Differential Equations Test

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Differential Equations Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

The solution of the differential equation $$(3xy+y^{2})dx+(x^{2}+xy)dy=0$$ is

A
$$x^2(2x+y)=3$$

B
$$y^2(2x+y)=3$$

C
$$x^2y(2x+y)=3$$

D
$$x^2y(2x+3)=9$$

Solution
Question 2
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The solution of the differential equation $$\dfrac{dy}{dx} = \dfrac{x^2 + y^2 + 1}{2xy}$$ satisfying $$y (1) = 1$$, is

A
a hyperbola

B
a circle

C
$$y^2 = x (1 + x) - 10$$

D
$$(x - 2)^2 + (y - 3)^2 = 5$$

Solution

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

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

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

$$y^2=u=2y\dfrac{dy}{dx}=\dfrac{du}{dx}$$

$$\dfrac{du}{dx}-\dfrac{u}{x}=x+\dfrac{1}{x}$$

$$\dfrac{du}{dx}+Pu=Q$$

$$\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=1+\dfrac{1}{x^2}$$

$$\int d\left ( \dfrac{u}{x} \right )=\int \left ( 1+ \dfrac{1}{x^2}\right )dx\Rightarrow \dfrac{u}{x} x-x^{-1}+C$$

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

$$\Rightarrow y^2=x^{2}-1+x$$
Question 3
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The solution of the differential equation $$\left ({x}^{2}+1\right)\dfrac {dy}{dx}+{y}^{2}+1=0,$$ is $$\left[ ify\left( 0 \right) =1 \right]$$

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

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

C
$$y=x\left (x-1\right)$$

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

Solution
Question 4
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The solution of differential equation $$\dfrac{dy}{dx} =\dfrac{x(2\ln{x+1})}{\sin{y}+y\cos{y}}$$ is

A
$$y\sin{y}=x\ln{x}+c$$

B
$$y\sin{y}=x^{2}\ln{x}+c$$

C
$$\sin{y}=x^{2}\ln{x}+c$$

D
$$y\cos{y}=x^{2}\ln{x}+c$$

Solution

Equation of family of circle is
$$(n+a)^2+(y-a)^2=a^2$$ or
$$x^2+y^2+2ax-2ay +a^2$$
$$=0$$ ------(1)
Diffentiating , we get
$$2n+2y \dfrac{dy}{dx}+2a-2a \dfrac{dy}{dx}=0$$
$$\Rightarrow x+y \dfrac{dy}{dx}=a\left(\dfrac{dy}{dx}-1\right)$$
$$\Rightarrow a=\dfrac{x+yy^1}{y^1-1}$$
where $$y^1=\dfrac{dy}{dx}$$
Substituting the value of a in eq $$(1)$$ and Simplifying
$$\Rightarrow (xy^1-x+x+yy^1)+(yy^1-y-x-yy^1)^2$$
$$=(x+(yy^1))^2$$
$$\Rightarrow (x+y)^2[(y^1)^2+1]=(x+yy^1)^2$$
The given differential equation is :
$$\dfrac{dy}{dx}=\dfrac{x(2\log x+1)}{\sim y +y \cos y}$$
$$\Rightarrow (\sin y+y \cos y)dy =x(2\log x+1)dx$$
$$\Rightarrow \int (\sin y +y \cos y )dy=\int x(2\log x+1)dx$$ integrating both sides
$$\Rightarrow -\cos y +y \sin y - \int \sin y \, dy$$
$$=2\log ^x \dfrac{x^2}{2}-2 \int \dfrac{1}{x} \dfrac{x^2}{2}dx +\dfrac{x^2}{2}$$
$$\Rightarrow -\cos y + y \sin y + \cos y =x^2 \log x- \dfrac{x^2}{2}+\dfrac{x^2}{2}+c$$
$$\Rightarrow y \sin y =x^2 \log x +c$$ .......(2)
It is given that when $$y=\dfrac{\pi}{2},x=1$$
So, putting $$y=\dfrac{\pi}{2}, x=1$$ in eq (2),
$$\Rightarrow \dfrac{\pi}{2} \sin \dfrac{\pi }{2}=1.\log 1+c$$
$$\Rightarrow c=\dfrac{\pi}{2}$$
Putting $$c=\dfrac{\pi}{2}$$ in eq (2) we oftain
$$\boxed{y \sin y =x^2\log x+\dfrac{\pi}{2}}$$
Question 5
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The order of the differential equation satisfying $$\sqrt { 1-{ x }^{ 4 } } +\sqrt { 1-{ y }^{ 4 } } =a\left( { x }^{ 2 }-{ y }^{ 2 } \right) $$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 6
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Solution of the differential equation $$ye^{x/y}dx=(xe^{\frac {x}{y}}+y^{2})dy(y \neq 0)$$ is ?

