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

The solution of $$ ye^{-x/y}dx-(xe^{(-x/y)} + y^{3}) dy=0 $$ is
  • $$ e^{-x/y} + y^{2} = C $$
  • $$ xe^{-x/y} + y = C $$
  • $$ 2e^{-x/y} + y^{2} = C $$
  • $$ e^{-x/y} + 2y^{2} = C $$
The solution of the following differential equation $$ [1+x\sqrt{(x^{2}+y^{2})}]dx+[\sqrt{(x^{2}+y^{2})-1}]ydy=0 $$ is equal to
  • $$ x^{2}+\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c $$
  • $$ x-\dfrac{y^{3}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c $$
  • $$ x-\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c $$
  • None of these
The solution of the differential equation $$\dfrac{dy}{dx}=\dfrac{3x^{2}y^{4}+2xy}{x^{2}-2x^{3}y^{3}} $$ is
  • $$ \dfrac{y^{2}}{x}-x^{3}y^{2}=c $$
  • $$ \dfrac{x^{2}}{y^{2}}+x^{3}y^{3}=c $$
  • $$ \dfrac{x^{2}}{y}+x^{3}y^{2}=c $$
  • $$ \dfrac{x^{2}}{3y}-2x^{3}y^{2}=c $$
The solution of the differential equation $$ x^{2}\dfrac{dy}{dx}\cos \dfrac{1}{x} - y\sin\dfrac{1}{x} = -1, $$ where $$ y \rightarrow -1 $$ as $$ x \rightarrow \infty $$ is
  • $$ y=\sin\dfrac{1}{x}-\cos \dfrac{1}{x} $$
  • $$ y= \dfrac{x+1}{x\sin \dfrac{1}{x}} $$
  • $$ y= \cos \dfrac{1}{x} + \sin\dfrac{1}{x} $$
  • $$ y= \dfrac{x+1}{x\cos 1/x} $$
The solution of the differential equation $$ y(2x^{2}+y)\dfrac{dy}{dx}=(1-4xy^{2})x^{2} $$ is given by
  • $$ 3(x^{2}y)^{2}+y^{3}-x^{3}=c $$
  • $$ xy^{2}+\dfrac{y^{3}}{3}-\dfrac{x^{3}}{3}= 0 $$
  • $$ \dfrac{2}{5}yx^{5}+\dfrac{y^{3}}{3}=\dfrac{x^{3}}{3}-\dfrac{4xy^{3}}{3}+c $$
  • None of these
Solution of the differential equation $$ (y+x\sqrt{xy}(x+y))dx + (y\sqrt{xy}(x+y)-x)dy = 0 $$ is
  • $$ \dfrac{x^{2}+y^{2}}{2}+\tan^{-1}\sqrt{\dfrac{y}{x}} =c $$
  • $$ \dfrac{x^{2}+y^{2}}{2}+\tan^{-1}\sqrt{\dfrac{x}{y}} =c $$
  • $$ \dfrac{x^{2}+y^{2}}{2}+2\cot^{-1}\sqrt{\dfrac{x}{y}} =c $$
  • None of these
Number of values of $$m\in N$$ for which $$y=e^{mx}$$ is a solution of the differential equation $$\dfrac{d^3y}{dx^3}-3\dfrac{d^2y}{dx^2}-4\dfrac{dy}{dx}+12y=0 $$
  • $$0$$
  • $$1$$
  • $$2$$
  • More than $$2$$
The solution of, $$ydx-xdy+(1+x^2)dx+x^2 \sin y dy=0$$,
  • $$x+1-y^2+ \cos y+C=0$$
  • $$y+1-x^2+ \cos y+C=0$$
  • $$\dfrac{x}{y}+\dfrac{1}{y}- y+ \cos y +C=0$$
  • $$\dfrac {y} {x} +\dfrac{1}{x}- x+ \cos y +C=0$$
  • Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
  • Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
  • Assertion is correct but Reason is incorrect
  • Both Assertion and Reason are incorrect
