Differential Equations Test

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Differential Equations Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The solution of equation $$\dfrac{dy}{dx}=\dfrac{ax+b}{cy+d}$$ represents :

A
A straight line if $$a=c=0$$ and $$b,d \neq 0$$

B
A parabola if $$a=2, c=0, d \neq 0$$

C
A huberbola if $$b=d=0, a$$ and $$c\neq 0$$

D
A rectangular hyberbola of $$b=d=0, a=c=2$$

Solution

Given Differential equation is $$\cfrac { dy }{ dx } =\cfrac { ax+b }{ cy+d } $$,

Separating the variables and integrating gives,

$$\int { (cy+d)dy } =\int { (ax+b)dx } $$,

$$\Longrightarrow \cfrac { c{ y }^{ 2 } }{ 2 } +dy=\cfrac { a{ x }^{ 2 } }{ 2 } +bx+K$$,

Where $$K$$ is the integration constant,

If $$a=c=0$$ and $$b,d\neq 0$$, the equation becomes,

$$dy-bx=K$$ which representss a straight line.
Question 2
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The solution of the differential equation $$3xy'-3y+{ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 1/2 }=0$$, satisfying the condition $$y(1)=1$$ is

A
$$3\cos ^{ -1 }{ \left( \cfrac { y }{ x } \right) } =\ln { \left| x \right| } $$

B
$$3\cos { \left( \cfrac { y }{ x } \right) } =\ln { \left| x \right| } $$

C
$$3\cos ^{ -1 }{ \left( \cfrac { y }{ x } \right) } =2\ln { \left| x \right| } $$

D
$$3\sin ^{ -1 }{ \left( \cfrac { y }{ x } \right) } =\ln { \left| x \right| } $$

Solution

The differential equation is
$$3x\dfrac{dy}{dx}-3y+(x^{2}-(y)^2)^{\frac{1}{2}}=0$$
$$3xdy-3ydx+(x^{2}-(y)^2)^{\frac{1}{2}}dx=0$$

Let $$y=vx$$
$$\implies dy=vdx+xdv$$
Substituting these values in the differential equation, we get
$$3x(vdx+xdv)-3vxdx+(x^{2}-(vx)^2)^{\frac{1}{2}}dx=0$$
$$3x(vdx+xdv)-3vxdx+x(1-v^2)^{\frac{1}{2}}dx=0$$
$$3x^{2}dv+x(1-v^2)^{\frac{1}{2}}dx=0$$
$$3xdv+(1-v^2)^{\frac{1}{2}}dx=0$$

$$-\dfrac{3dv}{(1-v^2)^{\frac{1}{2}}}=\dfrac{dx}{x}$$

Integrating the above equation, we get

$$3\int -\dfrac{dv}{(1-v^2)^{\frac{1}{2}}}=\int \dfrac{dx}{x}$$

$$3cos^{-1}v=ln|x| +c$$

$$3cos^{-1}(\dfrac{y}{x})=ln|x| +c$$
Using the goundary condition y(1)=1, we get
$$3cos^{-1}1=ln|1|+c$$
$$c=0$$
$$\implies3cos^{-1}\dfrac{y}{x}=ln|x|$$ is the solution of the differential equation.

Hence, the answer is option (A).
Question 3
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The solution of differential equation $$x \dfrac {dy}{dx} + y=y^2$$ is:

A
$$y=1+ cxy$$

B
$$y=\ln(cxy)$$

C
$$y+1=cxy$$

D
$$y=c+xy$$

Solution

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

$$\therefore \dfrac{dy}{y^2-y}=\dfrac{dx}{x}$$

$$\therefore \dfrac{dy}{y-1}-\dfrac{dy}{y}=\dfrac{dx}{x}$$

$$\therefore \ln(y-1)-\ln y=\ln x+C^{'}$$

$$\therefore \ln\left(\dfrac{y-1}{yx}\right)=C^{'}$$

$$\therefore \dfrac{y-1}{yx}=e^{C^{'}}$$

$$\therefore \dfrac{y-1}{yx}=c$$ ........ $$[e^{C^{'}}=c]$$

$$\therefore y=1+cxy$$
Question 4
1 / -0

The solution of the differential equation $$\left( x+3{ y }^{ 2 } \right) \dfrac { dy }{ dx } =y,\ y>0$$ is

