Differential Equations Test

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Differential Equations Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Solution of $$y-x\displaystyle \frac{dy}{dx}=3[1-x^{2}\frac{dy}{dx}]$$ is:

A
$$(y+3)(1+3x)=cx$$

B
$$(y-3)(1-3x)=cx$$

C
$$(y-3)(1+3x)=cx$$

D
$$(y+3)(1-3x)=cx$$

Solution

$$y-x\dfrac{dy}{dx}=3-3x^{2}\dfrac{dy}{dx}$$
$$\Rightarrow (y-3)=(x-3x^2)\dfrac{dy}{dx}$$
$$\Rightarrow \dfrac{dx}{x-3x}=\dfrac{dy}{y-3}$$

$$\dfrac{dx}{(1-3x)}=\dfrac{dy}{y-3}$$
$$\Rightarrow \int \dfrac{3}{x-3x}+\dfrac{1}{x} dx=\int \dfrac{dy}{y-3} -log c$$
$$\Rightarrow -\log(1-3x)+\log x=\log(y-3)-\log c$$
$$\Rightarrow \log x c=\log(y-3)(1-3x)$$
$$\Rightarrow xc=(y-3)(1-3x)$$
Question 2
1 / -0

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

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

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

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

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

Solution

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

$$\Rightarrow \int e^{2x}dx+ \int e^{2y}=c_{1}$$

$$\Rightarrow \dfrac{e^{2x}}{2}+\dfrac{e^{2y}}{2}=\dfrac{c_{1}}{2}$$

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

General solution of $$\displaystyle \frac{dy}{dx}=\frac{1}{\log_{x}e}$$ is given as $$y = $$

A
$$\displaystyle \frac{1}{x}+c$$

B
$$\displaystyle \frac{x^{2}}{2}+c$$

C
$$x\log_{e}x-x+c$$

D
$$\displaystyle \frac{x}{2}+c$$

Solution

$$\dfrac{dy}{dx} = \dfrac{log x}{log e}$$
$$\Rightarrow {dy} = \dfrac{log x}{log e} dx$$
$$\Rightarrow \int dy = \int log_{e} x + c$$
$$\Rightarrow y = x (log_{e} x-1) + c$$ $$ [\because\int logx=x(logx-1)]$$
Question 4
1 / -0

The solution of: $$e^{x}\displaystyle \sqrt{1-y^{2}}dx+\frac{y}{x}dy=0$$

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

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

C
$$x.\ e^{x}=\sqrt{1-y^{2}}+c$$

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

Solution

$$e^{x} \sqrt{1-y^{2}} dx + \dfrac{y}{x} dy = 0$$
$$\Rightarrow xe^{x} dx = -\dfrac{y}{\sqrt{1-y^{2}}} dy$$
$$\Rightarrow \int xe^{x} dx = \int \dfrac{-y}{\sqrt{1-y^{2}}} dy+c$$
$$\Rightarrow (xe^{x}-e^{x}) = \sqrt{1-y^{2}} + c$$
$$\Rightarrow e^{x} (x-1) = \sqrt{1-y^{2}} +c$$
Question 5
1 / -0

Solution of $$xe^{x^{2}+y}.dx=y.dy$$ is:

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

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

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

D
$$x.e^{x^{2}}y=c$$

Solution

$$xe^{x^{2}} dx = ye^{-y} dy$$
$$ \Rightarrow \dfrac{1}{2} \int 2xe^{x^{2}} dx = \int ye^{-y} dy$$
$$\Rightarrow \dfrac{e^{x^{2}}}{2} = -ye^{-y} + (e-y_{dy})$$
$$\dfrac{e^{x^{2}}}{2} = -ye^{-y} -e^{-y} + c$$
$$\Rightarrow e^{x^{2}} + e^{-y} (1+y)2= c$$
Question 6
1 / -0

The solution of $$\cos \mathrm{x}\cos \mathrm{y}\mathrm{d}\mathrm{x}+ \sin \mathrm{x} \sin \mathrm{y} d\mathrm{y} =0$$ is

A
$$cosx=c\sin y$$

B
$$\sin x=c\cos y$$

C
$$\sec x-\sec y=c$$

D
$$\tan x=c$$

Solution

$$\Rightarrow \sin x \sin y dy = -\cos x \cos y dx$$

$$\Rightarrow \int \tan y dy = -\int \cot x dx$$

$$ \Rightarrow log \sec y=- \log \sin x + \log c'$$

$$ \Rightarrow log \sec y= \log cosec x + \log c'$$

$$\Rightarrow log \sec y= \log ( cosec x \times c')$$

$$\Rightarrow \dfrac{1}{\cos y} = \dfrac{1}{c' \sin x}$$

$$\sin x = c \cos y$$
Question 7
1 / -0

Solution of $$\displaystyle \frac{dy}{dx}=4+4x-3y-3xy$$ is:

