Differential Equations Test

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Differential Equations Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Equation of the curve passing through the point (3, 9) and which satisfies $$y_{1}=x+\dfrac{1}{x^{2}}$$ is:

A
$$6xy=3x^{3}-6x+29$$

B
$$6xy=3x^{3}-29x+6$$

C
$$6xy=3x^{3}+29x-6$$

D
$$6xy=3x^{3}-29x-6$$

Solution

$$\dfrac{dy}{dx}= x+\ {1}/{x^{2}}$$

$$dy= xdx+\ {1}/{x^{2}}dx$$

$$\Rightarrow y=\dfrac{x^{2}}{2}-\ {1}/{x}+c$$

$$(x_{0},y_{0})= (3,9)$$

$$\Rightarrow 9= \ {9}/{2}-\ {1}/{3}+c$$

$${1}/{3}+\ {9}/{2}= c$$

$$c= \dfrac{29}{6}$$

$$\Rightarrow y= {x^{2}}/{2}-\ {1}/{x}+\ {29}/{6}$$

$$\Rightarrow 6xy = 3x^{3}+29x-6$$
Question 2
1 / -0

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

A
$$ytan^{-1}x=c$$

B
$$ysin^{-1}x=c$$

C
$$ycos^{-1}x=c$$

D
$$xsin^{-1}x=c$$

Solution

$$\sqrt{1-x^{2}}sin^{-1}xdy+ydx= 0$$
$$\Rightarrow \dfrac{dy}{y}+\dfrac{dx}{sin^{-1}x\sqrt{1-x^{2}}}= 0$$
let $$ sin^{-1}x= t$$
$$\Rightarrow \dfrac{1}{\sqrt{1-x^{2}}}\ dx= dt$$
$$\Rightarrow \int \dfrac{dy}{y}+\int \dfrac{dt}{t}= log \ c$$
$$\Rightarrow log \ yt= log\ c$$
$$yt= c$$
$$\Rightarrow y\ sin^{-1}x= c$$
Question 3
1 / -0

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

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

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

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

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

Solution

$$\int -\dfrac{x}{\sqrt{1-x^{2}}}dx=\int \dfrac{-y}{\sqrt{1-y^{2}}}dy+c$$
$$\Rightarrow -\sqrt{1-x^{2}}=\sqrt{1-y^{2}}+c$$ $$\therefore \frac{d}{dx}\sqrt{1-x^{2}}=\dfrac{-x}{\sqrt{1-x^{2}}}$$
$$\Rightarrow \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=c$$
Question 4
1 / -0

The solution of $$xdx+ydy=x^{2}ydy-xy^{2}dx$$ is

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

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

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

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

Solution

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

The solution of $$xcos^{2} y(dx) + tan y(dy)=0$$ is:

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

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

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

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

Solution

$$x\ cos^{2}y\ dx+tan\ ydy=0$$
$$\Rightarrow xdx+tan\ y\ sec^{2}ydy=0$$
Let tan y = t
$$\Rightarrow sec^{2}y\ dy=dt$$
$$\int xdx+\int tdt=c_{1}$$
$$\Rightarrow \dfrac{x^{2}}{2}+\dfrac{t^{2}}{2}=2c_{1}$$
$$x^{2}+tan^{2}y=2c_{1}$$
$$x^{2}+sec^{2}y=2c_{1}+1$$ put $$(2c_{1}+1)=c$$
$$\Rightarrow x^{2}+sec^{2}y=c$$
Question 6
1 / -0

The solution of

$$3e^{x}\cos^{2}y\ dx+(1-e^{x}) \cot y\ dy=0$$

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

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

C
$$\cot y=c(e^{x}-1)^{2}$$

D
$$\tan y=c(e^{x}-1)^{3}$$

Solution

$$3e^{x}\cos^{2}ydx=(e^{x}-1)\cot\ ydy$$
$$\Rightarrow \dfrac{3e^{x}}{e^{x}-1}dx=\dfrac{1}{\tan y}\sec^{2}ydy$$
Put $$e^{x}-1=t$$ ; $$tan\ y=k$$
$$\Rightarrow e^{x}dx=dt$$ ; $$ \sec^{2}ydy=dk$$
$$\Rightarrow \int \dfrac{3dt}{t}+log\ c=\int \dfrac{dk}{k}$$
$$\Rightarrow log\ t^{3}c=log\ k$$
$$\Rightarrow k=ct^{3}$$
$$\Rightarrow \tan y=c(e^{x}-1)^{3}$$
Question 7
1 / -0

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

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

B
$$1-y^{2}=cx$$

C
$$1+x^{2}=cy$$

D
$$1-x^{2}=cy$$

Solution

$$\dfrac{2y\ dy}{1+y^{2}}=dx$$
$$Let\ t\ =1+y^{2}$$
$$dt=2ydy$$
$$\Rightarrow \dfrac{dt}{t}=\dfrac{dx}{x}$$
$$\Rightarrow log\ t\ =log\ x+log\ c$$
$$\Rightarrow t=x\ c$$
$$\Rightarrow (1+y^{2})=cx$$
Question 8
1 / -0