A
$$e^{x/y}=x+C$$

B
$$e^{x/y}+x=C$$

C
$$e^{y/x}=x+C$$

D
$$e^{x/y}=y+C$$

Solution
Question 7
1 / -0

The D.E whose solution is $$y=a\cos(3x+b)$$ is?

A
$$y_2+3y=0$$

B
$$y_2+y=0$$

C
$$y_2+9y=0$$

D
$$y_2+6y=0$$

Solution

$$y=a\, \cos \left( { 3x+b } \right) \\ a,b\, \, are\, \, two\, \, arbitary\, \, \\ { { constant } }{ { .Then,requiredD-E } } \\ { { are } }\, \, { { insecondorder } }{ { . } } \\ \dfrac { y }{ a } =\cos \left( { 3x+b } \right) ,\, \, { \cos ^{ -1 } }\left( { \dfrac { y }{ a } } \right) =3x+b \\ differentiate\, \, w.r\, \, to\, \, x\, \, \\ \Rightarrow \dfrac { { d\left( { { { \cos }^{ -1 } }\dfrac { y }{ a } } \right) } }{ { dx } } =\dfrac { { 3dx } }{ { dx } } +\dfrac { { db } }{ { dx } } \\ \Rightarrow \sqrt { 1-\dfrac { { { y^{ 2 } } } }{ { { a^{ 2 } } } } } \times \dfrac { 1 }{ a } \dfrac { { dy } }{ { dx } } =3 \\ \Rightarrow \dfrac { { dy } }{ { dx } } =-3a\sqrt { 1-\dfrac { { { y^{ 2 } } } }{ { { a^{ 2 } } } } } \\ \Rightarrow \dfrac { { dy } }{ { dx } } =\dfrac { { -3a } }{ a } \sqrt { { a^{ 2 } }-{ y^{ 2 } } } \\ \Rightarrow \dfrac { { dy } }{ { dx } } =-3\sqrt { { a^{ 2 } }-{ y^{ 2 } } } \\ \Rightarrow { \left( { \dfrac { { dy } }{ { dx } } } \right) ^{ 2 } }={ \left( { -3 } \right) ^{ 2 } }{ \left( { \sqrt { { a^{ 2 } }-{ y^{ 2 } } } } \right) ^{ 2 } } \\ \Rightarrow { \left( { \dfrac { { dy } }{ { dx } } } \right) ^{ 2 } }=9\sqrt { { a^{ 2 } }-{ y^{ 2 } } } \\ \Rightarrow { \left( { \dfrac { { dy } }{ { dx } } } \right) ^{ 2 } }+9{ y^{ 2 } }=9{ a^{ 2 } } \\ again\, \, differentiate\, \, w.r\, to\, \, x \\ \dfrac { { d\left[ { { { \left( { \dfrac { { dy } }{ { dx } } } \right) }^{ 2 } } } \right] } }{ { dx } } +9\dfrac { { d{ y^{ 2 } } } }{ { dx } } =0 \\ \Rightarrow 2\dfrac { { dy } }{ { dx } } .d{ y^{ 2 } }\left( { d{ x^{ 2 } }+18y\dfrac { { dy } }{ { dx } } } \right) =0 \\ \Rightarrow 2\dfrac { { dy } }{ { dx } } \left( { { d^{ 2 } }y\left( { d{ x^{ 2 } }+9 } \right) } \right) =0 \\ It\, \, is\, \, required\, \, D.E\, \, $$
Question 8
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The solution of the differential equation $$\dfrac{dy}{dx}+\dfrac{1}{x}\tan y=\dfrac{\tan y \sin y}{{x}^{2}}$$ is