A
$$2\log(4-3y)+3x^{2}+6x=c$$

B
$$\log(3-4y)+3y^{2}+6y=c$$

C
$$(4-3y)(1+x)=c$$

D
$$\log(4-3y)+x^{2}+3x=c$$

Solution

$$\dfrac{dy}{dx} = 4(1+x)-3y(1+x)$$
$$\Rightarrow\displaystyle \int \dfrac{dy}{4-3y} = \int (1+x)dx + c$$
$$\Rightarrow -\dfrac{1}{3} log (\dfrac{4}{3}-y) = x+\dfrac{x^{2}}{2} + c$$
$$\Rightarrow c = \dfrac{2x+x^{2}}{2} + \dfrac{1}{3} log (\dfrac{4-3y}{3})$$
$$\Rightarrow c= 6x+3x^{2} + 2log (4-3y)$$
Question 8
1 / -0

The solution of $$\displaystyle \frac{dy}{dx}=xy+x+y+1$$

A
$$y=\displaystyle \frac{x^{2}}{2}+xc$$

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

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

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

Solution

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

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

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

B
$$y=c\sin^{-1}x$$

C
$$y=\displaystyle \sin(\frac{c}{x})$$

D
$$x=c\mathrm{s}in\mathrm{y}$$

Solution

Given $$\sin^{-1}ydx+\displaystyle \dfrac{x}{\sqrt{1-y^{2}}}dy=0$$
$$-\dfrac{dx}{x} = \dfrac{dy}{sin^{-1}y \sqrt{1-y^{2}}}$$
$$let \sin ^{-1}y = t$$
$$\Rightarrow \dfrac{1}{\sqrt{1-y^{2}}}dy = dt$$
$$\Rightarrow -\int \dfrac{dx}{x} = \int \dfrac{dt}{t} - log c$$
$$ \Rightarrow log c = log xt$$
$$\Rightarrow c=xt$$
$$\Rightarrow t = \dfrac{c}{x}$$
$$sin ^{-1}y = \dfrac{c}{x}$$
$$\Rightarrow y = sin (\dfrac{c}{x})$$
Question 10
1 / -0

lf the primitive of $$\displaystyle \frac{1}{f(x)}$$ is equal to $$\log\{f(x)\}^{2}+c$$, then $$f(x)$$ is:

A
$$ x+d$$

B
$$ \displaystyle \frac{x}{2}+d$$

C
$$\displaystyle \frac{x^{2}}{2}+d$$

D
$$ x^{2}+d$$

Solution

Given $$\int { \frac { 1 }{ f\left( x \right) } } dx={ \log { \{ f(x)\} } }^{ 2 }+C$$
$$\displaystyle \Rightarrow \int { \frac { 1 }{ f\left( x \right) } } dx=2{ \log { \{ f(x)\} } }+C$$
$$\displaystyle\Rightarrow \frac{1}{f(x)}=2 \times \frac{1}{f(x)} \times f'(x) $$
$$\Rightarrow f'(x)=\dfrac{1}{2}$$
$$\displaystyle \Rightarrow f(x) =\frac{1}{2}\int dx $$
$$\displaystyle \Rightarrow f(x) =\frac{x}{2}+d$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 23

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Differential Equations Test - 23

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SHARING IS CARING

Question 1

$$(y+3)(1+3x)=cx$$

$$(y-3)(1-3x)=cx$$

$$(y-3)(1+3x)=cx$$

$$(y+3)(1-3x)=cx$$

Solution

Question 2

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

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

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

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

Solution

Question 3

$$\displaystyle \frac{1}{x}+c$$

$$\displaystyle \frac{x^{2}}{2}+c$$

$$x\log_{e}x-x+c$$

$$\displaystyle \frac{x}{2}+c$$

Solution

Question 4

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

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

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

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

Solution

Question 5

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

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

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

$$x.e^{x^{2}}y=c$$

Solution

Question 6

$$cosx=c\sin y$$

$$\sin x=c\cos y$$

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

$$\tan x=c$$

Solution

Question 7

$$2\log(4-3y)+3x^{2}+6x=c$$

$$\log(3-4y)+3y^{2}+6y=c$$

$$(4-3y)(1+x)=c$$

$$\log(4-3y)+x^{2}+3x=c$$

Solution

Question 8

$$y=\displaystyle \frac{x^{2}}{2}+xc$$

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

$$y+1=c.e(\displaystyle \frac{x^{2}+2x}{2})$$

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

Solution

Question 9

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

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

$$y=\displaystyle \sin(\frac{c}{x})$$

$$x=c\mathrm{s}in\mathrm{y}$$

Solution

Question 10

$$ x+d$$

$$ \displaystyle \frac{x}{2}+d$$

$$\displaystyle \frac{x^{2}}{2}+d$$

$$ x^{2}+d$$

Solution

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