Differential Equations Test

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Differential Equations Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

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

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

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

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

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

Solution

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

Let $$1+x^{2}=t ; 1+e^{y}=k$$

$$\Rightarrow 2xdx=dt$$ and $$e^{y}dy=dk$$

$$\Rightarrow\displaystyle \int {\frac{dk}{k}}=\int {\frac{dt}{t}}$$

$$\Rightarrow\displaystyle \log\frac{k}{t}=\log C$$

Where, $$\log C$$ is the constant of integration.
$$\Rightarrow \dfrac kt = C$$

$$\Rightarrow \dfrac{(1+e^{y})}{1+x^{2}}=C$$

Hence, option A.
Question 2
1 / -0

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

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

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

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

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

Solution

$$x\sqrt{1+y^{2}}dx+y\sqrt{1+x^{2}}dy=0$$
let $$t=\sqrt{1+x^{2}}$$ ;$$k=\sqrt{1+y^{2}}$$
$$\Rightarrow dt=\dfrac{x}{\sqrt{1+x^{2}}}dx$$ ;$$dk=\dfrac{y}{\sqrt{1+y^{2}}}dy$$
$$\Rightarrow \int dt+\int dk=0$$
$$\Rightarrow t+k=c$$
Hence, Option B is correct
Question 3
1 / -0

I.F. of $$\dfrac { dy }{ dx } =\dfrac { x+y+1 }{ x+1 } $$ is:

A
$$\dfrac{1}{y+1}$$

B
$$\dfrac{1}{x+1}$$

C
$$log (x + 1)$$

D
$$log (y + 1)$$

Solution

$$\dfrac{dy}{dx}=\dfrac{x+y+1}{x+1}$$
$$\dfrac{dy}{dx}-\dfrac{y}{x+1}=1$$
$$\Rightarrow\ I.F=e^{\int \dfrac{-1}{x+1}}dx$$
$$=e^{-log(x+1)}$$
$$=\ \begin{matrix}1/{x+1}\\ \end{matrix}$$
Question 4
1 / -0

Equation of the curve whose polar sub tangent $$r^{2}\displaystyle \frac{d\theta}{dr}$$ is constant

A
$$r(\theta+c)+k=0$$

B
$$r^{2}(\theta+c)=2k$$

C
$$r(\theta-c)=k^{2}$$

D
$$r\theta=c$$

Solution

Let the constant length of the polar subtangent be $$a$$
Therefore as given in the question $$\displaystyle { r }^{ 2 }\frac { d\theta }{ dr } =a$$
which is the differential equation of the required curve
To solve, put it into the form
$$\displaystyle d\theta =a\left( \frac { 1 }{ { r }^{ 2 } } \right) dr\Rightarrow \int { d\theta } =\int { a\left( \frac { 1 }{ { r }^{ 2 } } \right) dr } $$
$$\displaystyle \Rightarrow \theta +c=-\frac { a }{ r } \Rightarrow r\left( \theta +c \right) +a=0$$
Question 5
1 / -0

$$\dfrac{dy}{dx}=e^{-2y},x=5\Rightarrow y=0,$$ then $$y=3,find\ the\ value\ of\ 2x$$.

A
$$e^{5}+9$$

B
$$e^{6}+9$$

C
$$e^{8}+9$$

D
$$e^{4}+9$$

Solution

$$\displaystyle\Rightarrow \int e^{2y}dy=\int dx +c$$
$$\Rightarrow \dfrac{e^{2y}}{2}=x+c$$
$$\Rightarrow e^{2y}=2x+c$$
put $$(x,y)=(5,0)$$
$$\Rightarrow 1=10+c$$ $$\Rightarrow c=-9$$
$$\Rightarrow e^{2y}=2x-9$$
$$\Rightarrow 2x=e^{2y}+9$$
$$\Rightarrow$$ $$ for\ y=3$$
$$2x=e^{6}+9$$
Question 6
1 / -0

