Differential Equations Test

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Differential Equations Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

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

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

B
$$\displaystyle \sin^{-1}(\frac{2x+1}{\sqrt{3}})+\sin^{-1}(\frac{2y+1}{\sqrt{3}})=c$$

C
$$\displaystyle \sinh^{-1}(\frac{2x+1}{\sqrt{3}})+\sinh^{-1}(\frac{2y+1}{\sqrt{3}})=c$$

D
$$\displaystyle \sin^{-1}(\frac{2x+1}{\sqrt{3}})=c$$

Solution

$$\Rightarrow \dfrac{-dy}{(y+\dfrac{1}{2})^{2}+(\dfrac{\sqrt{3}}{2})^{2}} = \dfrac{dx}{(x+\dfrac{1}{2})^{2}+ (\dfrac{\sqrt{3}}{2})^{2}}$$

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

$$\Rightarrow c= tan^{-1} \left(\dfrac{(2x+1)}{\sqrt{3}}\right) + tan^{-1} \left(\dfrac{2y+1}{\sqrt{3}}\right)$$
Question 2
1 / -0

The order of differential equation $$\displaystyle \frac{d^{2}y}{dx^{2}}-\left ( \frac{dy}{dx} \right )^{2}=1$$ is:

A
one

B
two

C
four

D
zero

Solution

The order of differential equation is the order of the highest derivative in the equation
$$\therefore $$ the above given equation is of second order
Question 3
1 / -0

A Function $$y=f(x)$$ stasfies $$\displaystyle {f}''(x)=-\displaystyle \frac {1}{x^2}-\pi^2\sin(\pi x)$$; $$\displaystyle {f}'(2)=\pi+ \displaystyle \frac{1}{2}$$ and $$f(1)=0$$. The value of $$f\left (\displaystyle \frac{1}{2} \right )$$ is:

A
$$\displaystyle \ln 2$$

B
$$1$$

C
$$\displaystyle \frac{\pi}{2}-\ln 2$$

D
$$1-\ln 2$$

Solution

$$\displaystyle {f}''(x)=-\displaystyle \dfrac {1}{x^2}-\pi^2\sin(\pi x)$$
Integrating w.r.t x on both sides,
$$f'(x)=\displaystyle \dfrac { 1 }{ x } +\pi ^{ 2 }\dfrac { \cos { \pi x } }{ \pi } +C$$
$$\Rightarrow f'(x)=\dfrac { 1 }{ x } +\pi \cos { \pi x } +C$$
$$\displaystyle f'(2)=\dfrac { 1 }{ 2 } +\pi +C$$
$$\displaystyle \pi +\dfrac { 1 }{ 2 } =\dfrac { 1 }{ 2 } +\pi +C$$
$$\Rightarrow C=0$$
Hence, $$f'(x)=\dfrac { 1 }{ x } +\pi \cos { \pi x } $$
Again , integrating w.r.t. x
$$\Rightarrow f(x)=ln|x|+\sin {\pi x}+C_{1}$$
Put $$x=1$$
$$0=0+0+C_{1}$$
$$\Rightarrow C_{1}=0$$
Hence, $$f(x)=\ln { | } x|+\sin { \pi x } $$
$$f(\dfrac{1}{2})=\ln |\dfrac{1}{2}|+1$$
$$\Rightarrow f(\dfrac{1}{2})=-\ln 2 +1$$
Hence, option 'D' is correct.
Question 4
1 / -0

If $$f(x) = x+2; g(x) = x^{2}-x-2$$, then $$\dfrac{g(1)+g(2)+g(3)}{f(-4)+f(-2)+f(2)}=$$

A
$$0$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{1}{6}$$

Solution
Question 5
1 / -0

f:R r be defined by $$f(x)=3x-5,$$ the formula that defines the reverse $$f^n f^{-1}$$ is

A
$$3y-5$$

B
$$5-3y$$

C
$$\dfrac{y+5}{3}$$

D
None of these

Solution
Question 6
1 / -0

If $$\displaystyle f\left ( 0 \right )={f}'\left ( 0 \right ) = 0$$ and $$\displaystyle {f}''\left ( x \right )=\tan ^{2}x$$ then $$\displaystyle f\left ( x \right )$$ is:

A
$$\displaystyle \log \sec x-\frac{1}{2}x^{2}$$

B
$$\displaystyle \log \cos x+\frac{1}{2}x^{2}$$

C
$$\displaystyle \log |\sec x|-\frac{1}{2}x^{2}$$

D
none of these

Solution

$$\displaystyle {f}''\left ( x \right )=\tan ^{2}x$$
$$\displaystyle f''(x)=(\sec^{2} x -1)$$
Integrating w.r.t x
$$f'(x)=\tan x - x+C$$ .....(1)
$$f'(0)=C$$
Integrating (1) w.r.t. x, we get
$$f(x)=\log |\sec x|-\frac{x^2}{2}+Cx+C_{1}$$
$$f(0)=C_{1}$$
Now since $$ f'(0) = f(0) = 0$$ we get $$ C_1 = C = 0$$
Question 7
1 / -0

The differential equation $$\displaystyle \frac{dy}{dx}=\frac{\sqrt{1-y^{2}}}{y}$$determines a family of circles with:

