Differential Equations Test

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Differential Equations Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Verify that $$y=Cx^3$$ is a solution of the differential equation $$xy'-3y=0$$ for any value of C. Then

find the particular solution determined by the initial condition $$y=2$$ when $$x=-3$$.

A
$$y=\dfrac{2}{27}x^2$$

B
$$y=-\dfrac{2}{27}x^3$$

C
$$y=-\dfrac{2}{25}x^3$$

D
None of these

Solution

Given that $$y=C{ x }^{ 3 }$$ is the solution of the differential equation$$x{ y }^{ ' }=3y$$
So to get a particular solution we need to find the value of $$C$$ for a given initial value condition,
i.e.,$$y=2$$ when $$x=-3$$,
Substituting this in the general solution gives,
$$2=C(-27)$$
$$\Longrightarrow C=-\cfrac { 2 }{ 27 } $$
$$\therefore$$ the particular solution is $$y=-\cfrac { 2 }{ 27 } { x }^{ 3 }$$
Question 2
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The solution for the differential equation $$\cfrac { dy }{ y } +\cfrac { dx }{ x } =0$$ is:

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

B
$$\log { x } .\log { y } =c$$

C
$$xy=C$$

D
$$x+y=c$$

Solution

We have, $$\cfrac { dy }{ y } +\cfrac { dx }{ x } =0$$
Integrate both sides,
$$\displaystyle \int\left(\cfrac { dy }{ y } +\cfrac { dx }{ x } \right)=\int 0$$
$$\displaystyle\Rightarrow \int\cfrac { dy }{ y } +\int \cfrac { dx }{ x } =c$$
, since differentiation any constant is 0
$$\Rightarrow \ln x+\ln y=c$$, use basic formula
$$\Rightarrow \ln(xy)=c$$, product rule of logarithm $$\log a+\log b=\log(ab)$$
$$\Rightarrow xy=e^c=C$$
Question 3
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The solution of differential equation $$x \dfrac {dy}{dx} + 2y= x^{2}$$ is ____

A
$$y = \dfrac {x^{2} + C}{4x^{2}}$$

B
$$y = \dfrac {x^{2}}{4} + C$$

C
$$y = \dfrac {x^{4} + C}{x^{2}}$$

D
$$y = \dfrac {x^{4} + C}{4x^{2}}$$

Solution

We have, $$x \dfrac {dy}{dx} + 2y= x^{2}$$
Divide each sides by $$x$$
$$\Rightarrow \dfrac{dy}{dx}+\dfrac{2}{x}y=x$$
Integrating factor $$=e^{ \int \frac{2}{x}dx}=e^{2\ln x}=e^{\ln x^2}=x^2$$
Hence solution is given by,
$$\displaystyle y(x^2)=\int x^2\cdot x dx+c$$
$$\displaystyle \Rightarrow x^2y=\int x^3dx+C=\dfrac{x^4}{4}+c$$
$$\Rightarrow y=\dfrac{x^4+4c}{x^2}=\dfrac{x^4+C}{4x^2}$$
Question 4
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The solution of the differential equation $$y\sin\left(\dfrac{x}{y}\right)dx=\left(x \sin\left(\dfrac{x}{y} \right)-y \right) dy$$ satisfying $$y(\dfrac{\pi}{4})=1$$ is

