Differential Equations Test

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Differential Equations Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

The solution of the differential equation $$ y(2x^{2}+y)\dfrac{dy}{dx}=(1-4xy^{2})x^{2} $$ is given by

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

B
$$ xy^{2}+\dfrac{y^{3}}{3}-\dfrac{x^{3}}{3}= 0 $$

C
$$ \dfrac{2}{5}yx^{5}+\dfrac{y^{3}}{3}=\dfrac{x^{3}}{3}-\dfrac{4xy^{3}}{3}+c $$

D
None of these

Solution

$$ y(2x^{4}+y)\dfrac{dy}{dx}=91-4xy^{2})x^{2} $$
$$ \implies 2x^{4}ydy+y^{2}dy+4x^{3}y^{2}dx-x^{2}dx=0 $$
$$ \implies 2x^{2}y(x^{2}dy+2xydx)+y^{2}dy-x^{2}dx=0 $$
$$ \implies 2x^{2}yd(x^{2}y)+y^{2}dy-x^{2}dx=0$$
Integrating, we get $$ (x^{2}y^{2})^{2}+\dfrac{y^{3}}{3}-\dfrac{x^{3}}{3}=c $$
or $$ 3(x^{2}y)^{2}+y^{3}-x^{3}=c $$
Question 2
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The solution of the following differential equation $$ [1+x\sqrt{(x^{2}+y^{2})}]dx+[\sqrt{(x^{2}+y^{2})-1}]ydy=0 $$ is equal to

A
$$ x^{2}+\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c $$

B
$$ x-\dfrac{y^{3}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c $$

C
$$ x-\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c $$

D
None of these

Solution

$$ [1+x\sqrt{(x^{2}+y^{2})}]dx+[\sqrt{(x^{2}+y^{2})-1}]ydy=0 $$

$$ \implies dx-ydy+\sqrt{(x^{2}+y^{2}})(xdx+ydy)=0 $$

$$ \implies dx-ydy+\dfrac{1}{2}\sqrt{(x^{2}+y^{2})}d(x^{2}+y^{2})=0 $$

Integrating, we have

$$ x-\dfrac{y^{2}}{2}+\dfrac{1}{2} \int \sqrt{t} dt=c, (t=\sqrt{(x^{2}+y^{2})}) $$

or $$ x-\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2)}=c $$
Question 3
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The solution of the differential equation $$\dfrac{dy}{dx}=\dfrac{3x^{2}y^{4}+2xy}{x^{2}-2x^{3}y^{3}} $$ is

A
$$ \dfrac{y^{2}}{x}-x^{3}y^{2}=c $$

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

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

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

Solution

Re-write the D.E. as
$$(2xydx-x^{2}dy)+y^{2}(3x^{2}y^{2}dx+2x^{3}ydy)=0 $$

Dividing by $$y^{2}$$, we get

$$ \dfrac{y2xdx-x^{2}dy}{y^{2}} +y^{2}3x^{2}dx+x^{3}2ydy=0 $$

or $$ d(\dfrac{x^{2}}{y})+d(x^{3}y^{2})=0 $$

Integrating, we get the solution

$$ \dfrac{x^{2}}{y}+x^{3}y^{2}=c $$
Question 4
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Number of values of $$m\in N$$ for which $$y=e^{mx}$$ is a solution of the differential equation $$\dfrac{d^3y}{dx^3}-3\dfrac{d^2y}{dx^2}-4\dfrac{dy}{dx}+12y=0 $$

A
$$0$$

B
$$1$$

C
$$2$$

D
More than $$2$$

Solution

$$ y=e^{mx} $$ satisfies $$ \dfrac{d^3y}{dx^3}-3\dfrac{d^2y}{dx^2}-4\dfrac{dy}{dx}+12y=0 $$
then $$ e^{mx}(m^3-3m^2-4m+12)=0 $$
$$ \implies m=\pm 2,3 $$
$$ m\in N $$ hence $$ m\in (2,3) $$
Question 5
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The solution of the differential equation $$ \dfrac{x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+ \dots}{1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dots}=\dfrac{dx-dy}{dx+dy} $$ is

A
$$2ye^{2x}=Ce^{2x}+1 $$

B
$$2ye^{2x}=Ce^{2x}-1$$

C
$$ ye^{2x}=Ce^{2x}+2 $$

D
None of these

Solution

Applying componendo and divedendo we get
$$\dfrac{dy}{dx}=\dfrac{e^{-x}}{e^x}=e^{-2x} $$
$$ \implies 2y=-e^{-2x}+C $$
$$ \implies 2ye^{2x}=Ce^{2x}-1 $$
Question 6
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The solution of the differential equation $$\dfrac {dy}{dx}=\dfrac {1+y^2}{1+x^2}$$ is :

