Differential Equations Test

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Differential Equations Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

Solve: $$\dfrac{{dy}}{{dx}} = 1 + x + y + xy$$

A
$${1+y=k{e}^\dfrac{x^2+2x}{4}}$$

B
$${1+y=k{e}^\dfrac{x^2+2x}{2}}$$

C
$${1+y=k{e}^\dfrac{x^2+4x}{2}}$$

D
$${1+y=k{e}^\dfrac{x^2+4x}{4}}$$

Solution

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

$$\dfrac{dy}{dx}=y(x+1)+(x+1)$$

$$\dfrac{dy}{dx}=(x+1)(y+1)$$

$$\dfrac{dy}{y+1}=(x+1)dx$$

Now, integrating both sides,

$$\int\dfrac{dy}{y+1}=\int(x+1)dx$$
$$\Rightarrow$$ $$ln(y+1)=\dfrac{x^2}{2}+x+c$$

Now, taking exponential both sides,

$$\Rightarrow$$ $$y+1=e^{\dfrac{x^2}{2}+x+c}$$
$$\therefore$$ $$y=e^{\dfrac{x^2}{2}+x+c}-1$$
Question 2
1 / -0

The function f(x) satisfying the equation, $$f^2(x)+4f(x)\cdot f(x)+[f(x)]^2=0$$.where Cis an arbitrary constant.

A
$$f(x)=c\cdot e^{(2-\sqrt{3})x}$$

B
$$f(x)=c\cdot e^{(2+\sqrt{3})x}$$

C
$$f(x)=c\cdot e^{(\sqrt{3}-2)x}$$

D
$$f(x)=c\cdot e^{-(2+\sqrt{3})x}$$

Solution
Question 3
1 / -0

$$\frac{{dy}}{{dx}} + \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}} = 0$$

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

B
$${\tan ^{ - 1}}\frac{{2x + 1}}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{{2y + 1}}{{\sqrt 3 }} = c$$

C
$${\tan ^{ - 1}}\frac{{2x}}{{\sqrt 5 }} + {\tan ^{ - 1}}\frac{{2y}}{{\sqrt 3 }} = c$$

D
$${\tan ^{ - 1}}\frac{{2x - 1}}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{{2y - 1}}{{\sqrt 3 }} = c$$

Solution

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

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

On integrating, we get
$$\displaystyle\int{\dfrac{dy}{{y}^{2}+y+1}}=-\displaystyle\int{\dfrac{dx}{{x}^{2}+x+1}}$$

$$\displaystyle\int{\dfrac{dy}{{y}^{2}+2\times\dfrac{1}{2}y+\dfrac{1}{4}-\dfrac{1}{4}+1}}=-\displaystyle\int{\dfrac{dx}{{x}^{2}+2\times\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+1}}$$

$$\displaystyle\int{\dfrac{dy}{{\left(y+\dfrac{1}{2}\right)}^{2}+\dfrac{3}{4}}}=-\displaystyle\int{\dfrac{dx}{{\left(x+\dfrac{1}{2}\right)}^{2}+\dfrac{3}{4}}}$$

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

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

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

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

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

or $${\tan}^{-1}{\left(\dfrac{2y+1}{\sqrt{3}}\right)}+{\tan}^{-1}{\left(\dfrac{2x+1}{\sqrt{3}}\right)}=c$$ where $$c=\dfrac{\sqrt{3}}{2}k$$
Question 4
1 / -0

If $$y={e}^{4x}+2{e}^{-x}$$ satisfies the relation $$\dfrac {{d}^{3}y}{{dx}^{3}}+A\dfrac {dy}{dx}+By=0$$,then value of $$A$$ and $$B$$ respectively are

