Differential Equations Test

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Differential Equations Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

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

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

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

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

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

Solution

$$ \displaystyle \frac{dy}{dx}=\frac{(1+y^{2})}{(1+x^{2})}$$
$$ \displaystyle \Rightarrow \int \frac{dy}{1+y^{2}}=\int \frac{dx}{1+x^{2}}+c_{1}$$
$$\Rightarrow tan^{-1}{y}=tan^{-1}x+c_{1}$$
$$\Rightarrow \displaystyle tan^{-1}(\frac{y-x}{1+xy})=c_{1}$$
$$\displaystyle \Rightarrow tan\ c_{1}=(\frac{y-x}{1+xy})$$
$$let \ tan\ c_{1}=c$$
$$ \displaystyle \Rightarrow c=\frac{y-x}{1+xy}$$
$$ \displaystyle \Rightarrow y-x=(1+xy)c$$
Question 2
1 / -0

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

A
$$x^{3}-y^{3}-y^{2}-x^{2}=c$$

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

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

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

Solution

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

$$\Rightarrow \int (y+y^{2})dy=\int (x+x^{2})dx+c$$

$$\Rightarrow \dfrac{y^{2}}{2}+\dfrac{y^{3}}{3}=\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+c$$

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

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

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

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

B
$$x=cy^{2}$$

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

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

Solution

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

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

A
$$e^{y}= c \sin x$$

B
$$e^{y}= c \cos x$$

C
$$e^{y}= c \csc x$$

D
$$e^{y}= c \sec x$$

Solution

$$\dfrac{dy}{dx}=tanx$$

$$dy=tanx.dx$$

$$dy=\dfrac{sinx}{cosx}.dx$$

Let $$cosx=t$$
Then
$$sinx.dx=-dt$$

Hence

$$dy=\dfrac{-1}{t}dt$$

$$\int dy =\int \dfrac{-dt}{t}$$

$$y=-lnt+lnC$$

$$y=-ln(cosx)+lnC$$

$$y=ln(secx)+lnC$$

$$y=lnCsecx$$

$$e^{y}=C.secx$$
Question 5
1 / -0

The solution of $$x^{2}dy-y^{2}dx=0$$ is:

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

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

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

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

Solution

Given, $$x^{2}dy-y^{2}dx=0$$
$$\Rightarrow \dfrac{dy}{y^{2}}=\dfrac{dx}{x^{2}}$$
$$\Rightarrow \dfrac{ -1 }{y} = \dfrac{ -1 }{x} +c$$
$$\Rightarrow \dfrac { 1 }{ x } -\dfrac { 1 }{ y } =c$$
Question 6
1 / -0

The solution of $$\displaystyle \dfrac{dy}{dx}=2y\tanh x $$ is

A
$$cy=\sinh^{2}x$$

B
$$cy=\sec h^{2}x$$

C
$$cy=\cosh^{2}x$$

D
$$cy=\coth^{2}x$$

Solution

$$\dfrac{dy}{dx}=2 y \tan hx$$
$$\Rightarrow \int \dfrac{dy}{y}= 2 \int \tan hx dx$$
$$\Rightarrow \log y = 2 \log(\cos hx)-\log c$$
$$\Rightarrow \log cy=\log \cosh^2x$$
$$cy=\cosh^2x$$
Question 7
1 / -0

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

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

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

C
$$(1-y)(1-x)=c$$

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

Solution

$$\dfrac{dy}{dx}=\dfrac{1+y}{1-x}$$

$$\dfrac{dy}{1+y}=\dfrac{dx}{1-x}$$

which is variable seprable form

take integration both side

we get,

$$\Rightarrow \log(1+y)=-\log(1-x)+logc$$

$$\log(1+y)(1-x)=logc$$

$$\Rightarrow (1+y)(1-x)=c$$
Question 8
1 / -0

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

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

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

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

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

Solution

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

The solution of $$\displaystyle \frac{dy}{dx}=e^{\displaystyle (y-x)}$$ is

A
$$e^{\displaystyle y}+e^{\displaystyle x}=c$$

B
$$e^{\displaystyle -x}=e^{\displaystyle -y}+c$$

C
$$e^{\displaystyle y-x}=c$$

D
$$e^{\displaystyle y/x}=c$$

Solution

$$\dfrac{dy}{dx}=\dfrac{e^{y}}{e^{x}}$$
$$\dfrac{dy}{e^{y}}=\dfrac{dx}{e^{x}}$$
$$\Rightarrow \dfrac{e^{-y}}{-1}=\dfrac{e^{-x}}{-1}+c$$
$$\Rightarrow e^{-x}=e^{-y}+c$$
Question 10
1 / -0

The solution of $$\dfrac{dy}{dx}+y=1$$ is:

A
$$x+\log\left | 1-y \right |=c$$

B
$$y+\log\left | 1-x \right |=c$$

C
$$x(1-y)=c$$

D
$$y(1-x)=c$$

Solution

$$\dfrac{dy}{dx}+y=1$$

$$\dfrac{dy}{dx}=1-y$$

$$\Rightarrow \dfrac{dy}{1-y}=dx$$

$$\Rightarrow - \log(1-y)=x+c$$
$$\Rightarrow x+\log \left | 1-y \right |=c$$

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Differential Equations Test - 17

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Differential Equations Test - 17

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Question 1

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

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

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

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

Solution

Question 2

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

$$2(x^{3}-y^{3})+3(x^{2}-y^{2})=c$$

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

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

Solution

Question 3

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

$$x=cy^{2}$$

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

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

Solution

Question 4

$$e^{y}= c \sin x$$

$$e^{y}= c \cos x$$

$$e^{y}= c \csc x$$

$$e^{y}= c \sec x$$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$cy=\sinh^{2}x$$

$$cy=\sec h^{2}x$$

$$cy=\cosh^{2}x$$

$$cy=\coth^{2}x$$

Solution

Question 7

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

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

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

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

Solution

Question 8

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

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

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

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

Solution

Question 9

$$e^{\displaystyle y}+e^{\displaystyle x}=c$$

$$e^{\displaystyle -x}=e^{\displaystyle -y}+c$$

$$e^{\displaystyle y-x}=c$$

$$e^{\displaystyle y/x}=c$$

Solution

Question 10

$$x+\log\left | 1-y \right |=c$$

$$y+\log\left | 1-x \right |=c$$

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

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

Solution

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