A
variable radii and a fixed centre at $$\displaystyle (1,1)$$

B
variable radii and a fixed centre at $$\displaystyle (0, -1)$$

C
fixed radius 1 and variable centres along the x-axis

D
fixed radius 1 and variable centres along the y-axis

Solution

$$\cfrac { dy }{ dx } =\cfrac { \sqrt { 1-y^{ 2 } } }{ y } \\ \Rightarrow \cfrac { dy }{ dx } \cfrac { y }{ \sqrt { 1-y^{ 2 } } } =1\\ \Rightarrow \int { \cfrac { y }{ \sqrt { 1-y^{ 2 } } } dy } =\int { dx } \\ \Rightarrow -\sqrt { 1-y^{ 2 } } =x+c\Rightarrow -y^{ 2 }+1=\left( x+c \right) ^{ 2 }\\ \Rightarrow \left( x+c \right) ^{ 2 }+y^{ 2 }=1$$
Question 8
1 / -0

If $$\frac{{dy}}{{dx}} = {e^{ - 2y}}$$ and $$y = 0$$ when $$x = 5$$, then value of x for $$y = 3$$ is

A
$$e^5$$

B
$$e^6+1$$

C
$$\frac{{{e^6} + 9}}{2}$$

D
$$log_e6$$

Solution

$$\frac {dy}{dx}=e^{-2y}$$
$$\int e^{2y}dy=\int dx$$
$$\frac {e^{2y}}{2}=x+c$$
$$\frac {2}{2}=5+c$$
$$c=\frac {-9}{2}$$
$$x=\frac {e^6+9}{2}$$
Question 9
1 / -0

The general solution of differential equation is $$y=ae^{bx+c}$$ where $$a, b, c$$ are arbitrary constant. The order of differential equation is:

A
$$3$$

B
$$2$$

C
$$1$$

D
none of these

Solution

$$\displaystyle=ae^{bx}.e^{c}$$ $$a,b,c$$ are three constants
in order to solve the above given equation we have to differentiate it twice so the given equation is
$$\dfrac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=ab^{2}e^{bx+c} $$
=> Thus, order is $$2$$
Hence, option 'B' is correct.
Question 10
1 / -0

Find the solution of $$\displaystyle \frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}$$.

A
$$\displaystyle e^{y}=e^{x}-\frac{x^{3}}{3}+c$$

B
$$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{4}+c$$

C
$$\displaystyle e^{y}=e^{x}-\frac{x^{3}}{4}+c$$

D
$$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{3}+c$$

Solution

Given $$\displaystyle \frac{dy}{dx}=e^{x-y}+x^{2}e^{-y}$$
$$\Rightarrow \dfrac {dy}{dx}=\dfrac {e^x}{e^y}+\dfrac {x^2}{e^y}$$
$$\Rightarrow \displaystyle e^{y}dy=\left ( e^{x}+x^{2} \right )dx.$$
Now integrating both sides.
$$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{3}+c.$$

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Differential Equations Test - 24

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Differential Equations Test - 24

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SHARING IS CARING

Question 1

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

$$\displaystyle \sin^{-1}(\frac{2x+1}{\sqrt{3}})+\sin^{-1}(\frac{2y+1}{\sqrt{3}})=c$$

$$\displaystyle \sinh^{-1}(\frac{2x+1}{\sqrt{3}})+\sinh^{-1}(\frac{2y+1}{\sqrt{3}})=c$$

$$\displaystyle \sin^{-1}(\frac{2x+1}{\sqrt{3}})=c$$

Solution

Question 2

one

two

four

zero

Solution

Question 3

$$\displaystyle \ln 2$$

$$1$$

$$\displaystyle \frac{\pi}{2}-\ln 2$$

$$1-\ln 2$$

Solution

Question 4

$$0$$

$$1$$

$$\dfrac{1}{2}$$

$$\dfrac{1}{6}$$

Solution

Question 5

$$3y-5$$

$$5-3y$$

$$\dfrac{y+5}{3}$$

None of these

Solution

Question 6

$$\displaystyle \log \sec x-\frac{1}{2}x^{2}$$

$$\displaystyle \log \cos x+\frac{1}{2}x^{2}$$

$$\displaystyle \log |\sec x|-\frac{1}{2}x^{2}$$

none of these

Solution

Question 7

variable radii and a fixed centre at $$\displaystyle (1,1)$$

variable radii and a fixed centre at $$\displaystyle (0, -1)$$

fixed radius 1 and variable centres along the x-axis

fixed radius 1 and variable centres along the y-axis

Solution

Question 8

$$e^5$$

$$e^6+1$$

$$\frac{{{e^6} + 9}}{2}$$

$$log_e6$$

Solution

Question 9

$$3$$

$$2$$

$$1$$

none of these

Solution

Question 10

$$\displaystyle e^{y}=e^{x}-\frac{x^{3}}{3}+c$$

$$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{4}+c$$

$$\displaystyle e^{y}=e^{x}-\frac{x^{3}}{4}+c$$

$$\displaystyle e^{y}=e^{x}+\frac{x^{3}}{3}+c$$

Solution

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