A
$$-13, 14$$

B
$$-13, -12$$

C
$$-13, 12$$

D
$$12, -13$$
Question 5
1 / -0

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

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

B
$$(x-2)=ke^{-\tan^{-1}y}$$

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

D
$$xe^{\tan^{-1}y}=\tan^{-1}y+k$$
Question 6
1 / -0

The solution of $$\dfrac {dy}{dx}+\dfrac {y}{x}=x^{2}y^{6}$$ is

A
$$\dfrac{1}{x^{2}y^{5}}=\dfrac{5}{2x^{2}}+c$$

B
$$\dfrac{1}{x^{5}y^{5}}=\dfrac{5}{x^{2}}+c$$

C
$$\dfrac{1}{x^{5}y^{5}}=\dfrac{5}{2x^{2}}+c$$

D
$$\dfrac {1}{x^{2}y^{2}}=\dfrac {5}{2x^{2}}+c$$
Question 7
1 / -0

The solution of the differential equation $$x^2\dfrac{dy}{dx}\cos\left(\dfrac{1}{x}\right) - y \sin\left(\dfrac{1}{x}\right) = -1 ; where \ y \rightarrow -1 \ as \ x \rightarrow \infty$$ is

A
$$y = \sin\left(\dfrac{1}{x}\right) - \cos\left(\dfrac{1}{x}\right) $$

B
$$y = \dfrac{x+1}{x\sin\left(\dfrac{1}{x}\right)}$$

C
$$y = \cos\left(\dfrac{1}{x}\right) + \sin\left(\dfrac{1}{x}\right) $$

D
$$y = \dfrac{x+1}{x\cos\left(\dfrac{1}{x}\right)}$$

Solution

$$\displaystyle x^{2}\frac{dy}{dx}cos\left ( \frac{1}{x} \right )- ysin\left ( \frac{1}{x} \right )=-1$$
$$\displaystyle \frac{dy}{dx}-\frac{y}{x^{2}}tan\left ( \frac{1}{x} \right )= -\frac{1}{x^{2}}cos \left ( \frac{1}{x} \right )$$
$$\displaystyle \frac{dy}{dx}+\left ( \dfrac{-1}{x^{2}}tan\left ( \dfrac{1}{x} \right ) \right )y=\frac{-1}{x^{2}cos\left ( \dfrac{1}{x} \right )}$$
$$I.F = e^{\displaystyle\int -\frac{1}{x^{2}}tan\left ( \dfrac{1}{x} \right )dx}$$
$$\displaystyle \frac{1}{x}=t$$
$$\displaystyle -\frac{dx}{x^{2}}=dt$$
$$ =e^{\displaystyle\int tan(t)dt}$$
$$\displaystyle =e^{ln(sect)}$$
$$\displaystyle =sec\left ( \dfrac{1}{x} \right )$$
$$\displaystyle sec\left ( \frac{1}{x} \right ).y=\int \frac{1}{x^{2}cos\left ( \dfrac{1}{x} \right )}.sec\left ( \frac{1}{x} \right )$$
$$\displaystyle =-\int \frac{sec^{2}\left ( \dfrac{1}{x} \right )}{x^{2}}$$
Let $$\displaystyle \frac{1}{x}=t$$
$$\displaystyle \frac{-dx}{x^{2}}=dt$$
$$\displaystyle =\int sec^{2}(t)dt$$
$$\displaystyle \boxed{sec\left ( \frac{1}{x} \right ).y=tan\left ( \frac{1}{x} \right )+C}$$
$$\displaystyle 1.-1= 0+C$$ $$C=-1$$
$$\displaystyle sec\left ( \frac{1}{x} \right ).y= tan\left ( \frac{1}{x} \right )-1, $$
$$\displaystyle \boxed{y=sin\left ( \frac{1}{x} \right )-cos\left ( \frac{1}{x} \right )}$$
Question 8
1 / -0

The general solution of differential equation $$\dfrac{dy}{dx}=\sin^{3}{x}\cos^{2}{x}+xe^{x}$$

