Differential Equations Test

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Differential Equations Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

The differential equation $$\dfrac{dy}{dx} = 2$$ represents

A
A straight line of slope $$2$$ units

B
A circle with radius $$2$$ units

C
A parabola with foci $$2$$ units

D
None of these

Solution

$$\dfrac{dy}{dy}=2$$
$$\Rightarrow dy=2dx$$
Integrating on both sides
$$\int dy=\int 2dx$$
$$\Rightarrow y= 2x+c$$
straight line of slope $$2$$.
Question 2
1 / -0

The solution of the differential equation $$\left( 1+y^{ { 2 } } \right) +\left( x-e^{ { { \tan }^{ -1 }y } } \right) \dfrac { dy }{ dx } =0,$$ is

A
$$( x - 2 ) = k e ^ { - \tan ^ { - 1 } y }$$

B
$$2 x e ^ { 2 \tan ^ { - 1 } y } = e ^ { 2 \tan ^ { - 1 } y } = k$$

C
$$x e ^ { \tan ^ { - 1 } y } = \tan ^ { - 1 } y + k$$

D
$$xe^{ { 2\tan ^{ { -1 } } y } }=e^{ { \tan ^{ { -1 } } y } }+k$$

Solution

$$\begin{array}{l} \left( { 1+{ y^{ 2 } } } \right) \frac { { dx } }{ { dy } } ={ e^{ { { \tan }^{ -1y } } } }-x \\ \frac { { dx } }{ { dy } } +\frac{x}{\left( { 1+{ y^{ 2 } } } \right)} =\frac { { { e^{ { { \tan }^{ -1y } } } } } }{ { \left( { 1+{ y^{ 2 } } } \right) } } \end{array}$$
Its a linear differential equation with $$I.F. = {e^{{{\tan }^{ - 1}}y}}$$
So,
We would be $$x.{e^{{{\tan }^{ - 1}}y}}$$=integral of $$\frac{{{e^{{{\tan }^{ - 1}}y}}.{e^{{{\tan }^{ - 1}}y}}}}{{\left( {1 + {y^2}} \right)}}$$
Put $${\tan ^{ - 1y}} = t$$
Question 3
1 / -0

The solution of $$\dfrac{dy}{dx}=\dfrac{x+x^{2}}{y-y^{2}}$$ is:

A
$$3(y^{2}-x^{2}{})=2(x^{3}+y^{3})+c$$

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

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

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

Solution

$$\dfrac{dy}{dx}=\dfrac{x+x^{2}}{y-y^{2}}$$
$$\int (y-y^{2})dy=\int (x+x^{2})dx+c$$
$$\dfrac{y^{2}}{2}-\dfrac{y^{3}}{3}=\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+c$$
$$\Rightarrow 3(y^{2}-x^{2})=2(x^{3}+y^{3})+c$$
Question 4
1 / -0

The solution of $$\dfrac{d^{2}y}{dx^{2}}=0$$ represents

A
a straight line

B
a circle

C
a parabola

D
a point

Solution

$$\dfrac{d}{dx}y_{1}=0$$
$$\Rightarrow \int dy_{1}=\int 0.dx+c_{1}$$
$$\Rightarrow \int dy_{1}=c_{1}$$
$$\Rightarrow y_{1}=c_{1}$$
$$\dfrac{dy}{dx}=c_{1}$$
$$\int dy=\int c_{1}\ dx+c_{2}$$
$$\Rightarrow y=c_{1}x+c_{2}\ straight\ line$$
Question 5
1 / -0

The solution of $$\dfrac{dy}{dx}= \tan\ y$$ is

A
$$\log \sin x = 2x + c$$

B
$$\log \sin y = x + c$$

C
$$\log \sin y = 2x + c$$

D
$$\log \sin x = 2 y + c$$

Solution

$$\dfrac{dy}{dx}=\tan y$$
$$\int \cot\ y\ dy=\int dx+c$$
$$\log\ \sin y=x+c$$
Question 6
1 / -0

The differential equation $$y\dfrac{dy}{dx}+x=A$$ where A is a constant represents a set of:

A
circles centred on y axis

B
circles centred on x axis

C
parabolas

D
ellipses

Solution

$$y\ \dfrac{dy}{dx}= A-x$$
$$\Rightarrow y\ dy= (A-x)dx$$
$$\Rightarrow \dfrac{y^{2}}{2}= Ax-\dfrac{x^{2}}{2}+c$$
$$\dfrac{y^{2}}{2}= \dfrac{2Ax-x^{2}+c}{2}$$
$$y^{2}= -(x^{2}-2Ax+A^{2})+A^{2}+c$$
$$\Rightarrow (x-A)^{2}+y^{2}= A^{2}+c$$
$$\Rightarrow $$ center of circle = (A,0)
Question 7
1 / -0

The solution of $$\displaystyle \frac{dy}{dx}+y \tan x = 0$$ is:

A
$$y = a \cos x$$

B
$$y = a \sin x$$

C
$$y = \log \cos x + c$$

D
$$y = a \tan x + c$$

Solution

$$\dfrac{dy}{dx}+y\ \tan\ x=0$$
$$\int \dfrac{dy}{y}=-\int \tan\ x\ dx+c$$
$$\log\ y=\log\ c\ \cos\ x$$
$$\Rightarrow y=c\ \cos\ x$$
Question 8
1 / -0

