CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 11 - MCQExams.com

The solution of $$x^2dy-y^2dx+xy^2(x-y)dy=0$$, is?
  • $$log\left|\dfrac{x-y}{xy}\right|=\dfrac{y^2}{2}+C$$
  • $$log\left|\dfrac{xy}{x-y}\right|=\dfrac{x^2}{2}+C$$
  • $$\log \left|\dfrac{x-y}{xy}\right|=\dfrac{x^2}{2}+C$$
  • $$log\left|\dfrac{x-y}{xy}\right|=x+C$$
Solution of equation $$ \frac { d y } { d x } = \frac { y ^ { \frac { d } { d } ( \phi ( x ) ) } - y ^ { 2 } } { \phi ( x ) } $$ is
  • $$

    y = \frac { \phi ( x ) + c } { x }

    $$
  • $$

    y = \frac { \phi ( x ) } { x } + c

    $$
  • $$

    y = \frac { \phi ( x ) } { x + c }

    $$
  • $$

    y = \phi ( x ) + x + c

    $$
An equation containing an independent variable, dependent variable, and differential coefficients of the dependent variable with respect to the independent variable is called a/an _______
  • Differential equation
  • Continuous function
  • Discontinuous function
  • Integral equation
The integrating factor of the equation $$siny\dfrac{dy}{dx}=cosy\left(1-xcosy\right)$$ is 
  • $$e^{-x}$$
  • $$e^{x}$$
  • $$e^{secy}$$
  • $$None\ of\ these$$
Solution of differential equation $$\cfrac{dy}{dx}+x\sin^2{y}=\sin{y}\cos{y}$$ is-
  • $$\tan{y}=(x-1)+Ce-x$$
  • $$\cot{y}=(x-1)+Ce-x$$
  • $$\tan{y}=(x-1)ex+c$$
  • $$\cot{y}=(x-1)ex+c$$
Consider the differential equation $$y^{ 2 }dx + \left(x - \dfrac { 1 }{ y }\right) dy = 0$$
If $$y \left( 1 \right) = 1$$, then $$x$$ is given by:
  • $$4 - \dfrac { 2 }{ y } - \dfrac {e^{\dfrac { 1 }{ y }}}e$$
  • $$3 - \dfrac { 1 }{ y } + \dfrac {e^{\dfrac { 1 }{ y }}}e$$
  • $$1 + \dfrac { 1 }{ y } - \dfrac {e^{\dfrac { 1 }{ y }}}e$$
  • $$1 - \dfrac { 1 }{ y } + \dfrac {e^{\dfrac { 1 }{ y }}}e$$
Solution of differential equation $$  \frac{d y}{d x}+x \sin ^{2} y=\sin y \cos y  $$ is-
  • $$