A
$$y=\dfrac{1}{5}\cos^{5}{x}+\dfrac{1}{3}csc^{3}{x}+(x+1)e^{x}+c$$

B
$$y=\dfrac{1}{5}\cos^{5}{x}-\dfrac{1}{3}csc^{3}{x}+(x-1)e^{x}+c$$

C
$$y=-\dfrac{1}{5}\cos^{5}{x}-\dfrac{1}{3}csc^{3}{x}-(x-1)e^{x}-c$$

D
None of these

Solution
Question 9
1 / -0

The value of $$\displaystyle \lim_{x \rightarrow \infty}$$ y(x) obtained from the differential equation $$\dfrac{dy}{dx} = y - y^2$$, where y(0) = 2 is

A
1

B
-1

C
0

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

Solution

$$\dfrac{dy}{dx}=y-y^{2}$$ $$y(0)=2$$
$$\dfrac{dy}{y-y^{2}}=dx$$
$$\dfrac{-dy}{y^{2}-y}=dx$$
$$\dfrac{-dy}{y^{2}\left(\dfrac{1}{2}\right)^{2}-2\times \dfrac{1}{2}y-\dfrac{1}{4}}=dx$$
$$\dfrac{dy}{\left(y-\dfrac{1}{2}\right)^{2}-\left(\dfrac{1}{2}\right)^{2}}=-dx$$
$$\dfrac{1}{2\times \dfrac{1}{2}}\ln\left|\dfrac{y-\dfrac{1}{2}-\dfrac{1}{2}}{y-\dfrac{1}{2}+\dfrac{1}{2}}\right|=-x+c$$
$$\ln\left(\dfrac{y-1}{y}\right)=x+c$$
Given $$y(0)=2$$
$$\ln\left(\dfrac{2-1}{2}\right)=C$$
$$\boxed{C=\ln\left(\dfrac{1}{2}\right)}$$
$$\ln\left(\dfrac{y-1}{y}\right)=-x+\ln\left(\dfrac{1}{2}\right)$$
$$\ln\left(\dfrac{y}{y-1}\right)=x+\ln(2)$$
$$\dfrac{y}{y-1}=e^{x+\ln (2)}$$
$$\dfrac{y-1}{y}=\dfrac{1}{e^{x+\ln(2)}}$$
$$1-\dfrac{1}{y}=e^{-(x+\ln(2))}$$
$$\dfrac{1}{y}=1-e^{-(x+\ln (2)]}$$
$$y=\dfrac{1}{1-e^{-[x+\ln (2)]}}$$
$$\displaystyle\lim_{x\rightarrow \infty}\dfrac{1}{1-e^{-\infty}}=\dfrac{1}{1-0}=1$$
Question 10
1 / -0

Solution of $$\left(\dfrac{x+y-a}{x+y-b}\right)\left(\dfrac{dy}{dx}\right)=\left(\dfrac{x+y+a}{x+y+b}\right)$$.

A
$$log\left[(x+y)^2-ab\right]=\dfrac{2}{b-a}[x-y]+k$$

B
$$log\left[(x+y)^2+ab\right]=\dfrac{1}{b-a}[x+y]+k$$

C
$$\left(\dfrac{b-a}{2}\right)\left[log \left((x+y)^2-ab\right)\right]=x+c$$

D
$$2log(x+y)=\dfrac{x+y}{b-a}+k$$

Solution

$$\left( \cfrac { x+y-a }{ x+y-b } \right) \left( \cfrac { dy }{ dx } \right) =\left( \cfrac { x+y+a }{ x+y+b } \right) $$
Let $$x+y=z$$
$$\Rightarrow 1+\cfrac { dy }{ dx } =\cfrac { dz }{ dx } \Rightarrow \cfrac { dy }{ dx } =\cfrac { dz }{ dx } -1\quad $$
substituting
$$\Rightarrow \left( \cfrac { z-a }{ z-b } \right) \left( \cfrac { dz }{ dx } -1 \right) =\left( \cfrac { z+a }{ z+b } \right) $$
$$\Rightarrow \cfrac { dz }{ dx } =1+\cfrac { \left( z+a \right) \left( z-b \right) }{ \left( z-a \right) \left( z+b \right) } =\cfrac { { z }^{ 2 }+bz-az+{ z }^{ 2 }-bz+az-ab }{ \left( z-a \right) \left( z+b \right) } =\cfrac { 2\left( { z }^{ 2 }-ab \right) }{ \left( z-a \right) \left( z+b \right) } $$
$$\Rightarrow \int { \cfrac { \left( { z }^{ 2 }-ab \right) +z\left( b-a \right) }{ { z }^{ 2 }-ab } } dz=\int { 2 } dx=2x+C$$
$$\Rightarrow \int { \left\{ 1+\cfrac { z\left( b-a \right) }{ { z }^{ 2 }-ab } \right\} } dz=2x+C\quad $$
$$\Rightarrow z+\cfrac { b-a }{ 2 } \ln { \left( { z }^{ 2 }-ab \right) } =2x+C$$
$$\Rightarrow \cfrac { b-a }{ 2 } \ln { \left( { z }^{ 2 }-ab \right) } =2x+C-(x+y)=x-y+C$$
$$\Rightarrow \log { \left[ { \left( x+y \right) }^{ 2 }-ab \right] } =\cfrac { 2 }{ b-a } \left( x-y \right) +C$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 54