A
$$\dfrac { x }{ y } =3y+c$$

B
$$x=2{ y }^{ 3 }+3{ y }^{ 2 }+c$$

C
$$y=3{ x }^{ 2 }+c$$

D
$$y=3x+c$$

Solution

Given,

$$(x + 3y^2) \dfrac{dy}{dx} = y , y > 0$$

we can rewrite it as,

$$x + 3y^2 = y \dfrac{dx}{dy}$$

$$\Rightarrow \dfrac{dx}{dy} = \dfrac{x}{y} + 3y$$

$$\Rightarrow \dfrac{dx}{dy} - \dfrac{1}{y} x = 3y$$

As it is a linear equation of the form

$$\dfrac{dx}{dy} + px = Q$$

Here,
$$P = \dfrac{-1}{y} , Q = 3y$$

$$I. F = e^{\displaystyle\int p dy} = e^{\displaystyle-\int \dfrac{1}{y} dy}$$

$$= e^{-\ln \ y} = e^{\ln \ y^{-1}}$$

$$= y^{-1} = \dfrac{1}{y}$$

$$\therefore \dfrac{1}{y} . x = \displaystyle \int 3y \times \dfrac{1}{y} dy + c$$

$$\Rightarrow \dfrac{x}{y} = 3 \displaystyle \int dy + c$$

$$= \dfrac{x}{y} = 3y + c$$
Question 5
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The solution of the differential equation $$y'=\cfrac { 1 }{ { e }^{ -y }-x } $$, is

A
$$x={ e }^{ -y }\left( y+c \right) $$

B
$$y+{ e }^{ -y }=x+c$$

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

D
$$x+y={ e }^{ -y }+c$$

Solution

$$y^{'}=\dfrac{1}{e^{-y}-x}$$

$$y^{'}e^{-y}-y^{'}x=1$$

$$y^{'}e^{-y}=y^{'}x+1$$

$$y^{'}=e^{y}y^{'}x+e^{y}$$

$$\dfrac{dy}{dx}=e^{y}\dfrac{dy}{dx}x+e^{y}$$

$$\dfrac{dy}{dx}=\dfrac{d(xe^{y})}{dx}$$

Integrating the above equation, we get
$$y=xe^{y}-c$$
$$xe^{y}=y+c$$
$$x=e^{-y}(y+c)$$

Hence, the answer is option (A).
Question 6
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the general solution of differential equation $$x^4 \, \frac{dy}{dx} \, + \, x^3 y \, + cosec \, xy \, =0$$, is

A
2 cos ( x y ) + $$\dfrac{1}{x^2}$$ = C

B
2 cos ( x y ) + $$\dfrac{1}{y^2}$$ = C

C
2 sin y + $$\dfrac{1}{x^2}$$ = C

D
2 sin ( x y ) + $$\dfrac{1}{y^2}$$ = C

Solution

$$x^4dy \, + \, x^3 \, ydx \, + \, cosec (xy) \, dx \, = \, 0$$
$$x^3(x \, dx \, + \,y \, dx )+ \, + \, cosec (xy) \, dx \, = \, 0$$
$$\therefore \int \dfrac{d (xy)}{cosec (xy)} \, + \,\int \dfrac{dx}{x^3} \, = \, 0$$
$$-cos \, xy \, + \, \dfrac{x^{-3+1}}{-3 + 1} \, = \, C$$
$$\therefore \,2\, cos\,(xy)\,+\, \dfrac{1}{x^2 }\,=\,C$$
Question 7
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The differential equation $$\dfrac{dy}{dx} = e^x.e^y$$ has solution ____________

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

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

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

D
$$e^{-x} + e^{-y} = C$$

Solution

$$\dfrac{dy}{dx}=e^x.e^y$$
$$\implies e^{-y}dy=e^xdx$$
Integrating both sides-
$$\int e^{-y}dy=\int e^xdx$$
$$-e^{-y}+C=e^x$$
$$\implies e^x+e^{-y}=C$$
Question 8
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The particular solution of differential equation $$\cfrac { dy }{ dx } =-4x{ y }^{ 2 },y(0)=1$$ is ______

