Differential Equations Test

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Differential Equations Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

What is the curve which passes through the point (1, 1) and whose slope is $$\dfrac{2y}{x} $$?

A
Circle

B
Parabola

C
Ellipse

D
Hyperbola

Solution

Let the curve be $$y=f\left( x \right) $$,
$$\therefore$$ slope of the tangent drawn at any point on the curve is $$\dfrac { df\left( x \right) }{ dx } =f^{ ' }\left( x \right) $$,
Given that slope at any point on the curve is $$\dfrac { 2y }{ x } $$,
$$\Longrightarrow \dfrac { dy }{ dx } =\dfrac { 2y }{ x } \\ \Longrightarrow \int { \dfrac { 1 }{ y } } dy=\int { \dfrac { 2 }{ x } } dx$$
$$\Longrightarrow \ln { y } =2\ln { x } +c$$
Where $$c$$ is the integration constant,
Given that the curve passes through the point $$(1,1)$$,
$$\Longrightarrow c=0$$,
$$\therefore y={ x }^{ 2 }$$ is the equation of the curve which is a parabola.
Question 2
1 / -0

If $$x dy = y dx + y^2 dy, y > 0$$ and y(1) = 1, then what is y(-3) equal to?

A
3 only

B
-1 only

C
Both -1 and 3

D
Neither -1 nor 3

Solution

Given Differential equation is $$xdy=ydx+{ y }^{ 2 }dy$$,
$$\Longrightarrow \dfrac { dx }{ dy } =\dfrac { x }{ y } -y$$,
$$\Longrightarrow \dfrac { dx }{ dy } -\dfrac { x }{ y } =-y$$ this is a linear differential equation in $$y$$,
The integrating factor is $$\dfrac { 1 }{ y } $$,
Now the differential equation becomes,
$$\dfrac { 1 }{ y } \dfrac { dx }{ dy } -\dfrac { x }{ { y }^{ 2 } } =-1$$,
$$\Longrightarrow \dfrac { x }{ y } =\int { -1dy } \\ \Longrightarrow \dfrac { x }{ y } =-y+c\\ $$,
Given that $$y(1)=1$$,
$$c=2$$,
to find $$y(-3)$$,
$$\Longrightarrow { y }^{ 2 }-2y-3=0$$,
Solving the quadratic equation gives,
$$y=-1\ and\ y=3$$,
Given that $$y>0$$,
$$\therefore y=3\ only$$
Question 3
1 / -0

Solution of the differential equation $$\dfrac { dx }{ x } +\dfrac { dy }{ y } =0$$ is

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

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

C
$$xy=c$$

D
$$x+y=c$$

Solution

$$\dfrac { dx }{ x } +\dfrac { dy }{ y } =0$$
Integrating , we get
$$ \Rightarrow \int { \dfrac { dx }{ x } } +\int { \dfrac { dy }{ y } } =0$$
$$\Rightarrow \log(x)+\log(y)=k$$
$$\Rightarrow xy={ e }^{ k }=c$$
$$\Rightarrow xy=c$$
Hence, option C is correct.
Question 4
1 / -0

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

A
$$y=2$$

B
$$y=2x$$

C
$$y=2x-4$$

D
$$y=2{ x }^{ 2 }-4$$

Solution

Let $$p = \dfrac{dy}{dx}$$.
Therefore, the given equation becomes
$$p^2-xp+y=0\implies y = px-p^2$$
On differentiating wrt x, we get
$$\dfrac{dy}{dx} = p + x\dfrac{dp}{dx}-2p\dfrac{dp}{dx} $$
$$\Rightarrow p = p + x\dfrac{dp}{dx}-2p\dfrac{dp}{dx} $$
$$\Rightarrow \dfrac{dp}{dx}(x-2p) = 0 $$
$$\Rightarrow \dfrac{dp}{dx} = 0 $$ or $$x-2p = 0 $$
Now, $$x-2p = 0$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{2}$$
$$\Rightarrow y=\dfrac{x^2}{4}+c$$
Also from the other equality,
$$\dfrac{dp}{dx} = 0 \Rightarrow p = k $$ (constant)
Substituting $$p=k$$ in the given equation, we get
$$y = kx-k^2$$
On comparing the given options with the obtained solutions
$$y = kx-k^2$$ and $$y = \dfrac{x^2}{4}+c$$
We get that the only possible solution is $$y=2x-4$$.
Question 5
1 / -0

