Differential Equations Test

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Differential Equations Test - 42

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Weekly Quiz Competition

Question 1
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The solution of $$xdx+ydy=\frac { xdy-ydx }{ { x }^{ 2 }+{ y }^{ 2 } } $$ is ________________________.

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

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

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

D
$${ x }^{ 2 }-{ y }^{ 2 }=2\quad { tan }^{ -1 }\frac { x }{ y } +C$$
Question 2
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The solution of the differential equation $$(x^2-yx^2)\dfrac{y^3}{x}=k+y^2+xy^2=0$$ is?

A
$$log\left(\dfrac{x}{y}\right)=\dfrac{1}{x}+\dfrac{1}{y}+c$$

B
$$log\left(\dfrac{y}{x}\right)=\dfrac{1}{x}+\dfrac{1}{y}+c$$

C
$$log(xy)=\dfrac{1}{x}+\dfrac{1}{y}+c$$

D
$$log(xy)+\dfrac{1}{x}+\dfrac{1}{y}=c$$

Solution
Question 3
1 / -0

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

A
f\left( x \right) is differentable for all $$\quad x\epsilon R$$

B
f\left( x \right) is not differentable at$$ \quad x\epsilon 0$$

C
f\left( x \right) is not differentable if$$ x=n\pi .n\epsilon I$$

D
f\left( x \right) is not differentable if$$ x=\frac { n\pi }{ 2 } n\epsilon I$$

Solution
Question 4
1 / -0

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

A
y$$\dfrac{dy}{dx} + x \dfrac{d^2y}{dx^2} = 1$$

B
x$$\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} = 1$$

C
$$y^2\dfrac{dy}{dx} + 2x \dfrac{dy}{dx} = 0$$

D
$$x\dfrac{dy}{dx} -x^2 \dfrac{d^2y}{dx^2} = 0$$

Solution

Parabola with axis on y- axis is given by,
$$ x^{2}= 4a(y-c), $$ Differentiating, $$\left \{ center = (0,c) \right \}$$
$$ 2x = \frac{4 a dy}{dx} \Rightarrow a = \frac{x}{2y'}$$
$$ \Rightarrow x^{2} = 4(\frac{x}{2y'})y$$ $$ \Rightarrow x^{2}y' = 2xy$$
$$ \Rightarrow 2x = 4ay' $$ $$ \Rightarrow a = \frac{x}{2y'} $$
$$ \Rightarrow x^{2} = \frac{4x}{2y'}(y-c)$$ $$\Rightarrow 2y'x^{2}= 2xy-2cx$$
$$ \Rightarrow {y}''x^{2}+2xy'= 2xy'+2y-2c$$
$$ \Rightarrow c = \frac{2y-{y}''x^{2}}{2}$$
$$ \Rightarrow x^{2} = \frac{2x}{y'}(y-y+{y}''\frac{x^{2}}{2})$$
$$\Rightarrow x^{2}y'= x({y}''x^{2})$$
$$ \Rightarrow x \frac{dy}{dx} -x^{2}\frac{d^{2}y}{dx^{2}} = 0 $$
Question 5
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$$Let_y-y(x)$$ 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

A
$$-\dfrac{4}{9}\pi^2$$

B
$$-\dfrac{4}{9\sqrt3}\pi^2$$

C
$$-\dfrac{-8}{9\sqrt3}\pi^2$$

D
None of these

Solution

$$\Rightarrow \dfrac{dy}{dx}-\cot x\cdot y=\dfrac{4x}{\sin x}$$
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,
$$\dfrac{y}{\sin x}=4\left[x(-\cot x)-\displaystyle\int (-\cot x)dx\right]$$
$$=4\left[-x\cot x+ln \sin x\right]+c$$
$$\Rightarrow y=-4x\sin x+4\sin x\cdot ln \sin x+c\sin x$$
As $$y\left(\dfrac{\pi}{2}\right)=0$$
$$\Rightarrow 0$$
$$=-2\pi+0+c$$
$$\Rightarrow c=2\pi |ln|=0$$
$$\Rightarrow y=-4x\sin x+4\sin x\cdot ln \sin x+2\pi \sin x$$
$$y\left(\dfrac{\pi}{6}\right)\Rightarrow y=\dfrac{-2\pi}{3}\left(\dfrac{1}{2}\right)+4\left(\dfrac{1}{2}\right)(-ln2)+\pi$$
$$y=\dfrac{2\pi}{3}-2ln2$$.
Question 6
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Integrating factors of the differential equation $$\frac{{dy}}{{dx}} + y = \frac{{1 + y}}{x}$$ is

A
$$x/{e^x}$$

B
$${e^x}/x$$

C
$$x{e^x}$$

D
$${e^x}$$

Solution
Question 7
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Solution of the differential equation $$\left( { x }^{ 2 }+1 \right) y'+2xy=4{ x }^{ 2 }$$ is

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

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

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

D
None of these

Solution

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.
Question 8
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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:

A
$$16$$

B
$$36$$

C
$$64$$

D
$$9$$

Solution
Question 9
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The differential equation found by the elimination of the arbitrary constant K from the equation $$y=(x+K)e^{-x}$$ is

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

B
$$\dfrac{dy}{dx}-ye^x=1$$

C
$$\dfrac{dy}{dx}+ye^x=1$$

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

Solution
Question 10
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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

A
$$y=\sin \dfrac {1}{x}-\cos \dfrac {1}{x}$$

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

C
$$y=\cos \dfrac {1}{x}-\sin \dfrac {1}{x}$$

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

Solution