Differential Equations Test

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Differential Equations Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

If $$y=y\left( x \right) $$ and $$\displaystyle \frac { 2+\sin { x } }{ y+1 } \left( \frac { dy }{ dx } \right) =-\cos { x } ,y\left( 0 \right) =1$$, then find the value of $$\displaystyle y\left( \frac { \pi }{ 2 } \right) $$.

A
$$\displaystyle \frac{1}{3}$$

B
$$\displaystyle \frac{2}{3}$$

C
$$\displaystyle - \frac{1}{3}$$

D
$$\displaystyle 1$$

Solution

$$\displaystyle \left( \frac { dy }{ dx } \right) \frac { 2+\sin { x } }{ y+1 } =-\cos { x } ,y\left( 0 \right) =1$$
$$\displaystyle \Rightarrow \frac { dy }{ \left( 1+y \right) } =-\frac { \cos { x } }{ 2+\sin { x } } dx$$
Integrating both sides $$\Rightarrow \ln { \left( 1+y \right) } =-\ln { \left( 2+\sin { x } \right) } +c$$
Substitute $$x=0$$ and $$y=1$$
$$\Rightarrow \ln { 2 } =-\ln { 2 } +c\Rightarrow c=\ln { 4 } $$
Substitute $$\displaystyle x=\frac { \pi }{ 2 } ,\ln { \left( 1+y \right) } =\ln { 3 } +\ln { 4 } =\ln { \frac { 4 }{ 3 } } \Rightarrow y=\frac { 1 }{ 3 } $$
Question 2
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Solution of differential equation $$sec\, x\, dy - cosec\, y \,dx = 0$$ is

A
$$cos\, x+ sin\, y=c$$

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

C
$$sin\, y-cos\, x=c$$

D
$$cos\, y- sin\, x=c$$

Solution

$$sec\,x\,dy-cosec\,y-dx = 0 $$
$$ \dfrac{dy}{cos\,x} = \dfrac{dx}{sin\,y} $$
By variable separable from
$$\displaystyle sin\,y\,dy = cos\,x\,dx $$
integrating,
$$ \displaystyle \int siny\,dy = \int cosx\,dx $$
$$ -cos\,y = sin\,x+c $$
$$ \boxed { sinx+cosy+c =0} $$
or
$$ \boxed {sinx+cosy = c} $$
Question 3
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The solution of the differential equation $$\dfrac { dy }{ dx } =2{ e }^{ x-y }+{ x }^{ 2 }{ e }^{ -y }$$ is

A
$${ e }^{ y }=2{ e }^{ x }+\dfrac { { x }^{ 3 } }{ 3 } +C$$

B
$${ e }^{ -y }=2{ e }^{ x }+\dfrac { { x }^{ -3 } }{ 3 } +C$$

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

D
$${ e }^{ y }=2{ e }^{ -x }+\dfrac { { x }^{ 3 } }{ 3 } +C$$

Solution

Given, $$\dfrac { dy }{ dx } =2{ e }^{ x-y }+{ x }^{ 2 }{ e }^{ -y }$$

$$\Rightarrow \dfrac { dy }{ dx } =2{ e }^{ x }\cdot \dfrac { 1 }{ { e }^{ y } } +{ x }^{ 2 }\cdot \dfrac { 1 }{ { e }^{ y } }$$

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

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

On integrating both sides, we get

$$\displaystyle\int { { e }^{ y }dy } =\displaystyle\int { \left( 2{ e }^{ x }+{ x }^{ 2 } \right) dx }$$

$$ \Rightarrow { e }^{ y }=2{ e }^{ x }+\dfrac { { x }^{ 3 } }{ 3 } +C$$
which is the required solution.
Question 4
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General solution of $$y\dfrac{dy}{dx}+by^2 = a \cos x, 0 < x < 1$$ is:
(here $$c$$ is an arbitrary constant)

A
$$y^2 = 2a(2b\sin x + \cos x) + ce^{-2bx}$$

B
$$(4b^2+1)y^2 = 2a(\sin x + 2b \cos x)+ ce^{-2bx}$$

C
$$(4b^2+1)y^2 = 2a(\sin x + 2b \cos x)+ ce^{2bx}$$

D
$$y^2 = 2a(2b\sin x + \cos x) + ce^{2bx}$$

Solution

Let us take $$y^{2}=t$$ , which means $$dt=2ydy$$

Given equation will transform into $$\cfrac{dt}{dx}+2bt=2a\cos x$$
The integration factor is $$e^{2bx}$$