A
$$\cos\dfrac{x}{y}=-\log_ey+\dfrac{1}{\sqrt{2}}$$

B
$$\sin\dfrac{x}{y}=\log_ey+\dfrac{1}{\sqrt{2}}$$

C
$$\sin\dfrac{x}{y}=\log_ex-\dfrac{1}{\sqrt{2}}$$

D
$$\cos\dfrac{x}{y}=-\log_ex-\dfrac{1}{\sqrt{2}}$$

Solution

Given differential equation is $$\cfrac { y\sin { \frac { x }{ y } } }{ x\sin { \frac { x }{ y } } -y } =\cfrac { dy }{ dx } $$
Now replace $$x=vy$$ such that,$$\cfrac { dx }{ dy } =v+y\cfrac { dv }{ dy } $$
$$\Longrightarrow v+y\cfrac { dv }{ dy } =\cfrac { v\sin { v } -1 }{ \sin { v } } $$
$$\Longrightarrow y\cfrac { dv }{ dy } =\cfrac { -1 }{ \sin { v } } $$
Separating the variables and integrating gives,
$$\int { \sin { v } } dv=\int { \cfrac { 1 }{ y } } dy$$,
$$\Longrightarrow -\cos { v } +c=\ln { y } $$
Given $$y\left( \cfrac { \pi }{ 4 } \right) =1$$
$$\Longrightarrow c=\cfrac { 1 }{ \sqrt { 2 } } $$
$$\therefore$$ the particular solution is ,
$$\cos { \cfrac { x }{ y } } =-\ln { y } +\cfrac { 1 }{ \sqrt { 2 } } $$
Question 5
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Solution of $$\cfrac { dx }{ dy } +mx=0$$, $$m< 0$$ is

A
$$x=c{ e }^{ my }$$

B
$$x=c{ e }^{ -my }$$

C
$$x=my+c$$

D
$$x=c$$

Solution

Given differential equation is $$\dfrac{dx}{dy}+mx=0$$
$$\Rightarrow \dfrac{dx}{x}=-mdy$$
Now integrate both sides,
$$\ln x=-my+C$$
$$\Rightarrow x=e^{-my+C}=e^{-my}\cdot e^C$$
$$\Rightarrow x=ce^{-my}$$, where $$c=e^C$$
Question 6
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The solution of $$\dfrac {d^{2}x}{dy^{2}} - x = k$$, where $$k$$ is a non-zero constant, vanishes when $$y = 0$$ and tends of finite limit as $$y$$ tends to infinity, is

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

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

C
$$x = k(e^{-y} - 1)$$

D
$$x = k(e^{y} - 1)$$

Solution

We can write given differential equation as,
$$(D^{2} - 1)x = k ... (i)$$
where, $$D = \dfrac {d}{dy}$$
Its auxiliary equation is $$m^{2} - 1 = 0$$, so that
$$m = 1, -1$$
Hence, $$CF = C_{1}e^{y} + C_{2}e^{-y}$$
where $$C_{1}, C_{2}$$ are arbitrary constants
Now, also $$PI = \dfrac {1}{D^{2} - 1}k$$
$$= k.\dfrac {1}{D^{2} - 1} e^{0.y}$$
$$= k.\dfrac {1}{0^{2} - 1}e^{0.y} = -k$$
So, solution of eq. (i) is
$$x = C_{1}e^{y} + C_{2}e^{-y} - k ... (ii)$$
Given that $$x = 0$$, when $$y = 0$$
So, $$0 = C_{1} + C_{2} - k$$ (From (ii))
$$\Rightarrow C_{1} + C_{2} = k .... (iii)$$
Multiplying both sides of eq. (ii) by $$e^{-y}$$, we get
$$x.e^{-y} = C_{1} + C_{2}e^{-2y} - ke^{-y} ... (iv)$$
Given that $$x\rightarrow m$$ when $$y\rightarrow \infty, m$$ being a finite quantity.
So, eq (iv) becomes
$$x\times 0 = C_{1} + C_{2} \times 0 - (k\times 0)$$
$$\Rightarrow C_{1} = 0 ....(v)$$
From eqs. (iv) and (v), we get
$$C_{1} = 0$$ and $$C_{2} = k$$
Hence, eq. (ii) becomes
$$x = ke^{-y} - k = k(e^{-y} - 1)$$ which is the required solution.
Question 7
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What is the general solution of the differential equation $${ e }^{ x }\tan { y } dx+\left( 1-{ e }^{ x } \right) \sec ^{ 2 }{ y } dy=0$$?

