Differential Equations Test

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Differential Equations Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The solution of $$\frac{dy}{dx}=e^{x-y}+e^{2logx-y}$$

A
$$e^{y}=e^{x}+\frac{x^{2}}{3}+c$$

B
$$e^{y}=e^{x}+\frac{x^{3}}{3}+c$$

C
$$e^{y}=e^{x}+ log x + c$$

D
$$y=\dfrac{e^{3x+y}}{3}$$

Solution

$$\frac{dy}{dx}= e^{x-y}+e^{2log\ x-y}$$
$$\Rightarrow e^{y}dy= (e^{x}+x^{2})dx$$
$$\Rightarrow e^{y}= e^{x}+x^{3}/{3}+c$$
Question 2
1 / -0

The solution of $$(x^{2}+x)\frac{dy}{dx}=1+2x$$ is:

A
$$e^{y}=c(x^{2}+x)$$

B
$$y=x(x+1)+c$$

C
$$y=(1+2x)+c$$

D
$$xy=x^{2}+x+c$$

Solution

$$\dfrac{dy}{dx}=\dfrac{1+2x}{x^{2}+x}$$
$$dy=\left ( \dfrac{1+2x}{x^{2}+x} \right )dx$$
$$\Rightarrow y=log(x^{2}+x)c$$
$$\Rightarrow e^{y}=c(x^{2}+x)$$
Question 3
1 / -0

The solution of

$$(x^{2}-yx^{2})\dfrac{dy}{dx}+(y^{2}+x^{2}y^{2})=0$$

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

B
$$log y+\dfrac{1}{y}=x-\dfrac{1}{x}+c$$

C
$$y-\dfrac{1}{y}=x-\dfrac{1}{x}+c$$

D
$$log x+\dfrac{1}{x}=y-\dfrac{1}{y}+c$$

Solution

$$\left( x^ 2-yx^ 2 \right) \dfrac { dy }{ dx } +\left( y^ 2+x^ 2y^ 2 \right) =0$$

$$x^ 2\left( 1-y \right) \dfrac { dy }{ dx } +y^ 2\left( 1+x^ 2 \right) =0$$

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

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

$$\int { \dfrac { \left( 1-y \right) }{ y^ 2 } } dy=-\int { \dfrac { \left( 1+x^ 2 \right) }{ x^ 2 } } dx$$

$$\log { y } +\dfrac { 1 }{ y } =x-\dfrac { 1 }{ x } +c$$
Hence, option 'B' is correct.
Question 4
1 / -0

Solution of $$\tan y.\sec^{2}xdx + \tan x.\sec^{2} ydy=0$$ is:

A
$$\sec x \sec y=c$$

B
$$\tan x. \tan y=c$$

C
$$\sin x . \sin y=c$$

D
$$\cos x .\cos y=c$$

Solution

$$tan\ y\ sec^{2}xdx+tan\ x\ sec^{2}y\ dy= 0$$
$$\Rightarrow \dfrac{sec^{2}x\ dx}{tan\ x}+\dfrac{sec^{2}y}{tan\ y}\ dy= 0$$
let $$\begin{matrix}tan \ x = t ; &tan \ y = k \\\Rightarrow sec^{2}x\ dx = dt &\Rightarrow sec^{2}y\ dy = k \end{matrix}$$
$$\Rightarrow \int \dfrac{dt}{t}+\int \dfrac{dk}{k}= c$$
$$\Rightarrow log \ kt= c$$
$$\Rightarrow kt= c$$
$$\Rightarrow tan\ x\ tan\ y= c$$
Question 5
1 / -0

The solution of $$x\sqrt{1+x^{2}}dx+y\sqrt{1+y^{2}}dy=0$$ is:

A
$$\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$$

B
$$(\sqrt{1+x^{2}}$$x$$\sqrt{1+y^{2}}))=c$$

C
$$(1+x^{2})^{\frac{3}{2}}+(1+y^{2})^{\frac{3}{2}}=c$$

D
$$\frac{x}{\sqrt{1+x^{2}}}+\frac{y}{\sqrt{1+y^{2}}}=c$$

Solution

$$let x= tan\theta ; y= tan \delta$$

$$\Rightarrow dx= \sec^{2}\theta\ d \theta ; dy = sec^{2}\delta\ d\ \delta$$

$$\Rightarrow tan\theta \ sec^{3}\theta d\theta +tan\delta sec^{2}\delta d \delta= 0$$

let$$\begin{matrix}sec\theta =t ; &sec \delta =k \\\Rightarrow sec\theta d \theta = dt \ \ \ \ | &\Rightarrow sec\ \delta\ tan\ \delta\ d\ \delta = dk\end{matrix}$$