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

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

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

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

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

Solution

$$\Rightarrow \dfrac{dy}{dx} =\sqrt{1+xy+x+y}=\sqrt{(1+x)(1+y)}$$

$$\displaystyle\Rightarrow \int \dfrac{dy}{\sqrt{y+1}}=\int \sqrt{x+1}dx+c$$

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

$$\Rightarrow 3\sqrt{y+1}=(x+1)^{3/2}+c$$
Question 7
1 / -0

I.F. of $$\frac{dy}{dx}$$ $$+$$ $$y tan x $$$$=x^{2}cos^{2}x$$ is:

A
$$sec x$$

B
$$cos x$$

C
$$sec^{2}x$$

D
$$cos^{2}x$$

Solution

$$I.F=e^{\int P(x)dx}=e^{\int \tan xdx}=e^{log\sec x}=\sec x$$
Question 8
1 / -0

I.F of $$x\displaystyle \frac{dy}{dx}=(2y+2x^{4}+x^{2})$$ is:

A
$$x^{-2}$$

B
$$x^{ -1}$$

C
$$x$$

D
$$x^{ 2}$$

Solution

$$\cfrac{dy}{dx}-\dfrac{2}{x}y=2x^{3}+x$$
$$I.F=e^{\int P(x)dx}=e^{-2\int \tfrac{1}{x}dx}=e^{-2\ \log x}=e^{\log\ \frac1{x^{2}}}$$
$$=\cfrac {1}{x^{2}}$$
Question 9
1 / -0

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

A
$$y^{2}= x - \sin x \cos x + c$$

B
$$y^{2}=\dfrac{x^{2}}{2}- \sin 2x + c$$

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

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

Solution

$$\Rightarrow ydy=\sin^{2}xdx$$
$$\int ydy=\int \dfrac{1-\cos 2x}{2}dx+c$$
$$\Rightarrow \dfrac{y^{2}}{2}=\dfrac{x-\dfrac{\sin 2x}{2}}{2}+c$$
$$\Rightarrow 2y^{2}=2x-\sin 2x+c$$
$$\Rightarrow y^{2}=x-\sin x \cos x +c$$
Hence, option 'A' is correct.
Question 10
1 / -0

The solution of $$\displaystyle \frac{dy}{dx}=xy+2x-3y-6$$ is:

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

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

C
$$y+2=2(x-3)+c$$

D
$$(y+2)(x-3)=c$$

Solution

$$\dfrac{dy}{dx}-y(x-3)+2(x-3)$$
$$\dfrac{dy}{dx}=(x-3)(y+2)$$
$$\Rightarrow \int \dfrac{dy}{y+2}=\int (x-3)dx+c_{1}$$
$$\Rightarrow log(y+2)=\dfrac{x^{2}}{2}-3x+c_{1}$$
$$\Rightarrow 2log(y+2)=(x-3)^{2}+c[c=c_{1}-9]$$
$$\Rightarrow (y+2)^{2}=e^{(x-3)^{2}}e^{c}$$
$$(y+2)^{2}=ce^{(x-3)^{2}}$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 22

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Differential Equations Test - 22

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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

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

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

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

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

Solution

Question 3

$$\dfrac{1}{y+1}$$

$$\dfrac{1}{x+1}$$

$$log (x + 1)$$

$$log (y + 1)$$

Solution

Question 4

$$r(\theta+c)+k=0$$

$$r^{2}(\theta+c)=2k$$

$$r(\theta-c)=k^{2}$$

$$r\theta=c$$

Solution

Question 5

$$e^{5}+9$$

$$e^{6}+9$$

$$e^{8}+9$$

$$e^{4}+9$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$sec x$$

$$cos x$$

$$sec^{2}x$$

$$cos^{2}x$$

Solution

Question 8

$$x^{-2}$$

$$x^{ -1}$$

$$x$$

$$x^{ 2}$$

Solution

Question 9

$$y^{2}= x - \sin x \cos x + c$$

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

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

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

Solution

Question 10

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

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

$$y+2=2(x-3)+c$$

$$(y+2)(x-3)=c$$

Solution

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