Differential Equations Test

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Differential Equations Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Find the solution of $$\displaystyle \frac{dy}{dx}=\frac{xy+y}{xy+x}$$

A
$$\displaystyle y-x=\log \frac{kx}{y}$$

B
$$\displaystyle y+x=\log \frac{ky}{x}$$

C
$$\displaystyle y+x=\log \frac{y}{kx}$$

D
$$\displaystyle y-x=\log \frac{ky}{x}$$

Solution

$$\displaystyle \frac{dy}{dx}=\frac{xy+y}{xy+x}$$
or $$\displaystyle \frac{y+1}{y}dy=\frac{x+1}{x}dx,$$
or $$\displaystyle \left ( 1+\frac{1}{y} \right )dy=\left ( 1+\frac{1}{x} \right )dx$$
or $$\displaystyle \int \left ( 1+\frac{1}{y} \right )dy=\int\left ( 1+\frac{1}{x} \right )dx$$
$$\displaystyle y+\log y=x+\log x+\log k$$
$$\displaystyle \therefore y-x=\log \frac{kx}{y}$$
Question 2
1 / -0

$$\displaystyle x\cos ^{2}ydx=y\cos ^{2}x dy$$

A
$$\displaystyle x\tan x+\log \sec x= y\tan y+\log \sec y+c.$$

B
$$\displaystyle x\tan x+\log \sec x= y\tan y-\log \sec y+c.$$

C
$$\displaystyle x\tan x-\log \sec x= y\tan y+\log \sec y+c.$$

D
$$\displaystyle x\tan x-\log \sec x= y\tan y-\log \sec y+c.$$

Solution

Given $$\displaystyle x\cos ^{2}ydx=y\cos ^{2}x dy$$
$$\displaystyle x\sec 2 x dx=y \sec ^{2}y dy.$$
$$\displaystyle \int x\sec 2 x dx=\int y \sec ^{2}y dy.$$
Integrate by parts, we get
$$\displaystyle x\tan x-\int \tan x dx= y\tan y-\int \tan y dy+c$$
$$\Rightarrow \displaystyle x\tan x-\log \sec x= y\tan y-\log \sec y+c.$$
Question 3
1 / -0

Solve the diffrential equation: $$\displaystyle \log \frac{dy}{dx}=ax+by$$

A
$$\displaystyle -\frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$$

B
$$\displaystyle \frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$$

C
$$\displaystyle -\frac{1}{b}e^{by}=\frac{1}{a}e^{ax}+c$$

D
$$\displaystyle \frac{1}{b}e^{by}=\frac{1}{a}e^{ax}+c$$

Solution

$$\displaystyle \log \frac{dy}{dx}=ax+by$$
$$\displaystyle \frac{dy}{dx}=e^{ax+by}=e^{ax}.e^{by}$$
or $$\displaystyle e^{-by}dy=e^{ax}dx$$
integrating we get,
$$\displaystyle -\frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$$
Question 4
1 / -0

Find the solution of $$\displaystyle xy\frac{dy}{dx}=\frac{1+y^{2}}{1+x^{2}}\left ( 1+x+x^{2} \right )$$.

A
$$\displaystyle \frac{1}{2}\log \left ( 1+y^{2} \right )=\log x+\tan ^{-1}x+c$$

B
$$\displaystyle \frac{1}{4}\log \left ( 1+y^{2} \right )=\log x+\tan ^{-1}x+c$$

C
$$\displaystyle \frac{1}{2}\log \left ( 1-y^{2} \right )=\log x+\tan ^{-1}x+c$$

D
$$\displaystyle \frac{1}{4}\log \left ( 1-y^{2} \right )=\log x+\tan ^{-1}x+c$$

Solution

Given, $$\displaystyle xy\frac{dy}{dx}=\frac{1+y^{2}}{1+x^{2}}\left ( 1+x+x^{2} \right )$$

or $$\displaystyle \int \frac{y\:dy}{1+y^{2}}=\int \frac{1}{x}.\frac{1+x^{2}+x}{1+x^{2}}=\int \frac{1}{x}\left ( 1+\frac{x}{1+x^{2}} \right )dx=\int \dfrac 1 x +\dfrac {1}{1+x^2} \ dx$$

$$\displaystyle \int \frac{y\:dy}{1+y^{2}} =\int \dfrac 1 x +\dfrac {1}{1+x^2} \ dx$$

$$\Rightarrow \displaystyle \frac{1}{2}\log \left ( 1+y^{2} \right )=\log x+\tan ^{-1}x+c$$
Question 5
1 / -0

Find the solution of $$\displaystyle a\left ( x\frac{dy}{dx}+2y \right )=xy\frac{dy}{dx}$$.

