Differential Equations Test

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Differential Equations Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

Solution of the equation $$\dfrac {dy}{dx} = \dfrac {y\dfrac {d(\phi (x))}{dx} - y^{2}}{\phi (x)}$$ is

A
$$y = \dfrac {\phi(x) + c}{x}$$

B
$$y = \dfrac {\phi(x)}{x} + c$$

C
$$y = \dfrac {\phi(x)}{x + c}$$

D
$$y = \phi(x) + x + c$$

Solution
Question 2
1 / -0

The solution of the differential equation $$\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{\theta (y/x)}{\theta ' (y/x)}$$ is

A
$$X\theta (y/x)=k$$

B
$$\theta (y/x)=kx$$

C
$$y\theta (y/x)=k$$

D
$$\theta (y/x)=ky$$
Question 3
1 / -0

The solution of the differential equation
$$\left( 2x\cos { y } +3{ x }^{ 2 }y \right) dx+\left( { x }^{ 3 }-{ x }^{ 2 }\sin { y } -y \right) dy=0$$, is given by

A
$${ x }^{ 2 }\cos { y } +{ x }^{ 3 }y-{ y }^{ 2 }/2=c$$

B
$${ x }^{ 2 }\sin { y } +{ x }^{ 3 }y-{ y }^{ 2 }/2=c\quad $$

C
$${ x }^{ 2 }\sin { y } -{ x }^{ 3 }y-{ y }^{ 2 }/2=c$$

D
$${ x }^{ 2 }\cos { y } -{ x }^{ 3 }y+{ y }^{ 2 }/2=c$$
Question 4
1 / -0

The solution of differential equation $$x\cos^{2}y dx = y\cos^{2} x dy$$ is

A
$$x\tan x - y\tan y - \log\left(\dfrac{\sec x}{\sec y}\right) = c$$

B
$$y\tan x - x\tan y - \log (\sec x \cdot \sec y) = c$$

C
$$x\tan x - y\tan y + \log (\sec x\cdot \sec y) = c$$

D
None of the above

Solution

$$x\cos^{2}y dx = y\cos^{2} x dy$$
$$\Rightarrow \dfrac {x}{\cos^{2}x} dx = \dfrac {y}{\cos^{2}y} dy$$
$$\Rightarrow x\sec^{2}x dx = y\sec^{2}y dy$$
On integration, we get
$$\Rightarrow x\tan x - \int 1\cdot \tan x dx$$ $$= y \tan y - \int 1 \cdot \tan y dy$$
$$\Rightarrow x\tan x - \log \sec x = y\tan y - \log \sec y + c$$
$$\Rightarrow x\tan x - y\tan y - (\log \sec x- \log \sec y) = c$$
$$\Rightarrow x\tan x - y\tan y - \log\left(\dfrac{\sec x}{\sec y}\right) = c$$.
Question 5
1 / -0

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

A
$$y=\log _{ e }{ (x) } +C$$

B
$$y={ (\log _{ e }{ x } ) }^{ 2 }+C\quad $$

C
$$y=\pm \sqrt { { (\log _{ e }{ x } ) }^{ 2 }+2C } $$

D
$$xy={ x }^{ y }+K$$

Solution

Given Differential equation is the expansion of $$e^{xy\frac{dy}{dx}}$$
So, $$x=e^{xy\frac{dy}{dx}}$$
Taking log on both sides, We get
$$log x=xy\dfrac{dy}{dx}$$
$$\dfrac{log x}{x}dx=ydy$$
Integrating both sides
$$\Rightarrow \dfrac{(log x)^2}{2}+C=\dfrac{y^2}{2}$$
$$\Rightarrow y=\pm\sqrt{(log_ex)^2+2C}$$
Question 6
1 / -0

Let f(x) be differentiable on the interval $$(0,\infty)$$ such that $$f(1)=1$$ and $$\displaystyle \lim_{t \rightarrow x }\dfrac{t^2f(x)-x^2f(t)}{t-x}=1$$ for each $$x>0$$. Then f(x) is

A
$$\dfrac{1}{3x}+\dfrac{2x^2}{3}$$

B
$$\dfrac{-1}{3x}+\dfrac{4x^2}{3}$$

C
$$\dfrac{-1}{x}+\dfrac{2}{x^2}$$

D
$$\dfrac{1}{x}$$

Solution
Question 7
1 / -0

The solution of the differential equation, $$x^2\dfrac{dy}{dx}.cos\dfrac{1}{x}=-1$$, where $$y\rightarrow -1$$ as $$x\rightarrow \infty$$ is

A
$$y=log_{ e }{ \left| \sec { \dfrac { 1 }{ x } } +tan{ \dfrac { 1 }{ x } } \right| }-1$$.

B
$$y=\dfrac{x+1}{x\, sin\dfrac{1}{x}}$$

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

D
$$y=\dfrac{x+1}{x\, cos\dfrac{1}{x}}$$

Solution

Given,

$${ x }^{ 2 }\dfrac { dy }{ dx } .cos{ \dfrac { 1 }{ x } }=-1$$

$$\Rightarrow dy=\dfrac { 1 }{ cos{ \dfrac { 1 }{ x } } } .\dfrac { -1 }{ { x }^{ 2 } } dx$$

Let us assume $$\dfrac { 1 }{ x } =t$$.