The general solution is $$ te^{2bx} = \displaystyle \int { 2a\cos x{ e }^{ 2bx }dx } +c$$

$$\Rightarrow te^{2bx}=\cfrac{4ab\cos x+2a\sin x}{4b^{2}+1}e^{2bx}+c$$

$$\Rightarrow (4b^{2}+1)y^{2}=2a(\sin x+2b\cos x)+Ce^{-2bx}$$

Therefore the correct option is $$B$$
Question 5
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The solution of the differential equation $$\displaystyle\frac{d^{2}y}{dx^2}+3y=-2x$$ is.

A
$$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{2}{3}x$$

B
$$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{4}{5}$$

C
$$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-2x^2+\dfrac{4}{9}$$

D
$$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{2}{3}x^2+\dfrac{4}{9}$$

Solution

$$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +3y=-2x.......(i)$$

This is second order non homogeneous differential equation.

Its solution is given as $$CF+PI$$

For $$CF$$ part

$$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +3y=0....(ii)\\ y={ { { c }_{ 1 } }e }^{ mx }\\ \dfrac { dy }{ dx } ={ c }_{ 1 }m{ e }^{ mx }\\ \dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } ={ c }_{ 1 }{ m }^{ 2 }{ e }^{ mx }$$

Substituting in $$(ii)$$, we get
$${ m }^{ 2 }{ e }^{ mx }+3{ e }^{ mx }=0\\ { e }^{ mx }({ m }^{ 2 }+3)=0\\ { m }^{ 2 }+3=0\\ \Rightarrow m=\sqrt { -3 } \\ m=i\sqrt { 3 } \\ y=c_{1}{ e }^{ i\sqrt { 3 } x }={ c }_{ 1 }(\cos { \sqrt { 3 } x } +i\sin { \sqrt { 3 } x } )\\ y={ c }_{ 1 }\cos { \sqrt { 3 } x } +{ c }_{ 2 }\sin { \sqrt { 3 } x } \quad \quad $$

For $$PI$$ part

Let $$y=cx+d$$

$$\dfrac { dy }{ dx } =c\\ \dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0$$

Substituting in $$(i)$$

$$0+3(cx+d)=-2x\\ 3cx+3d=-2x$$

Comparing both sides

$$c=-\dfrac { 2 }{ 3 } ,d=0$$

So solution for $$PI$$ part is

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

General solution is $$CF+PI$$

$${ c }_{ 1 }\cos { \sqrt { 3 } x } +{ c }_{ 2 }\sin { \sqrt { 3 } x } -\dfrac { 2 }{ 3 } x\quad $$
Question 6
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The general solution of the differential equation $$\displaystyle \frac{dy}{dx}+\frac{y}{x}=3x$$ is.

A
$$y=x+\frac{c}{x}$$

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

C
$$y=x-\frac{c}{x}$$

D
$$y=x^2-\frac{c}{x^2}$$

Solution

Comparing above equation with first order linear differential equation $$\dfrac{dy}{dx}+Py=Q$$
We get $$P=\dfrac{1}{x}$$ and $$Q=3x$$

So integration factor $$I.F. =e^{\int Pdx}=e^{\int (1/x)dx}=e^{\ln x}=x$$
Hence solution is given by, $$y(I.F.)=\displaystyle \int Q(I.F.)d x$$
$$\Rightarrow xy=\displaystyle \int 3x^2dx$$
$$\Rightarrow xy=x^3+c$$
$$\Rightarrow y=x^2+\dfrac{c}{x}$$
Question 7
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The differential equation $$y\dfrac{dy}{dx}=a-x$$ represents

A
a family of circles with centre on y-axis

B
a family of circles with centre at origin

C
a family of circles with given radius

D
a family of circles with centre on x-axis

Solution

Given: $$y\dfrac { dy }{ dx } =a-x$$

Integrating on both sides

$$\int { ydx } =\int { adx } -\int { xdx } $$

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

$${ x }^{ 2 }+{ y }^{ 2 }-2ax+c=0$$

$${ (x-a) }^{ 2 }+{ y }^{ 2 }=r^{ 2 }$$

Hence family of circles with centre on x axis
Question 8
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A particle, initially at origin moves along $$x$$ axis according to the rule $$\dfrac{dx}{dt}=x+4$$. The time taken by the particle to traverse a distance of $$96$$ units is

