MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 11

The solution of

$x^2dy-y^2dx+xy^2(x-y)dy=0$ , is?

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$log\left|\dfrac{x-y}{xy}\right|=\dfrac{y^2}{2}+C$
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$log\left|\dfrac{xy}{x-y}\right|=\dfrac{x^2}{2}+C$
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$\log \left|\dfrac{x-y}{xy}\right|=\dfrac{x^2}{2}+C$
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$log\left|\dfrac{x-y}{xy}\right|=x+C$

Solution of equation

$\frac { d y } { d x } = \frac { y ^ { \frac { d } { d } ( \phi ( x ) ) } - y ^ { 2 } } { \phi ( x ) }$ is

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

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 _______

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Differential equation
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Continuous function
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Discontinuous function
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Integral equation

The integrating factor of the equation

$siny\dfrac{dy}{dx}=cosy\left(1-xcosy\right)$ is

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$e^{-x}$
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$e^{x}$
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$e^{secy}$
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$None\ of\ these$

Solution of differential equation

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

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$\tan{y}=(x-1)+Ce-x$
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$\cot{y}=(x-1)+Ce-x$
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$\tan{y}=(x-1)ex+c$
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$\cot{y}=(x-1)ex+c$

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:

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$4 - \dfrac { 2 }{ y } - \dfrac {e^{\dfrac { 1 }{ y }}}e$
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$3 - \dfrac { 1 }{ y } + \dfrac {e^{\dfrac { 1 }{ y }}}e$
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$1 + \dfrac { 1 }{ y } - \dfrac {e^{\dfrac { 1 }{ y }}}e$
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$1 - \dfrac { 1 }{ y } + \dfrac {e^{\dfrac { 1 }{ y }}}e$

Solution of differential equation

$\frac{d y}{d x}+x \sin ^{2} y=\sin y \cos y$ is-

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$\tan y=(x-1)+ce^{-x}$
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$\cot y=(x-1)+ ce^{-x}$
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$\tan y = (x-1) e^x +C$
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$\cot y=(x-1) e^x+C$

(where C is an arbitrary constant)

Solve the differential equation:

$(e^ {y}+1)\cos x\ dx+e^ {y}\sin x\ dy=0$

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$\sin x(e^{-y}+1)=c$
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$\sin x(e^{y}+1)=c$
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$\sin x(e^{2y}+1)=c$
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$\cos x(e^{y}+1)=c$

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:

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$600-500e^{t/2}$
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$400-300e^{-t/2}$
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$400-300e^{t/2}$
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$300-200e^{-t/2}$

The solution of

$y^5x+y-x\dfrac{dy}{dx}=0$ is

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$\dfrac{x^4}{4}+\dfrac{1}{5}\left(\dfrac{x}{y}\right)^5=C$
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$\dfrac{x^5}{4}+\dfrac{1}{5}\left(\dfrac{x}{y}\right)^4=C$
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$\left(\dfrac{x}{y}\right)^5+\dfrac{x^4}{4}=C$
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$(xy)^4+\dfrac{x^5}{5}=C$

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

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1
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2
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3
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5

$y = (c_{1} + c_{2}x + x^{2}) e^{x}$ is a solution of

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

An integrating factor of the differential equation

$x\frac{dy}{dx}-y=x^{3};x>0$ is ________

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$\frac{1}{x}$
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-x
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$-\frac{1}{x}$
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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

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k = 0
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k > 0
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k < 0
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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

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$y=in(x)+C$
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$y=(in\,x)^2+C$
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$y=\pm\sqrt{(in\,x)^2+C}$
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$x^y=xy+K$

Solution of differential equation

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

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

$y=3\sqrt{2x}$ Find

$\frac{dy}{dx}$

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$\frac{3}{\sqrt{2x}}$
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$\frac{2}{\sqrt{2x}}$
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$\frac{2}{\sqrt{3x}}$
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$\frac{2}{\sqrt{4x}}$

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:

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$\dfrac{\sqrt{3}}{2}$
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$\dfrac{1}{\sqrt{2}}$
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$-\dfrac{\sqrt{3}}{2}$
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$-\dfrac{1}{\sqrt{2}}$

Explanation

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

$\Rightarrow \dfrac{dy}{\sqrt{1-y^2}} + \dfrac{dx}{\sqrt{1-x62}} = 0$

Taking integration

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

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

$x = \dfrac{1}{2}$ ,

$y = \dfrac{\sqrt{3}}{2}$

$\Rightarrow\sin^{-1} \left(\dfrac{\sqrt{3}}{2} \right) + \sin^{-1}\left(\dfrac{1}{2}\right) = c$

$\Rightarrow\dfrac{\pi}3+\dfrac{\pi}6=c$

$\Rightarrow c = \dfrac{\pi}{2}$

Put in equation (1)

$\sin^{-1} y = \cos^{-1}x$

$y\left(\dfrac{-1}{\sqrt{2}}\right) = \sin\left(\cos^{-1}\left(\dfrac{-1}{\sqrt{2}}\right)\right) = \dfrac{1}{\sqrt{2}}$

$\therefore$

$\boxed{y\left(\dfrac{-1}{\sqrt{2}}\right) = \dfrac{1}{\sqrt{2}}}.....Answer$

Hence option

$'B'$ is the answer.

Equation of the curve in which the subnormal is twice the square of the ordinate is given by

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$log y = 2x +log c$
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$y = ce^{2x}$
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$log y = 2x - log c$
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None of these

The solution of

$\dfrac{dy}{dx}+\dfrac{x}{1-x^{2}}y=x\sqrt{y}$ , is given by

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$3\sqrt{y}+(1-x^{2})=C(1-x^{2})^{1/4}$
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$\dfrac{3}{2}\sqrt{y}+1(1-x^{2})=C(1-x^{2})^{3/2}$
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$3\sqrt{y}-(1-x^{2})=C(1-x^{2})^{3/2}$
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$None\ of\ these$

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.

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True
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False

The solution of

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

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$(x-2)^4 \left(1+\dfrac {2y}{x}\right)=ke^{2y/x}$
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$(x-2)^4 \left(1+\dfrac {2(y-2)}{(x+2)}\right)=ke^{\dfrac {2(y-2)}{(x+2)}}$
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$(x-2)^3 \left(1+2\dfrac {(y-2)}{x+2}\right)=ke^{\dfrac {2(y-2)}{(x+2)}}$
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$none\ of\ these$

$(y^2 -2x^2 y)dx+(2ky^2-x^3)dy=0$ then the value of

$xy\sqrt {y^2 -x^2}$ is

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$y^2 +x$
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$xy^2$
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any constant
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None of these

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

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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