Differential Equations Test

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Differential Equations Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Find general solution of $$\sqrt{1 + 4x^{2}} dy = y^{3} xdx$$.

A
$$\displaystyle - \frac{1}{2y^{2}} = \frac{1}{4}\sqrt{1 + 4x^{2}} + k$$

B
$$\displaystyle \frac{1}{2y^{2}} = \frac{1}{4}\sqrt{1 + 4x^{2}} + k$$

C
$$\displaystyle \frac{1}{2y^{2}} = -\frac{1}{4}\sqrt{1 + 4x^{2}} + k$$

D
$$\displaystyle \frac{1}{2y^{2}} = -\frac{1}{4}\sqrt{1 - 4x^{2}} + k$$

Solution

$$\sqrt { 1+4{ x }^{ 2 } } dy={ y }^{ 3 }xdx\\$$
$$ \Rightarrow \cfrac { dy }{ dx } =\cfrac { x{ y }^{ 3 } }{ \sqrt { 1+4{ x }^{ 2 } } } \\ \Rightarrow \cfrac { \cfrac { dy }{ dx } }{ { y }^{ 3 } } =\cfrac { x }{ \sqrt { 1+4{ x }^{ 2 } } } $$

Integrating both sides w.r.t x
$$\displaystyle \int { \cfrac { \frac { dy }{ dx } }{ { y }^{ 3 } } dx } =\displaystyle \int { \cfrac { x }{ \sqrt { 1+4{ x }^{ 2 } } } dx } $$

$$\displaystyle \int { \cfrac { 1 }{ { y }^{ 3 } } dy } =\displaystyle \int { \cfrac { x }{ \sqrt { 1+4{ x }^{ 2 } } } dx } $$

$$ \Rightarrow -\cfrac { 1 }{ 2{ y }^{ 2 } } =\cfrac { 1 }{ 4 } \sqrt { 1+4{ x }^{ 2 } } +c$$
Question 2
1 / -0

The solution of the differential equation $$\displaystyle \left( 1+\cos { x } \right) \frac { dy }{ dx } =1-\cos { x } $$ is

A
$$\displaystyle y=\tan { \frac { x }{ 2 } } +x+c$$

B
$$\displaystyle y=2\tan { \frac { x }{ 2 } } -x+c$$

C
$$\displaystyle y=\tan { \frac { x }{ 2 } } -x+c$$

D
$$\displaystyle y=x-2\tan { \frac { x }{ 2 } } +c$$

Solution

$$\displaystyle \left( 1+\cos { x } \right) \frac { dy }{ dx } =1-\cos { x } $$
$$\displaystyle \Rightarrow \int { dy } =\int { \frac { 1-\cos { x } }{ 1+\cos { x } } dx } $$
$$\displaystyle \Rightarrow \int { dy } =\int { \tan ^{ 2 }{ \frac { x }{ 2 } } dx } +c$$
$$\displaystyle \Rightarrow y=\int { \left( \sec ^{ 2 }{ \frac { x }{ 2 } } -1 \right) dx } +c$$
$$\displaystyle \Rightarrow y=2\tan { \frac { x }{ 2 } } -x+c$$
Question 3
1 / -0

The equation of the curve in which sub-normal varies as the square of the ordinate is ($$k$$ is constant of proportionality)

A
$$ y=Ae^{2kx} $$

B
$$ y=e^{kx} $$

C
$$ y^{2}/2+kx=A $$

D
$$ y^{2}+kx^{2}=A $$

Solution

$$|y\dfrac{dy}{dx}|=ky^{2}$$

$$\dfrac{dy}{dx}=ky$$

$$\dfrac{dy}{y}=k.dx$$

Integrating both sides
$$lny=kx+c$$
$$y=e^{kx}.e^{c}$$
Or
$$y=Ce^{kx}$$ where $$C=e^{c}$$.
Question 4
1 / -0

Solve : $$(\tan y)\displaystyle \frac{dy}{dx} = \sin(x + y) + \sin(x - y)$$

A
$$\sec y = -2 \cos x + C$$

B
$$\sec y = 2 \cos x + C$$

C
$$\sec y = - \cos x + C$$

D
$$\sec y = \cos x + C$$

Solution

Given, $$(\tan y)\displaystyle \frac{dy}{dx} = \sin(x + y) + \sin(x - y)=2\sin x\cos y$$

$$\Rightarrow (\sec y \tan y)dy = \sin x dx$$

$$\Rightarrow d(\sec y) =- 2d(\cos x)$$

Integrating on both sides we get,

$$\sec y = -2\cos x+c$$
Question 5
1 / -0

Find general solution of $$\displaystyle \frac{2dy}{dx} = \frac{y(x + 1)}{x}$$.