A
$$\sin { y } =c\left( 1-{ e }^{ x } \right) $$ where $$c$$ is the constant of integration

B
$$\cos { y } =c\left( 1-{ e }^{ x } \right) $$ where $$c$$ is the constant of integration

C
$$\cot { y } =c\left( 1-{ e }^{ x } \right) $$ where $$c$$ is the constant of integration

D
None of the above

Solution

$${ e }^{ x }\tan ydx+(1−{ e }^{ x })\sec ^{ 2 }{ y } dy=0$$
$$\Rightarrow \dfrac { { e }^{ x } }{ 1-{ e }^{ x } } dx+\dfrac {\sec ^{ 2 }{ y } }{ \tan y } dy=0$$ ..... $$(i)$$
Now, to calculate $$\displaystyle \int { \dfrac { { e }^{ x } }{ 1-{ e }^{ x } } dx }$$
Put $$1-{ e }^{ x }=t\Rightarrow -{ e }^{ x }dx=dt$$
$$\therefore \displaystyle \int { \frac { { e }^{ x } }{ 1-{ e }^{ x } } dx } =\int { -\frac { dt }{ t } } =-\ln t =-\ln(1-{ e }^{ x })$$
Also, $$\displaystyle \int { \frac {\sec ^{ 2 }{ y } }{ \tan y } dy }$$
Put $$p=\tan y\Rightarrow dp=\sec^{ 2 }{ y } dy$$
$$\therefore \displaystyle \int { \frac {\sec ^{ 2 }{ y } }{ \tan y } dy } =\int { \frac { dp }{ p } } =\ln p=\ln(\tan y)$$

On integrating $$(i)$$, we get
$$ -\ln(1-{ e }^{ x })+\ln(\tan y)=\ln k\\ \Rightarrow \ln(\tan y)=\ln(k(1-{ e }^{ x }))\\ \Rightarrow \tan y=k(1-{ e }^{ x })$$
Hence, D is correct.
Question 8
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What is the general solution of the differential equation $${x}^{2}dy+{y}^{2}dx=0$$?

A
$$x+y=c$$ where $$c$$ is the constant of integration

B
$$xy=c$$ where $$c$$ is the constant of integration

C
$$c(x+y)=xy$$ where $$c$$ is the constant of integration

D
None of the above

Solution

$${ x }^{ 2 }dy{ +y }^{ 2 }dx=0$$
$$\Rightarrow \dfrac { dy }{ { y }^{ 2 } } +\dfrac { dx }{ { x }^{ 2 } } =0$$
On integrating, we get
$$\Rightarrow -\dfrac { 1 }{ y } -\dfrac { 1 }{ x } +\dfrac { 1 }{ c } =0\\ \Rightarrow c(x+y)=xy$$
Option C is correct
Question 9
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What is the solution of $$\frac{dy}{dx}=2y-1$$ is :

A
$$y=\frac{1-e^{-2x}}{2}$$

B
$$y=\frac{1+e^{2x}}{2}$$

C
$$y=1+e^x$$

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

Solution

Given : $$\cfrac{dy}{dx}=2y-1$$ ..... $$(i)$$
Let $$2y-1=t$$ ..... $$(ii)$$
Differentiating both sides with respect to $$x$$, we get
$$2\cfrac{dy}{dx}=\cfrac{dt}{dx}$$
Substituting the value of $$\cfrac{dy}{dx}$$ in the original equation, we get
$$\cfrac{dt}{2 dx}=t$$
Rearranging terms, we get
$$\cfrac{dt}{t}=2dx$$
Integrating both sides, we get
$$\int \cfrac{dt}{t}=\int 2dx$$
$$\Rightarrow \ln{t}=2x$$
$$\Rightarrow t={e}^{2x}$$
Substituting $$t$$ in $$(ii),$$ we get
$$2y-1={e}^{2x}$$
$$y=\cfrac{{e}^{2x}+1}{2}$$
Question 10
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Directions For Questions

For the next three (03) items that follow :
The general solution of the differential equation $$(x^2 + x + 1)dy + (y^2 + y + 1)dx =0$$ is
$$(x + y + 1) = A(1 + Bx + Cy + Dxy)$$ where B, C and D are constants and A is parameter.

...view full instructions

What is B equal to ?