A
$$y=\left( 2{ x }^{ 2 }+1 \right) =1$$

B
$${ x }^{ 2 }=\cfrac { 1 }{ { y }^{ 2 } } $$

C
$$y={ x }^{ 2 }+\log { x } $$

D
$$4{ e }^{ x }+\cfrac { 1 }{ { y }^{ 2 } } =8\quad $$

Solution

$$y={ \left( 2{ x }^{ 2 }+1 \right) }^{ -1 }\quad $$
$$\cfrac { dy }{ dx } =-4x{ y }^{ 2 }$$
$$\therefore \cfrac { dy }{ { y }^{ 2 } } =-4xdx$$
$$\therefore \int { { y }^{ -2 } } dy=-4\int { x } dx+c\quad $$
$$\therefore \cfrac { { y }^{ -1 } }{ -1 } =-4\left( \cfrac { { x }^{ 2 } }{ 2 } \right) +c$$
$$\therefore \cfrac { -1 }{ y } =-2{ x }^{ 2 }+c$$
Take $$x=0$$ and $$y=1$$
$$ \therefore -1=-2(0)+c$$
$$\therefore c=-1$$
$$\therefore \cfrac { -1 }{ y } =-2{ x }^{ 2 }-1$$
$$\therefore \cfrac { 1 }{ y } =1+2{ x }^{ 2 }$$
$$\therefore y={ \left( 1+2{ x }^{ 2 } \right) }^{ -1 }$$
Question 9
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The order of the differential equation whose general solution is $$y \, = \, c, \, cos \, 2x \, + \, c_2 \, cos^2x \, + \, c_3 \, sin^2x \, + \, c_4$$

A
$$2$$

B
$$4$$

C
$$3$$

D
None of these

Solution

$$y=c_1\cos2x+c_2\cos^2x+c_3sin^2x+c_4$$
$$=c_1(2\cos^2x-1)+$$
$$=(2c_1+c_2)\cos^2x+c_3\sin^2x +(c_4-c_1)$$
$$=t_1\cos^2x+c_3\sin^2x+t_2$$
So order will be $$3$$ Because of three unknown present
Question 10
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The solution of the differential equation $$2x \dfrac{dy}{dx} = y; y(1) = 2$$ represents $$=$$ ____.

A
parabola

B
ellipse

C
circle

D
line

Solution

$$2x\dfrac{dy}{dx}=y$$
$$\implies 2\dfrac{dy}{y}=\dfrac{dx}{x}$$
Integrating both sides-
$$2\log y=\log x+c$$
For $$y(1)=2-$$
$$2\log 2=\log 1+c\implies c=2\log 2$$
Hence,
$$2\log y=\log x+2\log 2$$
$$\implies \log y^2=\log {4x}$$
$$\implies y^2=4x$$
This is the equation of parabola.

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 34

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Differential Equations Test - 34

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SHARING IS CARING

Question 1

A straight line if $$a=c=0$$ and $$b,d \neq 0$$

A parabola if $$a=2, c=0, d \neq 0$$

A huberbola if $$b=d=0, a$$ and $$c\neq 0$$

A rectangular hyberbola of $$b=d=0, a=c=2$$

Solution

Question 2

$$3\cos ^{ -1 }{ \left( \cfrac { y }{ x } \right) } =\ln { \left| x \right| } $$

$$3\cos { \left( \cfrac { y }{ x } \right) } =\ln { \left| x \right| } $$

$$3\cos ^{ -1 }{ \left( \cfrac { y }{ x } \right) } =2\ln { \left| x \right| } $$

$$3\sin ^{ -1 }{ \left( \cfrac { y }{ x } \right) } =\ln { \left| x \right| } $$

Solution

Question 3

$$y=1+ cxy$$

$$y=\ln(cxy)$$

$$y+1=cxy$$

$$y=c+xy$$

Solution

Question 4

$$\dfrac { x }{ y } =3y+c$$

$$x=2{ y }^{ 3 }+3{ y }^{ 2 }+c$$

$$y=3{ x }^{ 2 }+c$$

$$y=3x+c$$

Solution

Question 5

$$x={ e }^{ -y }\left( y+c \right) $$

$$y+{ e }^{ -y }=x+c$$

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

$$x+y={ e }^{ -y }+c$$

Solution

Question 6

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

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

2 sin y + $$\dfrac{1}{x^2}$$ = C

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

Solution

Question 7

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

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

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

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

Solution

Question 8

$$y=\left( 2{ x }^{ 2 }+1 \right) =1$$

$${ x }^{ 2 }=\cfrac { 1 }{ { y }^{ 2 } } $$

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

$$4{ e }^{ x }+\cfrac { 1 }{ { y }^{ 2 } } =8\quad $$

Solution

Question 9

$$2$$

$$4$$

$$3$$

None of these

Solution

Question 10

parabola

ellipse

circle

line

Solution

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