The solution of differential equation
$$4xy\cfrac { dy }{ dx } =\cfrac { 3{ \left( 1+x \right) }^{ 2 }\left( 1+{ y }^{ 2 } \right) }{ \left( 1+{ x }^{ 2 } \right) } $$ is

A
$$\log { (1+y) } =\log { x } +2\tan { x } +constant$$

B
$$\log { \left( 1+{ y }^{ 2 } \right) } =3\log { \left( \cfrac { 1 }{ x } \right) } +6\tan ^{ -1 }{ x } +constant$$

C
$$2\log { \left( 1+{ y }^{ 2 } \right) } =3\log { x } +6\tan ^{ -1 }{ x } +constant$$

D
None of the above

Solution

Given, $$4xy\cfrac { dy }{ dx } =\cfrac { 3{ \left( 1+x \right) }^{ 2 }\left( 1+{ y }^{ 2 } \right) }{ \left( 1+{ x }^{ 2 } \right) } $$

$$\Rightarrow \cfrac { 4ydy }{ 1+{ y }^{ 2 } } =\cfrac { 3{ \left( 1+x \right) }^{ 2 } }{ x({ 1+x }^{ 2 }) } dx\quad $$

$$\Rightarrow \cfrac { 4ydy }{ 1+{ y }^{ 2 } } =\left( \cfrac { 3 }{ x } +\cfrac { 6 }{ { 1+x }^{ 2 } } \right) dx$$

Integrating both sides

$$\displaystyle \Rightarrow \int\cfrac { 4ydy }{ 1+{ y }^{ 2 } } =\int \left( \cfrac { 3 }{ x } +\cfrac { 6 }{ { 1+x }^{ 2 } } \right) dx$$

$$\Rightarrow 2\log { \left( 1+{ y }^{ 2 } \right) } =3\log { x } +6\tan ^{ -1 }{ x } +C\quad $$
Question 6
1 / -0

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

A
$$\log { \tan { \left( \dfrac { y }{ 2 } \right) } } =C-2\sin { x } $$

B
$$\log { \tan { \left( \dfrac { y }{ 4 } \right) } } =C-2\sin { \left( \dfrac { x }{ 2 } \right) } $$

C
$$\log { \tan { \left( \dfrac { y }{ 2 } +\dfrac { \pi }{ 4 } \right) } } =C-2\sin { x } $$

D
$$\log { \tan { \left( \dfrac { y }{ 2 } +\dfrac { \pi }{ 4 } \right) } } =C-2\sin { \left( \dfrac { x }{ 2 } \right) } $$

Solution

Given differential equation can be rewritten as
$$\dfrac { dy }{ dx } =\sin { \left( \dfrac { x-y }{ 2 } \right) } -\sin { \left( \dfrac { x+y }{ 2 } \right) } $$
$$=2\cos { \left( \dfrac { x }{ 2 } \right) } \sin { \left( \dfrac { -y }{ 2 } \right) } $$
$$=-2\cos { \left( \dfrac { x }{ 2 } \right) } \sin { \left( \dfrac { y }{ 2 } \right) } $$
$$\Rightarrow \displaystyle\int { \dfrac { dy }{ 2\sin { \left( \dfrac { y }{ 2 } \right) } } } =-\displaystyle\int { \cos { \left( \dfrac { x }{ 2 } \right) } dx }$$
$$\Rightarrow \dfrac { 1 }{ 2 } \displaystyle\int { \csc { \left( \dfrac { y }{ 2 } \right) } dy } =-\dfrac { \sin { \left( \dfrac { x }{ 2 } \right) } }{ \left( \dfrac { 1 }{ 2 } \right) } +C$$
$$\Rightarrow \dfrac { 1 }{ 2 } \dfrac { \log { \left( \csc { \left( \dfrac { y }{ 2 } \right) } -\cot { \left( \dfrac { y }{ 2 } \right) } \right) } }{ \left( \dfrac { 1 }{ 2 } \right) } =-\dfrac { \sin { \left( \dfrac { x }{ 2 } \right) } }{ \left( \dfrac { 1 }{ 2 } \right) } +C$$
$$\Rightarrow \log { \tan { \left( \dfrac { y }{ 4 } \right) } } =-2\sin { \left( \dfrac { x }{ 2 } \right) } +C$$
Question 7
1 / -0

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

A
$$ \tan^{-1} x + \cos^{-1} x = C $$

B
$$ \sin^{-1} x + \sin^{-1} y = C $$

C
$$ \sec^{-1} x + \text{cosec} ^{-1} x = C $$

D
None of the above.