Therefore, $$\dfrac { -1 }{ { x }^{ 2 } } dx=dt$$.

$$\Rightarrow dy=\dfrac { 1 }{ cos{ t } } dt$$

$$\Rightarrow dy=\sec { t } dt$$

Integrating both sides, we get,

$$\Rightarrow \int { dy } =\int { \sec { t } } dt$$

$$\Rightarrow y=log_{ e }{ \left| \sec { t } +tan{ t } \right| }+C$$

$$\Rightarrow y=log_{ e }{ \left| \sec { \dfrac { 1 }{ x } } +tan{ \dfrac { 1 }{ x } } \right| }+C$$

Now, given that $$y\rightarrow -1\quad as\quad x\rightarrow \infty $$

$${ \Rightarrow lim }_{ x\rightarrow \infty }y=-1$$

$${ \Rightarrow lim }_{ x\rightarrow \infty }\left( log_{ e }{ \left| \sec { \dfrac { 1 }{ x } } +tan{ \dfrac { 1 }{ x } } \right| }+C \right) =-1$$

$${ \Rightarrow }\left( log_{ e }{ \left| \sec { 0 } +tan{ 0 } \right| }+C \right) =-1$$

$$\Rightarrow C=-1$$.

Therefore, the solution of given differential equation is:

$$y=log_{ e }{ \left| \sec { \dfrac { 1 }{ x } } +tan{ \dfrac { 1 }{ x } } \right| }-1$$.
Question 8
1 / -0

Solution of the differential equation
$$\frac{dy}{dx}$$ = $$\frac{3 x^2 y^4 + 2 xy}{x^2 - 2 x^3 y^3}$$ is

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

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

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

D
$$x^3 y^2 - {x^2}{y}$$ = c
Question 9
1 / -0

If x and y are independent variables and $$f(x)=(\int^x_0e^{x-y}f'(y)dy)-(x^2-x+1)e^x$$ where f(x) is a differentiable function, then $$f(\dfrac{-1}{2})$$ equals.

A
$$2\sqrt{e}$$

B
$$-2\sqrt{e}$$

C
$$\dfrac{-2}{\sqrt{e}}$$

D
$$\dfrac{2}{\sqrt{e}}$$

Solution
Question 10
1 / -0

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

A
$$y = x\cot (c - x)$$

B
$$\cos - t\left (\dfrac {y}{x}\right ) = (-x + c)$$

C
$$y = x\tan (c - x)$$

D
none of these

Solution

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 52

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Differential Equations Test - 52

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SHARING IS CARING

Question 1

$$y = \dfrac {\phi(x) + c}{x}$$

$$y = \dfrac {\phi(x)}{x} + c$$

$$y = \dfrac {\phi(x)}{x + c}$$

$$y = \phi(x) + x + c$$

Solution

Question 2

$$X\theta (y/x)=k$$

$$\theta (y/x)=kx$$

$$y\theta (y/x)=k$$

$$\theta (y/x)=ky$$

Question 3

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

$${ x }^{ 2 }\sin { y } +{ x }^{ 3 }y-{ y }^{ 2 }/2=c\quad $$

$${ x }^{ 2 }\sin { y } -{ x }^{ 3 }y-{ y }^{ 2 }/2=c$$

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

Question 4

$$x\tan x - y\tan y - \log\left(\dfrac{\sec x}{\sec y}\right) = c$$

$$y\tan x - x\tan y - \log (\sec x \cdot \sec y) = c$$

$$x\tan x - y\tan y + \log (\sec x\cdot \sec y) = c$$

None of the above

Solution

Question 5

$$y=\log _{ e }{ (x) } +C$$

$$y={ (\log _{ e }{ x } ) }^{ 2 }+C\quad $$

$$y=\pm \sqrt { { (\log _{ e }{ x } ) }^{ 2 }+2C } $$

$$xy={ x }^{ y }+K$$

Solution

Question 6

$$\dfrac{1}{3x}+\dfrac{2x^2}{3}$$

$$\dfrac{-1}{3x}+\dfrac{4x^2}{3}$$

$$\dfrac{-1}{x}+\dfrac{2}{x^2}$$

$$\dfrac{1}{x}$$

Solution

Question 7

$$y=log_{ e }{ \left| \sec { \dfrac { 1 }{ x } } +tan{ \dfrac { 1 }{ x } } \right| }-1$$.

$$y=\dfrac{x+1}{x\, sin\dfrac{1}{x}}$$

$$y=cos\dfrac{1}{x}+sin\dfrac{1}{x}$$

$$y=\dfrac{x+1}{x\, cos\dfrac{1}{x}}$$

Solution

Question 8

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

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

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

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

Question 9

$$2\sqrt{e}$$

$$-2\sqrt{e}$$

$$\dfrac{-2}{\sqrt{e}}$$

$$\dfrac{2}{\sqrt{e}}$$

Solution

Question 10

$$y = x\cot (c - x)$$

$$\cos - t\left (\dfrac {y}{x}\right ) = (-x + c)$$

$$y = x\tan (c - x)$$

none of these

Solution

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