$$\Rightarrow \int t^{2}dt+\int k^{2}dk= c$$

$$\Rightarrow \dfrac {t^{3}+k^{3}}{3}= c$$

$$\Rightarrow sec^{3}\theta +sec^{3}\delta= c$$

$$\Rightarrow (\sqrt{1+x^{2}})^{3}+(\sqrt{1+y^{2}})^{3}= c$$

$$\Rightarrow (1+x^{2})^{^{3}/_{2}}+(1+y^{2})^{^{3}/_{2}}= c$$
Question 6
1 / -0

If $$\dfrac{dy}{dx}=x^{2}+1$$ and $$y=12$$ when $$x=3$$. Then the function is:

A
$$3(y+x)=x^{3}$$

B
$$3(y-x)=x^{3}$$

C
$$(y+x)=x^{3}$$

D
$$(y-x)=x^{3}$$

Solution

Given, $$\dfrac {dy}{dx}=x^2+1$$
Thus $$dy= (x^{2}+1)dx$$
$$\Rightarrow y= \dfrac{x^{3}}{3}+x+c$$
Also given, $$(x,y)= (3,12)$$
$$\Rightarrow 12= 9+3+c$$
$$\Rightarrow c= 0$$
Therefor,e the function is $$ 3y= x^{3}+3x$$
$$\Rightarrow 3(y-x)= x^{3}$$
Question 7
1 / -0

If $$\dfrac{dy}{dx}=e^{x+y}$$ and at $$x=1;\> y=1$$, then for $$x=-1;\> y$$ ___.

A
$$-1$$

B
$$e$$

C
$$1$$

D
$$e+\dfrac{1}{e}$$

Solution

$$\dfrac{dy}{dx}= e^{x}\cdot e^{y}$$
$$\Rightarrow e^{-y}dy= e^{x}dx$$
Integrating both sides, we get
$$\Rightarrow \dfrac{e^{-y}}{-1}=e^{x}-c$$
$$\Rightarrow e^{x}+e^{-y}= c$$
Substitute $$ (x,y)= (1,1)$$
$$\Rightarrow e+\dfrac{1}{e}= c$$
Now, for $$x= -1$$
$$\Rightarrow \dfrac{1}{e}+e^{-y}= e+\dfrac{1}{e}$$
$$\Rightarrow e^{-y}= e$$
$$\Rightarrow y= -1$$
Question 8
1 / -0

The solution of $$\dfrac{dy}{dx}=e^{2x-y}+x^{3}e^{-y}$$ is:

A
$$e^{y}=e^{2x}+3x^{2}+c$$

B
$$e^{y}=e^{2x}+\dfrac{x^{4}}{4}+c$$

C
$$4e^{y}=2e^{2x}+x^{4}+c$$

D
$$2e^{y}=e^{2x}+3x^{4}+c$$

Solution

$$\dfrac {dy}{dx}= \dfrac{e^{2x}}{e^{y}}+\dfrac{x^{3}}{e^{y}}$$
$$\Rightarrow e^{y}dy= (e^{2x}+x^{3})dx$$
$$\Rightarrow e^{y}= \dfrac{e^{2x}}{2}+\dfrac{x^{4}}{4}+\dfrac{c}{4}$$
$$\Rightarrow 4e^{y}= 2e^{2x}+x^{4}+c$$
Question 9
1 / -0

Solution of $$(xy^{2}+x)dx+(yx^{2}+y)dy=0$$

A
$$(x^{2}+1)(y^{2}+1)=c$$

B
$$(xy+1)(xy-1)=c$$

C
$$(x^{3}+1)(y^{3}+1)=c$$

D
$$(1-x^{2})(1-y^{2})=c$$

Solution

$$\Rightarrow x(y^{2}+1)dx+y(x^{2}+1)dy= 0$$
$$\Rightarrow \left ( \dfrac{2x}{x^{2}+1} \right )dx+\left ( \dfrac{2y}{y^{2}+1} \right )dy= 0$$
let $$\begin{matrix}t= x^{2}+1 > &k= y^{2}+1 \\\Rightarrow dt= 2xdx &\Rightarrow dk=2ydy \end{matrix}$$
$$\Rightarrow \int \dfrac{dt}{t}+\int \frac{dk}{k}= log \ c$$
$$\Rightarrow kt = c$$
$$\Rightarrow (x^{2}+1)(y^{2}+1)= c$$
Question 10
1 / -0

Solve the differential equation:
$$\sqrt{1+x^{2}}dx+\sqrt{1+y^{2}}dy=0$$

A
$$x\sqrt{1+x^{2}}+y\sqrt{1+y^{2}}+log\left [ \left ( x+\sqrt{1+x^{2}} \right )\left ( y+\sqrt{1+y^{2}} \right ) \right ]=c$$