A
$$y=\tan^{-1}x$$

B
$$y-x=k(1+xy)$$

C
$$x=\tan^{-1}y$$

D
$$\tan ( xy)=k$$

Solution

Given that,
$$\dfrac{1+y^2}{1+x^2}$$
$$\Rightarrow \dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}$$
On integrating both sides, we get
$$\displaystyle \int{\dfrac{dy}{1+y^2}}=\displaystyle \int{\dfrac{dx}{1+x^2}}$$
$$\tan^{-1}y=\tan^{-1}x+C$$
$$\Rightarrow \tan^{-1}y-\tan^{-1}x=C$$
$$\Rightarrow \tan^{-1}\left( \dfrac{y-x}{1+xy}\right)=C$$
$$\Rightarrow \dfrac{y-x}{1+xy}=\tan C$$
$$\Rightarrow y-x=\tan C (1+xy)$$
$$\Rightarrow y-x=K(1+xy)$$
Where, $$k=\tan C$$
Question 7
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Directions For Questions

Differential equation $$\dfrac{dy}{dx}=f(x)g(x)$$ can be solved by
separating variable $$\dfrac{dy}{g(y)}=f(x)dx$$

...view full instructions

Solution of the diffrential equation $$\dfrac{dy}{dx}+ \dfrac {1+y^2}{\sqrt {1-x^2}}=0$$ is

A
$$ \tan^{-1} y + \sin ^{-1} x=C$$

B
$$ \tan^{-1} x + \sin ^{-1} y=C$$

C
$$ \tan^{-1} x . \sin ^{-1} y=C$$

D
$$ \tan^{-1} x - \sin ^{-1} y=C$$

Solution

$$\dfrac{dy}{dx}+ \dfrac {1+y^2}{\sqrt {1-x^2}}=0$$

$$ \displaystyle \Rightarrow \dfrac{dy}{1+y^2}+ \dfrac {dx}{\sqrt {1-x^2}}=0$$

$$ \Rightarrow \tan^{-1} y + \sin ^{-1} x=C$$
Question 8
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Differential equation $$\dfrac{dy}{dx}=f(x)g(x)$$ can be solved by
separating variable $$\dfrac{dy}{g(y)}=f(x)dx$$

If $$\dfrac{dy}{dx}=1+x+y+xy$$ and $$y(-1)=0$$, then $$y$$ is equal to

A
$$e^{\dfrac{(1-x)^1}{2}}$$

B
$$e^{\dfrac{(1+x)^2}{2}} -1$$

C
In $$(1+x)-1$$

D
$$1+x$$

Solution

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

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

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

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

at $$x=-1, y=0$$

$$\ln (1+0)=-1+\dfrac {1}{2}+c$$

$$0=\dfrac {-1}{2}+c \Rightarrow c=\dfrac12$$

$$\ln (1+y)=x+\dfrac {x^2}{2}+\dfrac 12$$

$$\ln (1+y)=\dfrac {(x+1)^2}{2}$$

$$y=e^{\dfrac {(x+1)^2}{2}}-1$$
Question 9
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Which of the following is the general solution of $$\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0$$?

A
$$y=(Ax+B) e^x$$

B
$$y=(Ax+B) e^{-x}$$

C
$$y=Ae^x +Be^{-x}$$

D
$$y=A\cos x+B\sin x$$

Solution

Given that, $$\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0$$
$$D^2y-2Dy+y=0$$,
Where, $$D=\dfrac{d}{dx}$$
$$(D^2-2D+1)y=0$$
The auxiliary equation is $$m^2-2m+1=0$$
$$(m-1)^2=0 \Rightarrow m=1, 1$$
Since, the roots are real and equal.
$$\therefore CF=(Ax+B)e^x\\ \Rightarrow y=(Ax+B)e^x$$
[ since, if roots of Auxiliary equation are real and equal say (m), then $$CF=(C_1x+C_2)e^{mx}]$$
Question 10
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Solution of $$\displaystyle \sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\frac{dy}{dx}=0$$, is:

A
$$\log\left(\displaystyle \frac{x}{1+\sqrt{1+x^{2}}}\right)+\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$$

B
$$\displaystyle \log\left(\frac{x}{\sqrt{1+x^{2}}}\right)+\sqrt{1-x^{2}}+\sqrt{1+y^{2}}=c$$

C
$$\displaystyle \log\left(\frac{x}{\sqrt{1+x^{2}}}\right)=c$$

D
$$\displaystyle \log\left(\sqrt{1+x^{2}}-\sqrt{1+y^{2}}\right)+\log\left(\frac{x}{\sqrt{1+x^{2}}}\right)=c$$