A
-1

B
1

C
2

D
None of the above

Solution

$${\textbf{Step-1: Finding the value of D. }}$$
$${\text{Given,}}$$
$$\Rightarrow(x^2+x+1)dy+(y^2+y+1)dx=0$$
$$\Rightarrow(x^2+x+1)dy=-(y^2+y+1)dx$$
$$\Rightarrow\dfrac{dy}{y^2+y+1}=-\dfrac{dx}{x^2+x+1}$$
$${\text{Integrating both sides}}$$
   $$\Rightarrow \int \dfrac{dy}{y^2+y+1}=-\int\dfrac{dx}{x^2+x+1}$$
$${\text{By using perfect square method}}$$
$$\Rightarrow\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}=-\int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$$

$$\Rightarrow \int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}+\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$$

$${\text{use formula}}$$ $$\int\dfrac{dx}{a^2+x^2}=\dfrac{1}{a}tan^{-1}\dfrac{x}{a}$$

$$\Rightarrow \dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2x+1}{2}]+\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2y+1}{2}]+\dfrac{2}{\sqrt3} tan^{-1}C=0$$

$${\text{[.... Where C is the integral constant]}}$$

  $${\text{use formula}}$$ $$tan^{-1}A+tan^{-1}B$$

$$\Rightarrow\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{\dfrac{2x+1}{3}+\dfrac{2y+1}{\sqrt3}}{1-(\dfrac {2x+1}{\sqrt3})(\dfrac{2y+1}{\sqrt3})}]=-\dfrac{2}{\sqrt3}tan^{-1} C$$

$$\Rightarrow \dfrac{2x+1+2y+1}{\sqrt 3}\times \dfrac{\sqrt3 \sqrt3}{3-4xy-2x-2y-1}=C$$

$$\Rightarrow \dfrac{\sqrt3(2x+2y+2)}{2-4xy-2x-2y}=C$$

$$\therefore \sqrt3\times2(x+y+1)=c\times 2(1-2xy-x-y)$$

$$\therefore (x+y+1)=\dfrac{C}{\sqrt3}(1-x-y-2xy)$$

$${\text{ Compare this equation with the given equation gives,}}$$

$$\therefore B=-1$$

$${\textbf{Hence option A is correct }}$$

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Differential Equations Test - 30

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Differential Equations Test - 30

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SHARING IS CARING

Question 1

$$y=\dfrac{2}{27}x^2$$

$$y=-\dfrac{2}{27}x^3$$

$$y=-\dfrac{2}{25}x^3$$

None of these

Solution

Question 2

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

$$\log { x } .\log { y } =c$$

$$xy=C$$

$$x+y=c$$

Solution

Question 3

$$y = \dfrac {x^{2} + C}{4x^{2}}$$

$$y = \dfrac {x^{2}}{4} + C$$

$$y = \dfrac {x^{4} + C}{x^{2}}$$

$$y = \dfrac {x^{4} + C}{4x^{2}}$$

Solution

Question 4

$$\cos\dfrac{x}{y}=-\log_ey+\dfrac{1}{\sqrt{2}}$$

$$\sin\dfrac{x}{y}=\log_ey+\dfrac{1}{\sqrt{2}}$$

$$\sin\dfrac{x}{y}=\log_ex-\dfrac{1}{\sqrt{2}}$$

$$\cos\dfrac{x}{y}=-\log_ex-\dfrac{1}{\sqrt{2}}$$

Solution

Question 5

$$x=c{ e }^{ my }$$

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

$$x=my+c$$

$$x=c$$

Solution

Question 6

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

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

$$x = k(e^{-y} - 1)$$

$$x = k(e^{y} - 1)$$

Solution

Question 7

$$\sin { y } =c\left( 1-{ e }^{ x } \right) $$ where $$c$$ is the constant of integration

$$\cos { y } =c\left( 1-{ e }^{ x } \right) $$ where $$c$$ is the constant of integration

$$\cot { y } =c\left( 1-{ e }^{ x } \right) $$ where $$c$$ is the constant of integration

None of the above

Solution

Question 8

$$x+y=c$$ where $$c$$ is the constant of integration

$$xy=c$$ where $$c$$ is the constant of integration

$$c(x+y)=xy$$ where $$c$$ is the constant of integration

None of the above

Solution

Question 9

$$y=\frac{1-e^{-2x}}{2}$$

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

$$y=1+e^x$$

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

Solution

Question 10

-1

1

2

None of the above

Solution

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