Solution

Given, equation is $$ \dfrac {dy}{dx} + \sqrt{ \left( \dfrac {1-y^2}{1-x^2} \right) } = 0 $$
Dividing the above equation by $$\sqrt{1-y^{2}}$$ on both sides

$$\implies \dfrac {dy}{\sqrt{1-y^2}} +\dfrac{1}{\sqrt{1 - x^2}} dx = 0 $$
On integrating, we get
$$\displaystyle \int \dfrac {dy}{\sqrt{1-y^2}} + \int \dfrac{1}{\sqrt{1 - x^2}} dx= 0 $$
$$\implies \sin^{-1} y + \sin^{-1} x = C $$
Question 8
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The solution of the differential equation $$\dfrac { dy }{ dx } ={ e }^{ x-y }\left( { e }^{ x }-{ e }^{ y } \right) $$ is

A
$${ e }^{ y }=\left( { e }^{ x }+1 \right) +C{ e }^{ -{ e }^{ x } }$$

B
$${ e }^{ y }=\left( { e }^{ x }-1 \right) +C$$

C
$${ e }^{ y }=\left( { e }^{ x }-1 \right) +C{ e }^{ -{ e }^{ x } }$$

D
None of the above

Solution

Given equation can be rewritten as
$$\dfrac { dy }{ dx } =\dfrac { { e }^{ x } }{ { e }^{ y } } \left( { e }^{ x }-{ e }^{ y } \right) $$
$$\Rightarrow { e }^{ y }\dfrac { dy }{ dx } ={ e }^{ 2x }-{ e }^{ x }{ e }^{ y }$$
$$\Rightarrow { e }^{ y }\dfrac { dy }{ dx } +{ e }^{ x }{ e }^{ y }={ e }^{ 2x }$$
Put $${ e }^{ y }=t\Rightarrow { e }^{ y }\dfrac { dy }{ dx } =\dfrac { dt }{ dx }$$
$$ \therefore \dfrac { dt }{ dx } +{ e }^{ x }t={ e }^{ 2x }$$
On comparing with $$\dfrac { dt }{ dx } +Pt=Q$$, we get
$$P={ e }^{ x }$$ and $$Q={ e }^{ 2x }$$
$$\therefore IF={ e }^{ \int { Pdx } }={ e }^{ \int { { e }^{ x }dx } }={ e }^{ { e }^{ x } }$$
Required solution is
$$t\cdot { e }^{ { e }^{ x } }=\displaystyle\int { { e }^{ 2x }{ e }^{ { e }^{ x } }dx } +C$$
$$\Rightarrow { e }^{ y }{ e }^{ { e }^{ x } }=\left( { e }^{ x }-1 \right) { e }^{ { e }^{ x } }+C$$
$$\Rightarrow { e }^{ y }=\left( { e }^{ x }-1 \right) +C{ e }^{ -{ e }^{ x } }$$
Question 9
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The solution of the differential equation $$\displaystyle (x^2-yx^2)\frac{dy}{dx}+y^2+xy^2=0$$ is?

A
$$\displaystyle \log\left(\frac{x}{y}\right)=\frac{1}{x}+\frac{1}{y}+C$$

B
$$\displaystyle \log\left(\frac{y}{x}\right)=\frac{1}{x}+\frac{1}{y}+C$$

C
$$\displaystyle \log(xy)=\frac{1}{x}+\frac{1}{y}+C$$

D
$$\displaystyle \log(xy)+\frac{1}{x}+\frac{1}{y}=C$$

Solution

The given differential equation is $$x^2(1-y)\displaystyle\frac{dy}{dx}+y^2(1+x)=0$$
$$\Rightarrow x^2(1-y)dy+y^2(1+x)dx=0$$
$$\Rightarrow \displaystyle\frac{1-y}{y^2}dy+\frac{1+x}{x^2}dx=0$$
On integrating, we get
$$-\displaystyle\frac{1}{y}-\log y-\frac{1}{x}+\log x=C$$
$$\Rightarrow \log\left(\displaystyle\frac{x}{y}\right)=\dfrac{1}{x}+\dfrac{1}{y}+C$$.
Question 10
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The solution of $$\dfrac{dy}{dx} = 1+y+y^2+x+xy+xy^2$$ is

