along the x-axis
  • fixed radius 1 and variable centres along the y-axis
Find the solution of $$\displaystyle \left ( 1-x \right )dy-\left ( 3+y \right )dx=0$$
  • $$\displaystyle \left ( 3+y \right )\left ( 1-x \right )=k$$
  • $$\displaystyle \left ( 3-y \right )\left ( 1-x \right )=k$$
  • $$\displaystyle \left ( 3+y \right )\left ( 1+x \right )=k$$
  • $$\displaystyle \left ( 3-y \right )\left ( 1+x \right )=k$$
The solution of the equation $$\displaystyle \frac{d^{2}y}{dx^{2}}=e^{-2x}$$ is:
  • $$\displaystyle \frac{1}{4}e^{-2x}$$
  • $$\displaystyle \frac{1}{4}e^{-2x}+cx+d$$
  • $$\displaystyle \frac{1}{4}e^{-2x}+cx^{2}+d$$
  • $$\displaystyle \frac{1}{4}e^{-2x}+c+d$$
Find the solution of $$\displaystyle \left ( 1-x^{2} \right )\left ( 1-y \right )dx=xy\left ( 1+y \right )dy.$$
  • $$\displaystyle \log x-\frac{x^{2}}{2}=-\frac{y^{2}}{2}-2y-2\log \left ( 1-y \right )+k$$
  • $$\displaystyle \log x+\frac{x^{2}}{2}=-\frac{y^{2}}{2}-2y-2\log \left ( 1-y \right )+k$$
  • $$\displaystyle \log x-\frac{x^{2}}{2}=-\frac{y^{2}}{2}+2y-2\log \left ( 1-y \right )+k$$
  • $$\displaystyle \log x-\frac{x^{2}}{2}=-\frac{y^{2}}{2}+y+2\log \left ( 1-y \right )+k$$
$$\displaystyle ydx-x dy=xy\:dx$$
Then the solution is:
  • $$\displaystyle x=ky\:e^{x}.$$
  • $$\displaystyle y=kx\:e^{y}.$$
  • $$\displaystyle x=y\:e^{x}+k$$
  • $$\displaystyle y=x\:e^{y}+k$$
Solve the given differential equation  $$\displaystyle \left ( xy^{2}+x \right )dx+\left ( yx^{2}+y \right )dy=0.$$
  • $$\displaystyle \left ( x^{2}+1 \right )\left ( y^{2}+1 \right )=c$$
  • $$log\left( x^{ 2 }+1 \right) log\left( y^{ 2 }+1 \right) =c$$
  • $$\displaystyle \left ( x^{2}+1 \right )+\left ( y^{2}+1 \right )=c$$
  • none of these
Find the solution of  $$\displaystyle \frac{dy}{dx}=\frac{xy+y}{xy+x}$$
  • $$\displaystyle y-x=\log \frac{kx}{y}$$
  • $$\displaystyle y+x=\log \frac{ky}{x}$$
  • $$\displaystyle y+x=\log \frac{y}{kx}$$
  • $$\displaystyle y-x=\log \frac{ky}{x}$$
Solve the diffrential equation:  $$\displaystyle \log \frac{dy}{dx}=ax+by$$
  • $$\displaystyle -\frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$$
  • $$\displaystyle \frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$$
  • $$\displaystyle -\frac{1}{b}e^{by}=\frac{1}{a}e^{ax}+c$$
  • $$\displaystyle \frac{1}{b}e^{by}=\frac{1}{a}e^{ax}+c$$
Find the solution of  $$\displaystyle xy\frac{dy}{dx}=\frac{1+y^{2}}{1+x^{2}}\left ( 1+x+x^{2} \right )$$.
  • $$\displaystyle \frac{1}{2}\log \left ( 1+y^{2} \right )=\log x+\tan ^{-1}x+c$$
  • $$\displaystyle \frac{1}{4}\log \left ( 1+y^{2} \right )=\log x+\tan ^{-1}x+c$$
  • $$\displaystyle \frac{1}{2}\log \left ( 1-y^{2} \right )=\log x+\tan ^{-1}x+c$$
  • $$\displaystyle \frac{1}{4}\log \left ( 1-y^{2} \right )=\log x+\tan ^{-1}x+c$$
Find the solution of  $$\displaystyle a\left ( x\frac{dy}{dx}+2y \right )=xy\frac{dy}{dx}$$.
  • $$\displaystyle yx^{2}=ke^{y/a}$$
  • $$\displaystyle xy^{2}=ke^{y/a}$$
  • $$\displaystyle yx^{3}=ke^{y/a}$$
  • $$\displaystyle xy^{3}=ke^{y/a}$$
Find the solution of  $$\displaystyle\frac{dy}{dx}+\sin\left ( \frac{x+y}{2} \right ) =\sin \left ( \frac{x-y}{2} \right )$$.
  • $$\displaystyle \log \tan \frac{y}{4}=c-2\sin \frac{x}{2}$$
  • $$\displaystyle \log \tan \frac{y}{2}=c+2\sin \frac{x}{2}$$
  • $$\displaystyle \log \tan \frac{y}{2}=c-2\sin \frac{x}{2}$$
  • $$\displaystyle \log \tan \frac{y}{4}=c+2\sin \frac{x}{2}$$
Find the solution of  $$\displaystyle \left ( x^{2}-yx^{2} \right )\frac{dy}{dx}+\left ( y^{2}+xy^{2} \right )=0$$
  • $$\displaystyle \log \frac{x}{ky}=\frac{x+y}{xy}$$
  • $$\displaystyle \log \frac{x}{ky}=\frac{x-y}{xy}$$
  • $$\displaystyle \log \frac{x}{ky}=\frac{x-y}{y}$$
  • $$\displaystyle \log \frac{x}{ky}=\frac{x+y}{y}$$
$$\displaystyle \frac{dy}{dx}+\sqrt{\left ( \frac{1-y^{2}}{1-x^{2}} \right )}=0.$$
Solve the equation.
  • $$\displaystyle \sin ^{-1}y+\sin ^{-1}x=c$$
  • $$\displaystyle \sin ^{-1}y+\sin ^{-1}x=-c$$
  • $$\displaystyle -\sin ^{-1}y+\sin ^{-1}x=c$$
  • $$\displaystyle \sin ^{-1}y-\sin ^{-1}x=c$$
$$y=ae^{-1/x}+b$$ is a solution of $$\displaystyle\frac{dy}{dx}=\frac{y}{x^{2}}$$ when
  • $$a=1,b=0$$
  • $$a=3,b=1$$
  • $$a=1,b=1$$
  • $$a=2,b=2$$
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