A
$$\dfrac{1}{x}\cos y = \dfrac{1}{{2x}^{2}}+K$$

B
$$\dfrac{1}{x}\cot y = \dfrac{1}{{2x}^{2}}+K$$

C
$$\dfrac{1}{x}\csc y = \dfrac{1}{{2x}^{2}}+K$$

D
$$\dfrac{1}{x}\cos x = \dfrac{1}{{2x}^{2}}+K$$

Solution
Question 9
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The equation of a curve passing through the point (0, 0) and whose differential equation is $${y'} = {e^x}\sin x$$ is

A
$$2y - 1 = {e^x}\left( {\sin x - \cos x} \right)$$

B
$$2y - 1 = {e^{ - x}}\left( {\cos x - \sin x} \right)$$

C
$$2y + 1 = {e^{ - x}}\left( {\cos x - \sin x} \right)$$

D
$$None\,of\,these$$

Solution

given, $$\dfrac {dy}{dx}=e^x\sin x$$
$$dy=e^x\sin x.dx$$
$$\displaystyle \int dy=\displaystyle \int e^x\sin x.dx\ \rightarrow $$ integrating both sides.
$$det,\ \displaystyle \int e^x\sin x\ dx=I----(1)$$
$$y=\displaystyle \int \underset{f(x)}{\underset{\downarrow}{e^x}}\underset{g(x)}{\underset{\downarrow }{\sin x}}\ dx=I$$
Since, $$\displaystyle \int f(x)\,g(x)dx =f(x).\displaystyle \int g(x)dx-\displaystyle \int \left [f'(x) \displaystyle \int g(x)dx\right]dx$$
$$\therefore \ \Rightarrow \ e^x.\displaystyle \int \sin x\ dx-\displaystyle \int \left [e^x.\displaystyle \int \sin x\ dx\right]$$
$$I=e^x.(-\cos x)-\displaystyle \int e^x.(-\cos x).dx$$
$$I=-e^x\cos x+\displaystyle \int \underset{f(x)}{\underset{\downarrow}{e^x}}.\underset{g(x)}{\underset{\downarrow}{\cos x}}\ dx$$
Now apply same formula as above
$$I=-e^x\cos x.+\left [e^x.\displaystyle \int \cos x.dx-\displaystyle \int (ex.\displaystyle \int \cos x.dx)dx\right]$$
$$I=-{ e }^{ x }.\cos { x } +\left[ { e }^{ x }.\sin { x } -\underbrace { \displaystyle \int { { e }^{ x }.\sin { x } .dx } }_{ \left( I \right) } \right] $$
$$I=-e^x\cos x+e^x\sin x-I+c$$
$$2y=e^x(\sin x-\cos x)+c$$
Since this curve passes through $$(0,0)$$ therefore
$$2\times 0=e^0(\sin 0\cos 0)+c$$
$$c=1$$
$$\therefore \ $$ equation of curve $$2y=e^x(\sin x-\cos x)+1$$
$$(2y-1)=e^x(\sin x-\cos x)$$
Question 10
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$$(2y+xy^{3})dx+(x+x^{2}y^{2})dy=0$$ solution of differential equation is

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

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

C
$$yx^{2}+\dfrac{xy}{3}=c$$

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

Solution