The solution of $$\dfrac{dy}{dx}+x\sin 2y=x^{3}\cos^{2}y$$, is
  • $$e^{x^{2}}=(x^{2}-1)e^{x^{2}}\tan y+C$$
  • $$e^{x^{2}}\tan y=\dfrac{1}{2}(x^{2}-1)e^{x^{2}}+C$$
  • $$e^{x^{2}}\tan y=(x^{2}-1)\tan y+C$$
  • $$None\ of\ these$$
Differential equation $$\dfrac{dy}{dx}=f(x)g(x)$$ can be solved by
separating variable $$\dfrac{dy}{g(y)}=f(x)dx$$

The equation of the curve to the point $$(1,0)$$ which
stratifies the differential equation $$(1+y^2)dx-xydy=0$$
  • $$x^2+y^2=1$$
  • $$x^2-y^2=1$$
  • $$x^2+y^2=2$$
  • $$x^2-y^2=2$$
The solution of the differential equation $$(e^{x^2}+e^{y^2})y\dfrac{dy}{dx}+e^{x^{2}}(xy^2-x)=0$$ is
  • $$ e^{x^2}(y^2-1)+e^{y^2}=C $$
  • $$ e^{y^2}(x^2-1)+e^{x^2}=C $$
  • $$ e^{y^2}(y^2-1)+e^{x^2}=C $$
  • $$ e^{x^2}(y-1)+e^{y^2}=C $$
$$y$$ satisfies the differential equation
  • $$\dfrac{dy}{dx}+y=e^x (\cos x- \sin x) - e^{-x} (\cos x- \sin x)$$
  • $$\dfrac{dy}{dx}-y=e^x (\cos x- \sin x) + e^{-x} (\cos x+ \sin x)$$
  • $$\dfrac{dy}{dx}+y=e^x (\cos x+ \sin x) - e^{-x} (\cos x- \sin x)$$
  • $$\dfrac{dy}{dx}-y=e^x (\cos x- \sin x) + e^{-x} (\cos x- \sin x)$$
Solution of the diffrential equation $$\dfrac{dy}{dx}+ \dfrac {1+y^2}{\sqrt {1-x^2}}=0$$ is
  • $$ \tan^{-1} y + \sin ^{-1} x=C$$
  • $$ \tan^{-1} x + \sin ^{-1} y=C$$
  • $$ \tan^{-1} x . \sin ^{-1} y=C$$
  • $$ \tan^{-1} x - \sin ^{-1} y=C$$
Differential equation $$\dfrac{dy}{dx}=f(x)g(x)$$ can be solved by
separating variable $$\dfrac{dy}{g(y)}=f(x)dx$$

If $$\dfrac{dy}{dx}=1+x+y+xy$$ and $$y(-1)=0$$, then $$y$$ is equal to 
  • $$e^{\dfrac{(1-x)^1}{2}}$$
  • $$e^{\dfrac{(1+x)^2}{2}} -1$$
  • In $$(1+x)-1$$
  • $$1+x$$
The solution of the differential equation $$ \dfrac{x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+ \dots}{1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dots}=\dfrac{dx-dy}{dx+dy} $$ is
  • $$2ye^{2x}=Ce^{2x}+1 $$
  • $$2ye^{2x}=Ce^{2x}-1$$
  • $$ ye^{2x}=Ce^{2x}+2 $$
  • None of these
Integrating factor of the differential equation $$(1-x^2)\dfrac {dy}{dx}-xy=1$$ is 
  • $$-x$$
  • $$\dfrac {x}{1+x^2}$$
  • $$\sqrt {1-x^2}$$
  • $$\dfrac 12 \log (1-x^2)$$
Which of the following is a second order differential equation?
  • $$(y')^2+x=y^2$$
  • $$y'y''+y=\sin x$$
  • $$y''' +(y'')^2+y=0$$
  • $$y'=y^2$$
$$y=x$$ is a particular solution of the differential equation $$\dfrac {d^2y}{dx^2}-x^2 \dfrac {dy}{dx}+xy=x$$.
  • True
  • False
$$x+y=\tan^{-1}y$$ is a solution of differential equation $$y^2 \dfrac {dy}{dx}+y^2+1=0$$.
  • True
  • False
State true or false. 