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Differential Equations Test - 54

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SHARING IS CARING

Question 1

$${1+y=k{e}^\dfrac{x^2+2x}{4}}$$

$${1+y=k{e}^\dfrac{x^2+2x}{2}}$$

$${1+y=k{e}^\dfrac{x^2+4x}{2}}$$

$${1+y=k{e}^\dfrac{x^2+4x}{4}}$$

Solution

Question 2

$$f(x)=c\cdot e^{(2-\sqrt{3})x}$$

$$f(x)=c\cdot e^{(2+\sqrt{3})x}$$

$$f(x)=c\cdot e^{(\sqrt{3}-2)x}$$

$$f(x)=c\cdot e^{-(2+\sqrt{3})x}$$

Solution

Question 3

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

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

$${\tan ^{ - 1}}\frac{{2x}}{{\sqrt 5 }} + {\tan ^{ - 1}}\frac{{2y}}{{\sqrt 3 }} = c$$

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

Solution

Question 4

$$-13, 14$$

$$-13, -12$$

$$-13, 12$$

$$12, -13$$

Question 5

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

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

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

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

Question 6

$$\dfrac{1}{x^{2}y^{5}}=\dfrac{5}{2x^{2}}+c$$

$$\dfrac{1}{x^{5}y^{5}}=\dfrac{5}{x^{2}}+c$$

$$\dfrac{1}{x^{5}y^{5}}=\dfrac{5}{2x^{2}}+c$$

$$\dfrac {1}{x^{2}y^{2}}=\dfrac {5}{2x^{2}}+c$$

Question 7

$$y = \sin\left(\dfrac{1}{x}\right) - \cos\left(\dfrac{1}{x}\right) $$

$$y = \dfrac{x+1}{x\sin\left(\dfrac{1}{x}\right)}$$

$$y = \cos\left(\dfrac{1}{x}\right) + \sin\left(\dfrac{1}{x}\right) $$

$$y = \dfrac{x+1}{x\cos\left(\dfrac{1}{x}\right)}$$

Solution

Question 8

$$y=\dfrac{1}{5}\cos^{5}{x}+\dfrac{1}{3}csc^{3}{x}+(x+1)e^{x}+c$$

$$y=\dfrac{1}{5}\cos^{5}{x}-\dfrac{1}{3}csc^{3}{x}+(x-1)e^{x}+c$$

$$y=-\dfrac{1}{5}\cos^{5}{x}-\dfrac{1}{3}csc^{3}{x}-(x-1)e^{x}-c$$

None of these

Solution

Question 9

1

-1

0

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

Solution

Question 10

$$log\left[(x+y)^2-ab\right]=\dfrac{2}{b-a}[x-y]+k$$

$$log\left[(x+y)^2+ab\right]=\dfrac{1}{b-a}[x+y]+k$$

$$\left(\dfrac{b-a}{2}\right)\left[log \left((x+y)^2-ab\right)\right]=x+c$$

$$2log(x+y)=\dfrac{x+y}{b-a}+k$$

Solution

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