A
$$\log y^{2} = x + \log |x| + k$$

B
$$\log y^{2} = x + \log (x) + k$$

C
$$\log y = x + \log (x) + k$$

D
$$2\log y = x - \log |x| + k$$

Solution

Given,
$$2\dfrac { dy }{ dx } =\dfrac { y\left( x+1 \right) }{ x } $$

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

Integrating both sides w.r.t $$x$$

$$\displaystyle \int { \dfrac { 2 dy }{ y } } =\displaystyle \int { \dfrac { x+1 }{ x } dx } $$

$$\displaystyle \int { \dfrac { 2 dy }{ y } } =\displaystyle \int { 1+\dfrac { 1 }{ x } dx } $$

$$ \Rightarrow 2\log { y } =x+\log { |x| } +c$$

$$ \Rightarrow \log { { y }^{ 2 } } =x+\log {| x| } +c$$
Question 6
1 / -0

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

A
$$y\, =\, 2$$

B
$$y\, =\, 2x$$

C
$$y\, =\, 2x\, \, -4$$

D
$$y\, =\, 2x^2\, \,- 4$$

Solution

$$y=-{ \left( \cfrac { dy }{ dx } \right) }^{ 2 }+x\cfrac { dy }{ dx } \\ \Rightarrow \cfrac { dy }{ dx } =x\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } +\cfrac { dy }{ dx } -2\cfrac { dy }{ dx } \cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } \\ \Rightarrow \cfrac { dy }{ dx } =\cfrac { dy }{ dx } +\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } \left( x-2\cfrac { dy }{ dx } \right) \\ \Rightarrow \cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } \left( x-2\cfrac { dy }{ dx } \right) =0$$
$$\Rightarrow \cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0$$ or $$x-2\cfrac { dy }{ dx } =0$$
$$\Rightarrow y=cx-c^{ 2 }$$ or $$y=\cfrac { { x }^{ 2 } }{ 4 } $$
Question 7
1 / -0

Find the equation of the curve for which the normal at any point (x, y) passes through the origin. The curve represents a :

A
ellipse

B
rectange

C
circle

D
hyperbola

Solution

The equation of the normal at any point $$(x,y)$$ of the curve is
$$\cfrac { dy }{ dx } \left( Y-y \right) +\left( X-x \right) =0$$
Since the required curve passes through origin
$$\cfrac { dy }{ dx } \left( 0-y \right) +\left( 0-x \right) =0\Rightarrow 2\left( -y \right) dy+2\left( -x \right) dx=0$$
Integrating $${ x }^{ 2 }+{ y }^{ 2 }=c$$
Question 8
1 / -0

Which one of the following curves represents the solution of the initial value problem $$Dy = 100- y$$, where $$y(0) = 50$$

A

B

C

D

Solution

$$\displaystyle \frac {dy}{dx}\, =\, 100\, -\, y$$
$$\Rightarrow\, \displaystyle \int \frac {dy}{100-y}\, =\int \, dx\, \Rightarrow\, -\, ln\, (100-y)\, =\, x\, +\, c$$
since $$y(0)\, =\, 50\,\, \Rightarrow\, -\, ln\, 50\, =\, c$$
$$\therefore\, -\, ln(100\, -y)\, =\, x\, -\, ln\, 50$$
$$\Rightarrow\,

ln\, \left ( \displaystyle \frac {50}{100-y} \right )\, =\, x\,

\Rightarrow\, \displaystyle \frac {50}{10-y}\, =\, e^x$$
Question 9
1 / -0

Find general solution of $$y-x\, \cfrac {dy}{dx}\, =\, b(1+x^2\cfrac {dy}{dx})$$ is:

A
$$\quad b+kx=y(1+bx)$$

B
$$\quad b+ky=x(1+bx)$$

C
$$\quad b+ky=x(1+by)$$

D
$$\quad b+kx=x(1+by)$$

Solution

$$y-x\, \cfrac{dy}{dx}\, =\, b\left(1+x^2\cfrac{dy}{dx}\right)$$
$$\Rightarrow\quad

y- b = (x+bx^2)\, \displaystyle \frac {dy}{dx}\,

\Rightarrow\, \displaystyle \frac {dx}{x(1+bx)}\, =\, \displaystyle

\frac {dy}{y-b}$$
$$\Rightarrow\quad \int \left ( \displaystyle \frac

{1}{x}\, -\, \displaystyle \frac {b}{1+bx} \right )\, dx\, =\, \log\,

(y-b)\, +\, \log\, c$$
$$\Rightarrow \log x - \log (1 + bx) = \log (y - b) + \log c$$
$$\Rightarrow\quad \displaystyle \frac {x}{c}\, =\, (1+bx)(y-b)$$
$$\Rightarrow\quad b+ \left (\displaystyle \frac {1}{c}\, +\, b^2 \right ) x = y (1+bx)$$
$$\Rightarrow\quad b+kx=y(1+bx)$$
Question 10
1 / -0

The solution to the differential equation $$y\ln y \, +\, xy'\, =\, 0\,$$ where$$\, y(1)\, =\, e$$, is:

A
$$x(\ln y)\, =\, 1$$

B
$$xy(\ln y)\, =\, 1$$

C
$$(\ln y )^2\, =\, 2$$

D
$$\ln y\, +\, \left ( \displaystyle \frac {x^2}{2} \right )\, y\, =\, 1$$

Solution

$$y\ln { y } +xy'=0\\ \Rightarrow \cfrac { dy }{ dx } =-\cfrac { \left( \log { y } \right) y }{ x } \\ \Rightarrow \cfrac { dy }{ \left( \log { y } \right) y } =-\cfrac { 1 }{ x } $$
Integrating both sides w.r.t x
$$\int { \cfrac { dy }{ \left( \log { y } \right) y }} =\int { -\cfrac { 1 }{ x } dx } \\ \Rightarrow \log { \log { y } } =-\log { x } +c\\ \Rightarrow y={ e }^{ c/x }$$
Now $$y\left( 1 \right) =e\Rightarrow x\ln { y } =1$$

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Differential Equations Test - 27

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Differential Equations Test - 27

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Question 1

$$\displaystyle - \frac{1}{2y^{2}} = \frac{1}{4}\sqrt{1 + 4x^{2}} + k$$

$$\displaystyle \frac{1}{2y^{2}} = \frac{1}{4}\sqrt{1 + 4x^{2}} + k$$

$$\displaystyle \frac{1}{2y^{2}} = -\frac{1}{4}\sqrt{1 + 4x^{2}} + k$$

$$\displaystyle \frac{1}{2y^{2}} = -\frac{1}{4}\sqrt{1 - 4x^{2}} + k$$

Solution

Question 2

$$\displaystyle y=\tan { \frac { x }{ 2 } } +x+c$$

$$\displaystyle y=2\tan { \frac { x }{ 2 } } -x+c$$

$$\displaystyle y=\tan { \frac { x }{ 2 } } -x+c$$

$$\displaystyle y=x-2\tan { \frac { x }{ 2 } } +c$$

Solution

Question 3

$$ y=Ae^{2kx} $$

$$ y=e^{kx} $$

$$ y^{2}/2+kx=A $$

$$ y^{2}+kx^{2}=A $$

Solution

Question 4

$$\sec y = -2 \cos x + C$$

$$\sec y = 2 \cos x + C$$

$$\sec y = - \cos x + C$$

$$\sec y = \cos x + C$$

Solution

Question 5

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

$$\log y^{2} = x + \log (x) + k$$

$$\log y = x + \log (x) + k$$

$$2\log y = x - \log |x| + k$$

Solution

Question 6

$$y\, =\, 2$$

$$y\, =\, 2x$$

$$y\, =\, 2x\, \, -4$$

$$y\, =\, 2x^2\, \,- 4$$

Solution

Question 7

ellipse

rectange

circle

hyperbola

Solution

Question 8

Solution

Question 9

$$\quad b+kx=y(1+bx)$$

$$\quad b+ky=x(1+bx)$$

$$\quad b+ky=x(1+by)$$

$$\quad b+kx=x(1+by)$$

Solution

Question 10

$$x(\ln y)\, =\, 1$$

$$xy(\ln y)\, =\, 1$$

$$(\ln y )^2\, =\, 2$$

$$\ln y\, +\, \left ( \displaystyle \frac {x^2}{2} \right )\, y\, =\, 1$$

Solution

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