The solution of $$\dfrac{dy}{dx}=\left(\frac{y}{x}\right)^{\tfrac{1}{3}}$$ is $$y^{\tfrac{2}{3}}-x^{\tfrac{2}{3}}=C$$.
  • True
  • False
The order of the differential equation
$$(2 x^{2} \cfrac{d^{2} y}{d x^{2}}-3 \cfrac{d y}{d x}+y=0)$$
  • $$2$$
  • $$1$$
  • $$0$$
  • Not defined
The solution of the differential equation $$\dfrac {dy}{dx}=\dfrac {1+y^2}{1+x^2}$$ is :
  • $$y=\tan^{-1}x$$
  • $$y-x=k(1+xy)$$
  • $$x=\tan^{-1}y$$
  • $$\tan ( xy)=k$$
Which of the following is the general solution of $$\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0$$?
  • $$y=(Ax+B) e^x$$
  • $$y=(Ax+B) e^{-x}$$
  • $$y=Ae^x +Be^{-x}$$
  • $$y=A\cos x+B\sin x$$
Solution of $$\displaystyle \sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\frac{dy}{dx}=0$$, is:
  • $$\log\left(\displaystyle \frac{x}{1+\sqrt{1+x^{2}}}\right)+\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$$
  • $$\displaystyle \log\left(\frac{x}{\sqrt{1+x^{2}}}\right)+\sqrt{1-x^{2}}+\sqrt{1+y^{2}}=c$$
  • $$\displaystyle \log\left(\frac{x}{\sqrt{1+x^{2}}}\right)=c$$
  • $$\displaystyle \log\left(\sqrt{1+x^{2}}-\sqrt{1+y^{2}}\right)+\log\left(\frac{x}{\sqrt{1+x^{2}}}\right)=c$$
lf $$f (x)$$ and $$g (x)$$ are two solutions of the differential equation $$a\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$$, then $$f (x) - g (x)$$ is the solution of
  • $$a^{2}\displaystyle \frac{d^{2}y}{dx^{2}}+\displaystyle \frac{dy}{dx}+y=e^{\displaystyle x}$$
  • $$\displaystyle {a}^{2}\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$$
  • $$a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=0$$
  • $$a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$$
Solution of $$\dfrac{d^{2}y}{dx^{2}}$$= $$\log x$$ is:
  • $$y=\dfrac{1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$$
  • $$y=\dfrac{1}{2}x^{2}\log x+\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$$
  • $$y=\dfrac{-1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$$
  • $$y=3x^{2}+4x+c_{1}$$

lf the solution of the differential equation
$$\displaystyle \frac{1}{\sin^{-1}x}(\frac{dy}{dx})=1$$ is $$y=x \sin^{-1}x+f(x)+c$$ then f (x) is
  • $$1-x^{2}$$
  • $$\sqrt{1-x^{2}}$$
  • $$1+x^{2}$$
  • $$\sqrt{1+x^{2}}$$
The solution of $$\displaystyle \frac{dy}{dx}+\frac{x(1+y^{3})}{y^{2}(1+x^{2})}=0$$ is:
  • $$(1-x^{2})+(1+y^{3})=c$$
  • $$(1+y^{2})+(1+x^{3})=c$$
  • $$(1+x^{2})^{3}(1+y^{3})^{2}=c$$
  • $$(1+x^{2})+(1+y^{3})=cxy^{2}$$