A
$$\displaystyle yx^{2}=ke^{y/a}$$

B
$$\displaystyle xy^{2}=ke^{y/a}$$

C
$$\displaystyle yx^{3}=ke^{y/a}$$

D
$$\displaystyle xy^{3}=ke^{y/a}$$

Solution

Given $$\displaystyle a\left ( x\frac{dy}{dx}+2y \right )=xy\frac{dy}{dx}$$
or $$\displaystyle \frac{\left ( a-y \right )}{y}dy=\frac{2a}{x}dx$$
or $$\displaystyle \frac{a}{y}dy+\frac{2a}{x}dx=dy$$
Integrating,we get
$$\displaystyle \int\frac{a}{y}dy+\int\frac{2a}{x}dx=\int dy$$
$$a \displaystyle \log y+2a \log x=y+c$$
or $$\displaystyle \log yx^{2}=\frac{y+c}{a};$$
$$\displaystyle \therefore yx^{2}=e^{\left ( y+c \right )/a}=ke^{y/a}$$
Question 6
1 / -0

Find the solution of $$\displaystyle \left ( x^{2}-yx^{2} \right )\frac{dy}{dx}+\left ( y^{2}+xy^{2} \right )=0$$

A
$$\displaystyle \log \frac{x}{ky}=\frac{x+y}{xy}$$

B
$$\displaystyle \log \frac{x}{ky}=\frac{x-y}{xy}$$

C
$$\displaystyle \log \frac{x}{ky}=\frac{x-y}{y}$$

D
$$\displaystyle \log \frac{x}{ky}=\frac{x+y}{y}$$

Solution

$$\displaystyle x^{2}\left ( 1-y \right )\frac{dy}{dx}+y^{2}\left ( 1+x \right )=0$$
or $$\displaystyle \frac{1-y}{y^{2}}dy+\frac{1+x}{x^{2}}dx=0.$$
or $$\displaystyle \left ( \frac{1}{y^{2}}-\frac{1}{y} \right )dy+\left ( \frac{1}{x^{2}}+\frac{1}{x} \right )dx=0.$$
Integrating, we get
$$\displaystyle \log x-\log y-\left ( \frac{1}{x}+\frac{1}{y} \right )=c=\log k$$ (say)
or $$\displaystyle \log \frac{x}{ky}=\frac{x+y}{xy}$$
Question 7
1 / -0

$$y=ae^{-1/x}+b$$ is a solution of $$\displaystyle\frac{dy}{dx}=\frac{y}{x^{2}}$$ when

A
$$a=1,b=0$$

B
$$a=3,b=1$$

C
$$a=1,b=1$$

D
$$a=2,b=2$$

Solution

$$\displaystyle\frac{dy}{dx}=\frac{y}{x^{2}}$$
$$\Rightarrow \dfrac{dy}{y}=\dfrac{dx}{x^2}=x^{-2}dx$$
Integrating both sides, we get
$$\ln y = -x^{-1}+c=-\dfrac{1}{x}+c$$
$$\Rightarrow y = e^{-1/x+c}=e^ce^{-1/x}=ae^{-1/x}$$
Where $$e^c = a=$$ any constant.
Comparing it with given solution we get, $$b=0, a\in R$$
Question 8
1 / -0

Solution of the given differential equation $$\displaystyle \left ( 1-x^{2} \right )\frac{dy}{dx}+xy=xy^{2}$$ is

A
$$y(1-y)=c\displaystyle \sqrt{1-x^2}$$

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

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

D
$$x(x-1)=c\displaystyle \sqrt{1-y^2}$$
Question 9
1 / -0

The order of the differential equation whose general solution is given by $$y =(C_1+ C_2 )\, cos\, (x+ C_3 )\, -\, C_4e^{x+c_5}$$ where $$C_1, C_2, C_3, C_4, C_5$$ are arbitrary constants, is:

A
5

B
4

C
3

D
2

Solution

We can write $$y=A+\cos { \left( x+B \right) } -C{ e }^{ x }$$
where $$A={ C }_{ 1 }+{ C }_{ 2 },B={ C }_{ 3 }$$ and $$C={ C }_{ 4 }{ e }^{ { C }_{ 5 } }$$
$$\cfrac { dy }{ dx } =-A\sin { \left( x+B \right) } -C{ e }^{ x }\\ \cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-A\cos { \left( x+B \right) } -C{ e }^{ x }\\ \Rightarrow \cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=-2C{ e }^{ x }\\ \Rightarrow \cfrac { { d }^{ 3 }y }{ { dx }^{ 3 } } +\cfrac { dy }{ dx } =-2C{ e }^{ x }=\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y\\ \Rightarrow \cfrac { { d }^{ 3 }y }{ { dx }^{ 3 } } -\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } +\cfrac { dy }{ dx } -y=0$$
which is differential equation of order 3
Question 10
1 / -0

Find the solution of $$\displaystyle\frac{dy}{dx}+\sin\left ( \frac{x+y}{2} \right ) =\sin \left ( \frac{x-y}{2} \right )$$.