A
$$\tan^{-1}\left(\dfrac{2y+1}{\sqrt{3}}\right) = x+x^2+C$$

B
$$4\tan^{-1}\left(\dfrac{2y+1}{\sqrt{3}}\right) = \sqrt{3}(2x+x^2)+C$$

C
$$\sqrt{3}\tan^{-1}\left(\dfrac{3y+1}{3}\right)=4(1+x+x^2)+C$$

D
$$\tan^{-1}\left(\dfrac{2y+1}{3}\right)=4(2x+x^2)+C$$

Solution

Given, $$\dfrac{dy}{dx} = 1+y+y^2+x+xy+xy^2$$
$$\implies \dfrac{dy}{1+y+y^2} = (1+x)dx$$

$$\displaystyle \int\dfrac{dy}{\left(y+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}=\int(1+x)dx$$

$$\dfrac{1}{\sqrt{3}/2} \tan^{-1}\left(\dfrac{y+\dfrac{1}{2}}{\sqrt{3}/2}\right) = x+\dfrac{x^2}{2}+\dfrac{c}{2}$$

$$4\tan^{-1}\left(\dfrac{2y+1}{\sqrt{3}}\right)=\sqrt{3}(2x+x^2)+C$$

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Differential Equations Test - 32

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Differential Equations Test - 32

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SHARING IS CARING

Question 1

Circle

Parabola

Ellipse

Hyperbola

Solution

Question 2

3 only

-1 only

Both -1 and 3

Neither -1 nor 3

Solution

Question 3

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

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

$$xy=c$$

$$x+y=c$$

Solution

Question 4

$$y=2$$

$$y=2x$$

$$y=2x-4$$

$$y=2{ x }^{ 2 }-4$$

Solution

Question 5

$$\log { (1+y) } =\log { x } +2\tan { x } +constant$$

$$\log { \left( 1+{ y }^{ 2 } \right) } =3\log { \left( \cfrac { 1 }{ x } \right) } +6\tan ^{ -1 }{ x } +constant$$

$$2\log { \left( 1+{ y }^{ 2 } \right) } =3\log { x } +6\tan ^{ -1 }{ x } +constant$$

None of the above

Solution

Question 6

$$\log { \tan { \left( \dfrac { y }{ 2 } \right) } } =C-2\sin { x } $$

$$\log { \tan { \left( \dfrac { y }{ 4 } \right) } } =C-2\sin { \left( \dfrac { x }{ 2 } \right) } $$

$$\log { \tan { \left( \dfrac { y }{ 2 } +\dfrac { \pi }{ 4 } \right) } } =C-2\sin { x } $$

$$\log { \tan { \left( \dfrac { y }{ 2 } +\dfrac { \pi }{ 4 } \right) } } =C-2\sin { \left( \dfrac { x }{ 2 } \right) } $$

Solution

Question 7

$$ \tan^{-1} x + \cos^{-1} x = C $$

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

$$ \sec^{-1} x + \text{cosec} ^{-1} x = C $$

None of the above.

Solution

Question 8

$${ e }^{ y }=\left( { e }^{ x }+1 \right) +C{ e }^{ -{ e }^{ x } }$$

$${ e }^{ y }=\left( { e }^{ x }-1 \right) +C$$

$${ e }^{ y }=\left( { e }^{ x }-1 \right) +C{ e }^{ -{ e }^{ x } }$$

None of the above

Solution

Question 9

$$\displaystyle \log\left(\frac{x}{y}\right)=\frac{1}{x}+\frac{1}{y}+C$$

$$\displaystyle \log\left(\frac{y}{x}\right)=\frac{1}{x}+\frac{1}{y}+C$$

$$\displaystyle \log(xy)=\frac{1}{x}+\frac{1}{y}+C$$

$$\displaystyle \log(xy)+\frac{1}{x}+\frac{1}{y}=C$$

Solution

Question 10

$$\tan^{-1}\left(\dfrac{2y+1}{\sqrt{3}}\right) = x+x^2+C$$

$$4\tan^{-1}\left(\dfrac{2y+1}{\sqrt{3}}\right) = \sqrt{3}(2x+x^2)+C$$

$$\sqrt{3}\tan^{-1}\left(\dfrac{3y+1}{3}\right)=4(1+x+x^2)+C$$

$$\tan^{-1}\left(\dfrac{2y+1}{3}\right)=4(2x+x^2)+C$$

Solution

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