A tangent drawn to the curve $$\mathrm{y}=\mathrm{f}(\mathrm{x})$$ at $$\mathrm{P}(\mathrm{x}, \mathrm{y})$$cuts the $$\mathrm{x}$$-axis and $$\mathrm{y}$$-axis at A and $$\mathrm{B}$$ respectively such that BP: AP $$=3$$ : 1, given that $$\mathrm{f}(1)=1$$, then 

  • equation of curve is $$\displaystyle \mathrm{x}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-3\mathrm{y}=0$$
  • normal at (1, 1) is $$\mathrm{x}+3\mathrm{y}=4$$
  • curve passes through (2, 1/8)
  • equation of curve is $$\displaystyle \mathrm{x}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+3\mathrm{y}=0$$
The solution of $$(1-x^{2})\displaystyle \frac{dy}{dx}+xy=xy^{2}$$ is:
  • $$ y(c+\sqrt{1-x^{2}})=1-x^{2}$$
  • $$y(c-\sqrt{1-x^{2}})=1-x^{2}$$
  • $$y(c+\sqrt{1-x^{2}})=\sqrt{1-x^{2}}$$
  • $$y(c-\sqrt{1-x^{2}})=\sqrt{1-x^{2}}$$
Solution of $$\>y-x\displaystyle \frac{dy}{dx}=5\left(y^{2}+\frac{dy}{dx}\right)$$, is:
  • $$(5+x)(1-5y)=cy$$
  • $$\dfrac { 5+x }{ 1-5y } =cy$$
  • $$(5-x)(1+5y)=cy$$
  • $$(5+x)(5-x)=c$$
Solve : $$\dfrac{dy}{dx}=\dfrac{x\left ( 2\ln x+1 \right )}{\sin y+y\cos y}$$
  • $$y\sin y=x^{2}\ln x+c$$
  • $$y\sin x=y^{2}\ln x+c$$
  • $$y\sin y=x^{2}\ln x-c$$
  • $$y\sin x=y^{2}\ln x-c$$
If $$ \dfrac{dy}{dx}=xy+2x+3y+6$$, then find the value of $$y(-1)-e^2y(-3)$$.
  • $$e^2-1$$
  • $$e^2+1$$
  • $$2(e^2-1)$$
  • $$-2(e^2-1)$$
Solve: $$y-x\dfrac{dy}{dx}=a\left ( y^{2}+\dfrac{dy}{dx} \right )$$
  • $$y=c(1-ay)(x+a)$$
  • $$2y=c(1-ay)(x+a)$$
  • $$y=c(1-ay)(x+a)/2$$
  • $$2y=c(1+ay)(x+a)$$
$$\dfrac{dy}{dx}+\dfrac{\sqrt{\left ( x^{2}-1 \right )\left ( y^{2}-1 \right )}}{xy}=0$$
Solve the above equation:
  • $$\sqrt{x^{2}-1}-\sec ^{-1}x+\sqrt{y^{2}-1}=c$$
  • $$\sqrt{x^{2}-1}-\sec ^{-1}x+\sqrt{y^{2}-1}=-c$$
  • $$\sqrt{x^{2}-1}+\sec ^{-1}x+\sqrt{y^{2}-1}=c$$
  • $$\sqrt{x^{2}-1}-\sec ^{-1}x-\sqrt{y^{2}-1}=c$$
The general solution of differential equation $$x\left ( 1+y^{2} \right )\mathrm{d} x+y\left ( 1+x^{2} \right )\mathrm{d} y=0$$ is/are:
  • $$\displaystyle \frac{1}{2} \ln\left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=k$$
  • $$ \displaystyle \left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=c$$
  • $$ \displaystyle \left ( 1+y^{4} \right )=c\left ( 1+x^{2} \right )$$
  • $$ \displaystyle \ln\left [\frac{(1+x^{2})}{ (1+y^{2}) } \right ]= k $$
If the length of tangent at any point on the curve $$y=f(x)$$ intercepted between the points and the x-axis is of length $$1$$. Find the equation of the curve.