The solution of $$log\left (\displaystyle \frac{dy}{dx} \right )=ax+by$$

A
$$be^{\displaystyle ax}+ae^{\displaystyle -by}=k$$

B
$$e^{\displaystyle ax}+e^{\displaystyle -by}=c$$

C
$$e^{\displaystyle ax+by}=c$$

D
$$(ax+by)=cxy$$

Solution

$$log\left( \dfrac { dy }{ dx } \right) =ax+by\\ \dfrac { dy }{ dx } ={ e }^{ ax+by }\\ \dfrac { dy }{ dx } ={ e }^{ ax }.{ e }^{ by }\\ \int { { e }^{ -by }dy } =\int { { e }^{ ax }dx } +c\\ \Rightarrow \dfrac { { e }^{ -by } }{ -b } =\dfrac { { e }^{ ax } }{ a } +c$$
$$\Rightarrow c= \dfrac{e^{ax}}{a}+\dfrac{e^{-by}}{b}$$
$$\Rightarrow be^{ax}+ae^{-by}= k$$
Question 9
1 / -0

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

A
$$2y=x^{2}\left [ logx+\dfrac{1}{2} \right ]+c$$

B
$$2y=x^{2}\left [ logx-\dfrac{1}{2} \right ]+c$$

C
$$y=\dfrac{x^{2}}{2}(log2-x)+c$$

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

Solution

$$\Rightarrow \int dy=\int x\ log\ x+c$$
$$\int x\ log\ x\ dx=x^{2}(log\ x-1)-\int (x\ log\ x-x)dx$$
$$\Rightarrow 2\int x\ log\ x\ dx=x^{2}(log\ x-1)+\ ^\dfrac{x^2}{2}$$

$$\int x\ log\ x\ dx=\dfrac{x^{2}(\dfrac{log\ x^{-1}}{2})}{2}$$

$$\Rightarrow \int dy=\int x\ log\ x+c$$

$$\Rightarrow y=\dfrac{x^{2}(\dfrac{log\ x-1}{2})}{2}+c$$
$$\Rightarrow 2y=x^{2}(\dfrac{log\ x-1}{2})+c$$
Question 10
1 / -0

The solution of $$\frac{dy}{dx}=e^{3x+y}$$ given $$y=0$$ when $$x=0$$ is:

A
$$e^{3x}+3e^{-y}=4$$

B
$$e^{-y}=e^{3x}+4$$

C
$$3e^{-y}=e^{3x}+12$$

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

Solution

$$\int e\ ^{-y}dy=\int e\ ^{3x}dx+c$$
$$\Rightarrow \dfrac{e\ ^{-y}}{-1}=\dfrac{e\ ^{3x}}{3}+c$$
$$\Rightarrow 3e\ ^{-y}+e\ ^{3x}=c$$
Put $$(x,y)=(0,0)$$
$$\Rightarrow 3+1=4=c$$
$$\Rightarrow 3e\ ^{-y}+e\ ^{3x}=4$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 20

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Differential Equations Test - 20

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SHARING IS CARING

Question 1

$$6xy=3x^{3}-6x+29$$

$$6xy=3x^{3}-29x+6$$

$$6xy=3x^{3}+29x-6$$

$$6xy=3x^{3}-29x-6$$

Solution

Question 2

$$ytan^{-1}x=c$$

$$ysin^{-1}x=c$$

$$ycos^{-1}x=c$$

$$xsin^{-1}x=c$$

Solution

Question 3

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

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

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

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

Solution

Question 4

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

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

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

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

Solution

Question 5

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

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

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

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

Solution

Question 6

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

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

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

$$\tan y=c(e^{x}-1)^{3}$$

Solution

Question 7

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

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

$$1+x^{2}=cy$$

$$1-x^{2}=cy$$

Solution

Question 8

$$be^{\displaystyle ax}+ae^{\displaystyle -by}=k$$

$$e^{\displaystyle ax}+e^{\displaystyle -by}=c$$

$$e^{\displaystyle ax+by}=c$$

$$(ax+by)=cxy$$

Solution

Question 9

$$2y=x^{2}\left [ logx+\dfrac{1}{2} \right ]+c$$

$$2y=x^{2}\left [ logx-\dfrac{1}{2} \right ]+c$$

$$y=\dfrac{x^{2}}{2}(log2-x)+c$$

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

Solution

Question 10

$$e^{3x}+3e^{-y}=4$$

$$e^{-y}=e^{3x}+4$$

$$3e^{-y}=e^{3x}+12$$

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

Solution

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