A
$$3\ln 5$$

B
$$\ln 5$$

C
$$2\ln 5$$

D
None of these

Solution

Given $$\frac{dx}{dt}=x+4$$
$$\Rightarrow \frac{dx}{x+4}=dt$$
By applying integration on both sides , we get $$\ln{(x+4)}=t+k$$
$$\Rightarrow x=e^{t}-4+C$$
At $$t=0$$ , $$x=0$$ , which implies $$C=3$$
Therefore the equation becomes $$x=e^{t}-1$$
Given that $$x=96$$ , which implies $$t=\ln{97}$$
Question 9
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Consider the differential equation $${ y }^{ 2 }dx+\left( x-\cfrac { 1 }{ y } \right) dy=0$$. If $$y(A)=1$$, then $$x$$ is given by

A
$$1-\cfrac { 1 }{ y } +\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$

B
$$4-\cfrac { 2 }{ y } -\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$

C
$$3-\cfrac { 1 }{ y } +\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$

D
$$1+\cfrac { 1 }{ y } -\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$
Question 10
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The solution of $$e^{2x-3y}dx+e^{2y-3x}dy=0$$ is

A
$$e^{5x}+e^{5y}=C$$

B
$$e^{5x}-e^{5y}=C$$

C
$$e^{5x+5y}=C$$

D
None of these

Solution

$${ e }^{ 2x-3y }dx + { e }^{ 2x-3y }dy=0\\ \Rightarrow \cfrac { { e }^{ 2x } }{ { e }^{ 3y } } dx+\cfrac { { e }^{ 2y } }{ { e }^{ 3x } } dy=0 \\ \Rightarrow \cfrac { { e }^{ 5x }dx + { e }^{ 5y }dy }{ { e }^{ 3y }{ e }^{ 3x } } =0\\ \Rightarrow { e }^{ 5x }dx + { e }^{ 5y }dy=0$$
Integrating n both sides.
$$\cfrac { { e }^{ 5x } }{ 5 } +\cfrac { { e }^{ 5y } }{ 5 } =k\\ \Rightarrow { e }^{ 5x }+{ e }^{ 5y }=k$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 51

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Differential Equations Test - 51

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SHARING IS CARING

Question 1

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{2}{3}$$

$$\displaystyle - \frac{1}{3}$$

$$\displaystyle 1$$

Solution

Question 2

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

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

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

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

Solution

Question 3

$${ e }^{ y }=2{ e }^{ x }+\dfrac { { x }^{ 3 } }{ 3 } +C$$

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

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

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

Solution

Question 4

$$y^2 = 2a(2b\sin x + \cos x) + ce^{-2bx}$$

$$(4b^2+1)y^2 = 2a(\sin x + 2b \cos x)+ ce^{-2bx}$$

$$(4b^2+1)y^2 = 2a(\sin x + 2b \cos x)+ ce^{2bx}$$

$$y^2 = 2a(2b\sin x + \cos x) + ce^{2bx}$$

Solution

Question 5

$$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{2}{3}x$$

$$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{4}{5}$$

$$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-2x^2+\dfrac{4}{9}$$

$$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{2}{3}x^2+\dfrac{4}{9}$$

Solution

Question 6

$$y=x+\frac{c}{x}$$

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

$$y=x-\frac{c}{x}$$

$$y=x^2-\frac{c}{x^2}$$

Solution

Question 7

a family of circles with centre on y-axis

a family of circles with centre at origin

a family of circles with given radius

a family of circles with centre on x-axis

Solution

Question 8

$$3\ln 5$$

$$\ln 5$$

$$2\ln 5$$

None of these

Solution

Question 9

$$1-\cfrac { 1 }{ y } +\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$

$$4-\cfrac { 2 }{ y } -\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$

$$3-\cfrac { 1 }{ y } +\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$

$$1+\cfrac { 1 }{ y } -\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$

Question 10

$$e^{5x}+e^{5y}=C$$

$$e^{5x}-e^{5y}=C$$

$$e^{5x+5y}=C$$

None of these

Solution

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