B
$$\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$$

C
$$\dfrac{1}{\sqrt{1+x^{2}}}+\dfrac{1}{\sqrt{1+y^{2}}}=c$$

D
$$log\left \{ \left ( \sqrt{1+x^{2}} \right )+\left ( \sqrt{1+y^{2}}\right ) \right \}=x+c$$

Solution

Given, $$\sqrt{1+x^{2}}dx=-\sqrt{1+y^{2}}.dy$$

It is known that
$$\int \sqrt{x^{2}\pm a^{2}}dx=\dfrac{1}{2}x\sqrt{x^{2}\pm a^{2}}\pm\dfrac{a^{2}}{2}ln|x+\sqrt{x^{2}\pm a^{2}}$$

Applying this formula gives us

$$\int \sqrt{x^{2}+1}.dx=-\int \sqrt{y^{2}+1}.dy$$

$$\Rightarrow \dfrac{x}{2}\sqrt{x^{2}+1}+\dfrac{1}{2}ln|x+\sqrt{x^{2}+1}|=-[\dfrac{y}{2}\sqrt{y^{2}+1}+\dfrac{1}{2}ln|y+\sqrt{y^{2}+1}|]+C$$

$$\Rightarrow x\sqrt{x^{2}+1}+y\sqrt{y^{2}+1}+ln|x+\sqrt{x^{2}+1}|+ln|y+\sqrt{y^{2}+1}|=C$$

$$\Rightarrow x\sqrt{x^{2}+1}+y\sqrt{y^{2}+1}+ln|(x+\sqrt{x^{2}+1})(y+\sqrt{y^{2}+1})|=C$$

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Differential Equations Test - 19

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Differential Equations Test - 19

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Question 1

$$e^{y}=e^{x}+\frac{x^{2}}{3}+c$$

$$e^{y}=e^{x}+\frac{x^{3}}{3}+c$$

$$e^{y}=e^{x}+ log x + c$$

$$y=\dfrac{e^{3x+y}}{3}$$

Solution

Question 2

$$e^{y}=c(x^{2}+x)$$

$$y=x(x+1)+c$$

$$y=(1+2x)+c$$

$$xy=x^{2}+x+c$$

Solution

Question 3

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

$$log y+\dfrac{1}{y}=x-\dfrac{1}{x}+c$$

$$y-\dfrac{1}{y}=x-\dfrac{1}{x}+c$$

$$log x+\dfrac{1}{x}=y-\dfrac{1}{y}+c$$

Solution

Question 4

$$\sec x \sec y=c$$

$$\tan x. \tan y=c$$

$$\sin x . \sin y=c$$

$$\cos x .\cos y=c$$

Solution

Question 5

$$\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$$

$$(\sqrt{1+x^{2}}$$x$$\sqrt{1+y^{2}}))=c$$

$$(1+x^{2})^{\frac{3}{2}}+(1+y^{2})^{\frac{3}{2}}=c$$

$$\frac{x}{\sqrt{1+x^{2}}}+\frac{y}{\sqrt{1+y^{2}}}=c$$

Solution

Question 6

$$3(y+x)=x^{3}$$

$$3(y-x)=x^{3}$$

$$(y+x)=x^{3}$$

$$(y-x)=x^{3}$$

Solution

Question 7

$$-1$$

$$e$$

$$1$$

$$e+\dfrac{1}{e}$$

Solution

Question 8

$$e^{y}=e^{2x}+3x^{2}+c$$

$$e^{y}=e^{2x}+\dfrac{x^{4}}{4}+c$$

$$4e^{y}=2e^{2x}+x^{4}+c$$

$$2e^{y}=e^{2x}+3x^{4}+c$$

Solution

Question 9

$$(x^{2}+1)(y^{2}+1)=c$$

$$(xy+1)(xy-1)=c$$

$$(x^{3}+1)(y^{3}+1)=c$$

$$(1-x^{2})(1-y^{2})=c$$

Solution

Question 10

$$x\sqrt{1+x^{2}}+y\sqrt{1+y^{2}}+log\left [ \left ( x+\sqrt{1+x^{2}} \right )\left ( y+\sqrt{1+y^{2}} \right ) \right ]=c$$

$$\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$$

$$\dfrac{1}{\sqrt{1+x^{2}}}+\dfrac{1}{\sqrt{1+y^{2}}}=c$$

$$log\left \{ \left ( \sqrt{1+x^{2}} \right )+\left ( \sqrt{1+y^{2}}\right ) \right \}=x+c$$

Solution

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