The solution of $$\frac{dy}{dx}=e^{x-y}$$ is:

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

$$\dfrac{dy}{dx}=\dfrac{e^{x}}{e^{y}}$$
$$\Rightarrow \int e^{y}dy=\int e^{x}dx+c_{1}$$
$$e^{y}=e^{x}+c_{1}$$
$$e^{y}-e^{x}=c_{1}$$
$$e^{y}=e^{x}+c_{1}$$
$$c=e^{x}-e^{y}$$
Question 9
1 / -0

The differential equation of the family of circles whose center lies on $$x-$$axis and passing through origin is

A
$$x^{2}+y^{2}+\dfrac{dy}{dx}=0$$

B
$$(y^{2}-x^2)dx-2xydy=0$$

C
$$y^{2}dx+(x^{2}+2xy)dy=0$$

D
$$xdy+ydx+x^{2}dx+y^{2}dy=0$$

Solution

General equation of a circle can be given as
$$x^2+y^2+2gx+2fy+c=0$$
Since it is given that this circle passes through origin and centre lies on $$x-$$axis
$$\therefore$$ Center of circle is $$(-g,0)$$

So, the equation of a circle passing through origin and centre on $$x-$$axis is
$$x^2+y^2+2gx=0$$ ....(1)
$$\Rightarrow g = -\dfrac{x^2+y^2}{2x}$$

Differentiating (1) w.r.t $$x$$
$$2x+2y\dfrac{dy}{dx}+2g=0$$ ....(2)
Substituting $$g$$ in (2), we get
$$\Rightarrow \displaystyle 2x+2y\frac{dy}{dx}-\frac{x^2+y^2}{x}=0$$

$$\Rightarrow \displaystyle 2x+2y\frac{dy}{dx}=\frac{x^2+y^2}{x}$$

$$\Rightarrow \displaystyle 2x^2+2xy\frac{dy}{dx}=x^2+y^2$$
Re-arranging this, we get
$$\Rightarrow (y^{2}-x^2)dx-2xydy=0$$
which is the required differential equation.

Hence, option B.
Question 10
1 / -0

The solution of $$\displaystyle \frac{dy}{dx}+\displaystyle \frac{\sqrt{1+y^{2}}}{\sqrt{1+x^{2}}}=0$$ is:

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

B
$$\tan^{-1}x+\tan^{-1}y=c$$

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

D
$$\cot^{-1}x+\cot^{-1}y=c$$

Solution

$$\dfrac{dy}{\sqrt{1+y^{2}}}= -\dfrac{dx}{\sqrt{1+x^{2}}}$$
Let $$y= \tan\theta$$
$$dy= \sec^{2}\theta \ d\ \theta$$
$$\Rightarrow \int \dfrac{dy}{\sqrt{1+y^{2}}}= \dfrac{\sec^{2}\theta\ d\ \theta }{sec\theta }$$
$$\Rightarrow \int \sec\theta\ d\ \theta = \log| \sec\theta +\tan\theta |$$
$$\Rightarrow \log |\sqrt{1+y^{2}}+y|= -\log |(\sqrt{1+x^{2}}+x)|+\log \ c$$
$$\Rightarrow (y+\sqrt{1+y^{2}})(x+\sqrt{1+x^{2}})= c$$, which can also be written as
$$\displaystyle \sinh^{-1}x +\sinh^{-1}y=c$$
Hence, option 'C' is correct.

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 15

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Differential Equations Test - 15

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SHARING IS CARING

Question 1

A straight line of slope $$2$$ units

A circle with radius $$2$$ units

A parabola with foci $$2$$ units

None of these

Solution

Question 2

$$( x - 2 ) = k e ^ { - \tan ^ { - 1 } y }$$

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

$$x e ^ { \tan ^ { - 1 } y } = \tan ^ { - 1 } y + k$$

$$xe^{ { 2\tan ^{ { -1 } } y } }=e^{ { \tan ^{ { -1 } } y } }+k$$

Solution

Question 3

$$3(y^{2}-x^{2}{})=2(x^{3}+y^{3})+c$$

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

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

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

Solution

Question 4

a straight line

a circle

a parabola

a point

Solution

Question 5

$$\log \sin x = 2x + c$$

$$\log \sin y = x + c$$

$$\log \sin y = 2x + c$$

$$\log \sin x = 2 y + c$$

Solution

Question 6

circles centred on y axis

circles centred on x axis

parabolas

ellipses

Solution

Question 7

$$y = a \cos x$$

$$y = a \sin x$$

$$y = \log \cos x + c$$

$$y = a \tan x + c$$

Solution

Question 8

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

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

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

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

Solution

Question 9

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

$$(y^{2}-x^2)dx-2xydy=0$$

$$y^{2}dx+(x^{2}+2xy)dy=0$$

$$xdy+ydx+x^{2}dx+y^{2}dy=0$$

Solution

Question 10

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

$$\tan^{-1}x+\tan^{-1}y=c$$

$$\sinh^{-1}x+\sinh^{-1}y=c$$

$$\cot^{-1}x+\cot^{-1}y=c$$

Solution

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