  • $$\displaystyle \sqrt { 1-{ y }^{ 2 } } -\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =x+c$$
  • $$\displaystyle \sqrt { 1-{ y }^{ 2 } } -\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =-x+c$$
  • $$\displaystyle \sqrt { 1-{ y }^{ 2 } } +\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =x+c$$
  • $$\displaystyle \sqrt { 1-{ y }^{ 2 } } +\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =-x+c$$
The solution of differential equation $$(e^x + 1) y dy = (y + 1) e^x dx$$ is :
  • $$(e^x + 1) (y + 1) = ce^y$$
  • $$(e^x + 1) (y + 1) = ce^{-y}$$
  • $$(e^x + 1) (y + 1) = ce^{2y}$$
  • none of the above
The solution of differential equation  $$(e^x + 1)y dy = (y + 1) e^x  dx$$ is:
  • $$(e^x + 1) (y + 1) = c . e^y$$
  • $$(e^x + 1) |(y + 1)| = ce^{-y}$$
  • $$(e^x + 1) (y + 1) = c$$
  • none of these
Solution of $$\displaystyle \frac { dy }{ dx } =\frac { y\left( x\log { y } -y \right)  }{ x\left( y\log { x } -x \right)  } $$ is:
  • $${ x }^{ y }=c{ y }^{ x }$$
  • $$xy=c$$
  • $${ \left( xy \right) }^{ x }=c$$
  • None of these
If $$\int_{a}^{x} ty (t) dt = x^{2} + y(x)$$ then y as a function of x is:
  • $$y = 2 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$
  • $$y = 1 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$
  • $$y = 2 - (1 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}$$
  • none
Through any point $$\left ( x,y \right )$$ of a curve which passes through the origin, lines are drawn parallel to the coordiante axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of 
  • circles
  • parabolas
  • hyperbolas
  • straight lines
The solutions of $$v=u\displaystyle\frac{dv}{du}+\left (\displaystyle\frac{dv}{du} \right )^{2}$$ where $$u = y$$ and $$v=xy$$ are:
  • $$y=0$$
  • $$y=-4x$$
  • $$xy=cy+c^{2}$$
  • $$x^{2}y=cy+c^{2}$$
$$\displaystyle y-x\frac{dy}{dx}=b\left ( 1+x^{2}\frac{dy}{dx} \right ).$$
Solve the above differential  equation.
  • $$\displaystyle y=k\left ( y-b \right )\left ( 1+bx \right ).$$
  • $$\displaystyle y=k\left ( y+b \right )\left ( 1-bx \right ).$$
  • $$\displaystyle y=k\left ( y+b \right )\left ( 1+bx \right ).$$
  • $$\displaystyle y=k\left ( y-b \right )\left ( 1-bx \right ).$$
The solution of $$\displaystyle\frac{dy}{dx}+x=xe^{(n-1)y}$$ is:
  • $$\displaystyle\frac{1}{n-1}\log \left (\displaystyle\frac{e^{(n-1)y}-1}{e^{(n-1)y}} \right )=x^{2}/2+C$$
  • $$e^{(n-1)y}=Ce^{(n-1)y+(n-1)x^{2/2}}+1$$
  • $$\log \left (\displaystyle\frac{e^{(n-1)y}-1}{(n-1)e^{(n-1)y}} \right )=x^{2}+C$$
  • $$e^{(n-1)}y=ce^{(n-1)y^{2/2+x}}+1$$
The general solution of $$xy^{5}=y_{4}(y_{n}=\displaystyle\frac{d^{n}y}{dx^{n}})$$ is given by:
  • $${ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{3}+C_{3}x^{2}+C_{4}x+C_{5}$$
  • $${ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{4}+C_{3}x^{2}+C_{4}x+C_{5}$$
  • $${ y }^{ -1 }=C_{1}+C_{2}x+C_{3}x^{2}+C_{4}x^{3}+C_{5}x^{4}$$
  • none of these
The normal to a curve at $$P\left ( x,y \right ) $$ meets the $$x$$-axis at $$G$$. If the distance of $$G$$ from the origin is twice the abscicca of $$P$$, then the curve is a:
  • circle
  • hyperbola
  • ellipse
  • parabola
The differential equation $$\displaystyle \frac{dy}{dx}= \displaystyle \frac{\sqrt{1-y^{2}}}{y}$$ determines a family of circles with:
  • variable radii and a fixed centre $$\left ( 0,1 \right )$$
  • variable radii and a fixed centre $$\left ( 0,-1 \right )$$
  • fixed radius $$1$$ and variable centres along the $$x$$-axis
  • fixed radius $$1$$ and variable centres along the $$y$$-axis
The general solution of the equation $$\displaystyle \frac { dy }{ dx } =\frac { { x }^{ 2 } }{ { y }^{ 2 } } $$ is:
  • $$x^3-y^3=c$$
  • $$x^3+y^3=c$$
  • $$x^2+y^2=c$$
  • $$x^2-y^2=c$$
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