    \tan y=(x-1)+ce^{-x}

    $$
  • $$

     \cot y=(x-1)+ ce^{-x}

    $$
  • $$\tan y  = (x-1)  e^x  +C$$
  • $$\cot y=(x-1) e^x+C$$


    (where C is an arbitrary constant)
Solve the differential equation:
$$(e^ {y}+1)\cos x\ dx+e^ {y}\sin x\ dy=0$$
  • $$\sin x(e^{-y}+1)=c$$
  • $$\sin x(e^{y}+1)=c$$
  • $$\sin x(e^{2y}+1)=c$$
  • $$\cos x(e^{y}+1)=c$$
Let the population of rabbits surviving at a time t be governed by the differential equation $$\dfrac{dp(t)}{dt}=\dfrac{1}{2}p(t)-200.$$ If $$p(0)=100$$ , then p(t) equals:
  • $$600-500e^{t/2}$$
  • $$400-300e^{-t/2}$$
  • $$400-300e^{t/2}$$
  • $$300-200e^{-t/2}$$
The solution of $$y^5x+y-x\dfrac{dy}{dx}=0$$ is
  • $$\dfrac{x^4}{4}+\dfrac{1}{5}\left(\dfrac{x}{y}\right)^5=C$$
  • $$\dfrac{x^5}{4}+\dfrac{1}{5}\left(\dfrac{x}{y}\right)^4=C$$
  • $$\left(\dfrac{x}{y}\right)^5+\dfrac{x^4}{4}=C$$
  • $$(xy)^4+\dfrac{x^5}{5}=C$$
Suppose y=y(x) satisfies the differential equation $$ydx+y^{2}dy=xdy.$$ If $$y(x) > 0 \forall x\epsilon R$$ and y(1)=1, then y(-3) equals
  • 1
  • 2
  • 3
  • 5
$$y = (c_{1} + c_{2}x + x^{2}) e^{x}$$ is a solution of
  • $$\dfrac {d^{2}y}{dx^{2}} + y = 0$$
  • $$\dfrac {d^{2}y}{dx^{2}} + xy = 0$$
  • $$\dfrac {d^{2}y}{dx^{2}} = x$$
  • $$\dfrac {d^{2}y}{dx^{2}} - 2\dfrac {dy}{dx} + y = 2e^{x}$$
An integrating factor of the differential equation $$x\frac{dy}{dx}-y=x^{3};x>0$$ is ________
  • $$\frac{1}{x}$$
  • -x
  • $$-\frac{1}{x}$$
  • x
The solution of the differential equation $$ \frac { d y } { d x } - k y = 0 , y ( 0 ) = 1 $$, approaches zero when $$ x \rightarrow \infty , $$ if
  • k = 0
  • k > 0
  • k < 0
  • k may be any real value
Solution of the differential equation
$$x=1+\left(xy\dfrac{dy}{dx}\right)+\dfrac{(x^2y^2)}{2!}\left(\dfrac{dy}{dx}\right)^3+.........$$ is
  • $$y=in(x)+C$$
  • $$y=(in\,x)^2+C$$
  • $$y=\pm\sqrt{(in\,x)^2+C}$$
  • $$x^y=xy+K$$
Solution of differential equation $$2xy \dfrac {dy}{dx} = x^{2} + 3y^{2}$$.
  • $$x^{3} + y^{2} = px^{2}$$
  • $$\dfrac {x^{2}}{2} + \dfrac {y^{3}}{x} =y^{2} + p$$
  • $$x^{2} + y^{3} = px$$
  • $$x^{2} + y^{2} = px^{3}$$
$$y=3\sqrt{2x}$$  Find  $$\frac{dy}{dx}$$
  • $$\frac{3}{\sqrt{2x}}$$
  • $$\frac{2}{\sqrt{2x}}$$
  • $$\frac{2}{\sqrt{3x}}$$
  • $$\frac{2}{\sqrt{4x}}$$
Let $$y = y(x)$$ be a solution of the differential equation, $$\sqrt{1-x^2}\dfrac{dy}{dx} + \sqrt{1-y^2} = 0, |x| < 1$$.
If $$y\left(\dfrac{1}{2}\right) = \dfrac{\sqrt{3}}{2}$$, then $$y \left(\dfrac{-1}{\sqrt{2}}\right)$$ is equal to:
  • $$\dfrac{\sqrt{3}}{2}$$
  • $$\dfrac{1}{\sqrt{2}}$$
  • $$-\dfrac{\sqrt{3}}{2}$$
  • $$-\dfrac{1}{\sqrt{2}}$$
Equation of the curve in which the subnormal is twice the square of the ordinate is given by 
  • $$ log y = 2x +log c $$
  • $$ y = ce^{2x} $$
  • $$ log y = 2x - log c $$
  • None of these
The solution of $$\dfrac{dy}{dx}+\dfrac{x}{1-x^{2}}y=x\sqrt{y}$$, is given by
  • $$3\sqrt{y}+(1-x^{2})=C(1-x^{2})^{1/4}$$
  • $$\dfrac{3}{2}\sqrt{y}+1(1-x^{2})=C(1-x^{2})^{3/2}$$
  • $$3\sqrt{y}-(1-x^{2})=C(1-x^{2})^{3/2}$$
  • $$None\ of\ these$$
State true or false.
$$\dfrac {dy}{dx}+y=5$$ is a differential equation of the type $$\dfrac {dy}{dx}+PY=Q$$ but it can be solved using the variable separable method also.
  • True
  • False
The solution of $$\dfrac {dy}{dx}=\dfrac {(x+y)^2}{(x+2)(y-2)}$$ is given by
  • $$(x-2)^4 \left(1+\dfrac {2y}{x}\right)=ke^{2y/x}$$
  • $$(x-2)^4 \left(1+\dfrac {2(y-2)}{(x+2)}\right)=ke^{\dfrac {2(y-2)}{(x+2)}}$$
  • $$(x-2)^3 \left(1+2\dfrac {(y-2)}{x+2}\right)=ke^{\dfrac {2(y-2)}{(x+2)}}$$
  • $$none\ of\ these$$
If $$(y^2 -2x^2 y)dx+(2ky^2-x^3)dy=0$$ then the value of $$xy\sqrt {y^2 -x^2}$$ is 
  • $$y^2 +x$$
  • $$xy^2$$
  • any constant
  • None of these
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