Differential Equations Test

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Differential Equations Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

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

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

B
$$2^{x}-2^{y}=c$$

C
$$2^{x-y}=c$$

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

Solution

$$\dfrac{dy}{dx}=\dfrac { { 2 }^{ x } }{ { 2 }^{ y } } $$
$$\Rightarrow \int 2^{y}dy=\int 2^{x}dx+c$$
$$\dfrac{2^{y}}{log _{2}}=\dfrac{2^{x}}{log_{2}}+c$$
$$\Rightarrow 2^{x}-2^{y}=c$$
Question 2
1 / -0

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

A
$$sin^{-1}x+sin^{-1}y=c$$

B
$$cot^{-1}x+cot^{-1}y=c$$

C
$$Tan^{-1}x+Tan^{-1}y=c$$

D
$$sinh^{-1}x+sinh^{-1}y=c$$

Solution

$$\dfrac{dy}{dx}=\dfrac{-\sqrt{1-y^{2}}}{\sqrt{1-x{2}}}$$
$$\dfrac{dy}{\sqrt{1-y^{2}}}=\dfrac{-dx}{\sqrt{1-x{2}}}$$
$$\int \dfrac{dy}{\sqrt{1-y^{2}}}=-\int \dfrac{dx}{\sqrt{1-x{2}}}+c$$
$$\sin^{-1}y=-\sin^{-1}x+c$$
$$\Rightarrow \sin^{-1}x+\sin^{-1}y=c$$
Question 3
1 / -0

Solution of $$\dfrac{dy}{dx}=\dfrac{y+2}{x-1}$$ is:

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

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

C
$$log(y+2)=c(x-y)$$

D
$$log(x-1)=c(y+2)$$

Solution

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

The solution of $$\dfrac{dy}{dx}=\dfrac{x}{y}$$ is:

A
$$x^{2}-y^{2}=c$$

B
$$x^{2}+y^{2}=c$$

C
$$x^{2}-y=c$$

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

Solution

$$\dfrac{dy}{dx}=\dfrac{x}{y}$$
$$\Rightarrow ydy=xdx$$
$$\Rightarrow c+\dfrac{y^{2}}{2}=\dfrac{x^{2}}{2}$$
$$\Rightarrow x^{2}-y^{2}=c$$
Question 5
1 / -0

The solution of $$e^{ x }\tan ydx+(1-e^{ x })\sec^{ 2 }ydy=0$$ is

A
$$\tan y=c(1-e^{x})$$

B
$$\sec y=c(1-e^{x})$$

C
$$\tan y(1-e^{x})=c$$

D
$$\sec y=1-e^{x}$$

Solution

$$\dfrac{-e^{x}dx}{-e^{x}+1}= \dfrac{\sec^{2}ydy}{\tan\ y} ;\left [ \begin{matrix}put -(e^{x}-1)= K ; &\tan y = t \\ \Rightarrow -e^{x}dx= dK &\sec^{2}ydy= dt \end{matrix} \right ]$$
$$\Rightarrow \log(e^{x}-1)+\log\ c= \log\ tan\ y$$
$$\Rightarrow -c(e^{x}-1)= \tan\ y$$
$$\Rightarrow \tan\ y = c(1-e^{x})$$
Question 6
1 / -0

The solution of $$\tan x\ dy + \tan y\ dx$$ $$=$$ 0

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

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

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

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

Solution

$$tan x dy + tan y dx = 0$$
$$\Rightarrow \dfrac{-dy}{tan\ y}= \dfrac{dx}{tan\ x}$$
$$\Rightarrow \dfrac{-cos\ y\ dy}{sin\ y}= \dfrac{cos\ x\ dx}{sin\ x}$$
$$\Rightarrow \dfrac{-dt}{t}= \dfrac{dk}{k}\ [t= sin\ y ; k= cos\ x]$$
$$\Rightarrow log\ C= log\ Kt$$
$$\Rightarrow C= sin\ x\ sin\ y$$
Question 7
1 / -0

The solution of $$x+y\dfrac{dy}{dx}=5$$ is:

A
$$x^{2}-10x+y^{2}=c$$

B
$$x^{2}-5x+y^{2}=c$$

C
$$\dfrac{y^{2}}{2}=10x+x^{2}+c$$

D
$$y^2=10x+x^{2}+c$$

Solution

Given, $$x+y\dfrac{dy}{dx}=5$$
$$\Rightarrow y dy=(5-x)dx$$
$$\Rightarrow \dfrac{y^{2}}{2}=5x-\dfrac{x^{2}}{2}+c$$
$$\Rightarrow y^{2}=10 x -x^{2} +c$$
$$\Rightarrow x^{2}-10x+y^{2}=0+c$$
$$\Rightarrow x^{2}-10x+y^{2}=c$$
Question 8
1 / -0

The general solution of the differential equation $$\dfrac{dy}{dx}-\dfrac{2xy}{1+x^{2}}=0$$ is

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

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

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

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

Solution

$$\dfrac{dy}{dx}=\dfrac{2xy}{1+x^{2}}$$
$$\dfrac{dy}{y}=\dfrac{2x}{1+x^{2}}dx$$
$$\Rightarrow log y=log C(1+x^{2})$$
$$y=C(1+x^{2})$$
Question 9
1 / -0

The solution of $$x^{2}+y^{2}\dfrac{dy}{dx}=4$$ is

A
$$x^{2}+y^{2}=12x+c$$

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

C
$$x^{3}+y^{3}=3x+c$$

D
$$x^{3}+y^{3}=12x+c$$

Solution

$$y^{2}\dfrac{dy}{dx}= 4-x^{2}$$
$$y^{2}dy = 4dx-x^{2}dx$$
$$\Rightarrow \dfrac{y^{3}}{3}= 4x-\dfrac{x^3}{3}+ \ ^{c}/_{3}$$
$$\Rightarrow x^{3}+y^{3}= 12x+c$$
Question 10
1 / -0

The solution of $$\left ( 1+y^{2} \right )dx=xydy$$ is:

A
$$1+y^{2}=cx^{2}$$

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

C
$$1+y^{2}=cx$$

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

Solution

$$(1+y^{2})dx=xydy$$
$$2{dx}=\dfrac{2xydy}{1+y^{2}}$$
$$\Rightarrow log c + 2 logx=log(1+y^{2})$$
$$\Rightarrow log Cx^{2}=log(1+y^{2})$$
$$\Rightarrow (1+y^{2})=Cx^{2}$$

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Differential Equations Test - 18

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Differential Equations Test - 18

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SHARING IS CARING

Question 1

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

$$2^{x}-2^{y}=c$$

$$2^{x-y}=c$$

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

Solution

Question 2

$$sin^{-1}x+sin^{-1}y=c$$

$$cot^{-1}x+cot^{-1}y=c$$

$$Tan^{-1}x+Tan^{-1}y=c$$

$$sinh^{-1}x+sinh^{-1}y=c$$

Solution

Question 3

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

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

$$log(y+2)=c(x-y)$$

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

Solution

Question 4

$$x^{2}-y^{2}=c$$

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

$$x^{2}-y=c$$

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

Solution

Question 5

$$\tan y=c(1-e^{x})$$

$$\sec y=c(1-e^{x})$$

$$\tan y(1-e^{x})=c$$

$$\sec y=1-e^{x}$$

Solution

Question 6

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

$$\sec^{2}x+\sec^{2}y=c$$

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

$$\cot x.\cot y$$ $$= c$$

Solution

Question 7

$$x^{2}-10x+y^{2}=c$$

$$x^{2}-5x+y^{2}=c$$

$$\dfrac{y^{2}}{2}=10x+x^{2}+c$$

$$y^2=10x+x^{2}+c$$

Solution

Question 8

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

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

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

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

Solution

Question 9

$$x^{2}+y^{2}=12x+c$$

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

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

$$x^{3}+y^{3}=12x+c$$

Solution

Question 10

$$1+y^{2}=cx^{2}$$

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

$$1+y^{2}=cx$$

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

Solution

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