Solution

Given, $$\displaystyle \sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\dfrac{dy}{dx}=0$$
$$\Rightarrow \sqrt{1+y^{2}}\sqrt{1+x^{2}} = -xy\dfrac{dy}{dx}$$
Integrating both sides, we get
$$\Rightarrow \displaystyle\int {\dfrac{\sqrt{1+x^{2}}}{x}dx} = -\displaystyle\int{\dfrac{y}{\sqrt{1+y^{2}}}}dy$$
Substitute, $$ 1+ y^{2}=t^{2} $$ $$\Rightarrow$$ $$ 2y dy = 2t dt $$
$$\Rightarrow$$ $$\displaystyle\int {\dfrac{\sqrt{1+x^{2}}}{x}dx} = - \displaystyle\int dt=- t+c = -\sqrt{1+y^{2}} +c $$
Now substitute, $$ 1+x^{2}= k^{2}$$ $$\Rightarrow$$ $$ 2x dx = 2k dk $$
$$\Rightarrow \displaystyle\int\dfrac{k^{2}}{x^{2}}dk = -\sqrt{1+y^{2}} +c $$
$$\Rightarrow \displaystyle\int\dfrac{k^{2}}{k^{2}-1}dk = -\sqrt{1+y^{2}} +c $$
$$\Rightarrow \displaystyle\int\left[\dfrac{k^{2}-1}{k^{2}-1}+\dfrac{1}{k^{2}-1}\right]dk = -\sqrt{1+y^{2}} + c $$
$$\Rightarrow k +\dfrac{1}{2}\log{\dfrac{k-1}{k+1}} = -\sqrt{1+y^{2}} +C $$
Since , $$\displaystyle\int\dfrac{1}{x^{2}-d^{2}}dx=\dfrac{1}{2d}\log{\dfrac{x-d}{x+d}}+\mbox{constant}$$
$$\therefore \sqrt{1+x^{2}} +\dfrac{1}{2}\log{\dfrac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}}+\sqrt{1+y^{2}}= C$$
$$\Rightarrow \sqrt{1+x^{2}} +\log\left(\dfrac{x}{\sqrt{1+x^{2}}+1}\right)+\sqrt{1+y^{2}}= C$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 47

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Differential Equations Test - 47

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SHARING IS CARING

Question 1

$$ 3(x^{2}y)^{2}+y^{3}-x^{3}=c $$

$$ xy^{2}+\dfrac{y^{3}}{3}-\dfrac{x^{3}}{3}= 0 $$

$$ \dfrac{2}{5}yx^{5}+\dfrac{y^{3}}{3}=\dfrac{x^{3}}{3}-\dfrac{4xy^{3}}{3}+c $$

None of these

Solution

Question 2

$$ x^{2}+\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c $$

$$ x-\dfrac{y^{3}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c $$

$$ x-\dfrac{y^{2}}{2}+\dfrac{1}{3}(x^{2}+y^{2})^{3/2} = c $$

None of these

Solution

Question 3

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

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

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

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

Solution

Question 4

$$0$$

$$1$$

$$2$$

More than $$2$$

Solution

Question 5

$$2ye^{2x}=Ce^{2x}+1 $$

$$2ye^{2x}=Ce^{2x}-1$$

$$ ye^{2x}=Ce^{2x}+2 $$

None of these

Solution

Question 6

$$y=\tan^{-1}x$$

$$y-x=k(1+xy)$$

$$x=\tan^{-1}y$$

$$\tan ( xy)=k$$

Solution

Question 7

$$ \tan^{-1} y + \sin ^{-1} x=C$$

$$ \tan^{-1} x + \sin ^{-1} y=C$$

$$ \tan^{-1} x . \sin ^{-1} y=C$$

$$ \tan^{-1} x - \sin ^{-1} y=C$$

Solution

Question 8

$$e^{\dfrac{(1-x)^1}{2}}$$

$$e^{\dfrac{(1+x)^2}{2}} -1$$

In $$(1+x)-1$$

$$1+x$$

Solution

Question 9

$$y=(Ax+B) e^x$$

$$y=(Ax+B) e^{-x}$$

$$y=Ae^x +Be^{-x}$$

$$y=A\cos x+B\sin x$$

Solution

Question 10

$$\log\left(\displaystyle \frac{x}{1+\sqrt{1+x^{2}}}\right)+\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c$$

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

$$\displaystyle \log\left(\frac{x}{\sqrt{1+x^{2}}}\right)=c$$

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

Solution

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