A
$$\displaystyle \log \tan \frac{y}{4}=c-2\sin \frac{x}{2}$$

B
$$\displaystyle \log \tan \frac{y}{2}=c+2\sin \frac{x}{2}$$

C
$$\displaystyle \log \tan \frac{y}{2}=c-2\sin \frac{x}{2}$$

D
$$\displaystyle \log \tan \frac{y}{4}=c+2\sin \frac{x}{2}$$

Solution

$$\displaystyle\frac{dy}{dx}+\sin\left ( \frac{x+y}{2} \right ) =\sin \left ( \frac{x-y}{2} \right )$$
or $$\displaystyle\frac{dy}{dx} =\sin \left ( \frac{x-y}{2} \right )-\sin\left ( \frac{x+y}{2} \right )$$
or $$\displaystyle \frac{dy}{dx}=-2\cos \frac{x}{2}\sin \frac{y}{2}$$
or $$\displaystyle \int cosec\frac{y}{2}dy=\int -2\cos \frac{x}{2}dx$$
or $$\displaystyle \log \tan \frac{y}{4}=c-2\sin \frac{x}{2}$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 26

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Differential Equations Test - 26

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Question 1

$$\displaystyle y-x=\log \frac{kx}{y}$$

$$\displaystyle y+x=\log \frac{ky}{x}$$

$$\displaystyle y+x=\log \frac{y}{kx}$$

$$\displaystyle y-x=\log \frac{ky}{x}$$

Solution

Question 2

$$\displaystyle x\tan x+\log \sec x= y\tan y+\log \sec y+c.$$

$$\displaystyle x\tan x+\log \sec x= y\tan y-\log \sec y+c.$$

$$\displaystyle x\tan x-\log \sec x= y\tan y+\log \sec y+c.$$

$$\displaystyle x\tan x-\log \sec x= y\tan y-\log \sec y+c.$$

Solution

Question 3

$$\displaystyle -\frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$$

$$\displaystyle \frac{1}{b}e^{-by}=\frac{1}{a}e^{ax}+c$$

$$\displaystyle -\frac{1}{b}e^{by}=\frac{1}{a}e^{ax}+c$$

$$\displaystyle \frac{1}{b}e^{by}=\frac{1}{a}e^{ax}+c$$

Solution

Question 4

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

$$\displaystyle \frac{1}{4}\log \left ( 1+y^{2} \right )=\log x+\tan ^{-1}x+c$$

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

$$\displaystyle \frac{1}{4}\log \left ( 1-y^{2} \right )=\log x+\tan ^{-1}x+c$$

Solution

Question 5

$$\displaystyle yx^{2}=ke^{y/a}$$

$$\displaystyle xy^{2}=ke^{y/a}$$

$$\displaystyle yx^{3}=ke^{y/a}$$

$$\displaystyle xy^{3}=ke^{y/a}$$

Solution

Question 6

$$\displaystyle \log \frac{x}{ky}=\frac{x+y}{xy}$$

$$\displaystyle \log \frac{x}{ky}=\frac{x-y}{xy}$$

$$\displaystyle \log \frac{x}{ky}=\frac{x-y}{y}$$

$$\displaystyle \log \frac{x}{ky}=\frac{x+y}{y}$$

Solution

Question 7

$$a=1,b=0$$

$$a=3,b=1$$

$$a=1,b=1$$

$$a=2,b=2$$

Solution

Question 8

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

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

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

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

Question 9

5

4

3

2

Solution

Question 10

$$\displaystyle \log \tan \frac{y}{4}=c-2\sin \frac{x}{2}$$

$$\displaystyle \log \tan \frac{y}{2}=c+2\sin \frac{x}{2}$$

$$\displaystyle \log \tan \frac{y}{2}=c-2\sin \frac{x}{2}$$

$$\displaystyle \log \tan \frac{y}{4}=c+2\sin \frac{x}{2}$$

Solution

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