Differential Equations Test

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Differential Equations Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

If $$y = \sqrt{\dfrac{1-x}{1+x}}$$ then find $$(1 - x^2) \dfrac{dy}{dx} + y$$ =

A
$$1$$

B
$$-1$$

C
$$2$$

D
$$0$$

Solution

$$y= \sqrt{ \dfrac{1-x}{1+x} }$$
$$y^{2}= \dfrac{1-x}{1+x}$$
$$y.y = \dfrac{ (1+x)(-1)-(1-x)(1) }{ (1+x)^{2} }$$
$$y.y = \dfrac{-1-x-1+x}{(1+x)^{2}}$$
$$yy = \dfrac{-2}{ (1+x)^{2} }$$
$$y \dfrac{dy}{dx}= \dfrac{-2}{(1+x)^{2}}$$
$$ \dfrac{dy}{dx}= \dfrac{-2}{y (1+x)^{2}}$$
Given
$$(1-x^{2}) \left( \dfrac{dy}{dx} \right)+y$$
$$= (1-x^{2}) \dfrac{-2}{ y (1+x)^{2}}+ y$$
$$= \dfrac{-(1-x)}{(1+x)}+ y$$
$$= \left(1- \dfrac{(1-x)}{(1+x)}. \dfrac{1}{y^{2}} \right)$$
$$=y \left(1- \dfrac{(1-x)}{(1+x)} \times \dfrac{1+x}{1-x} \right)$$
$$=0$$
Question 2
1 / -0

The solution of $$x \dfrac{dy}{dx} + y logy = xy e^x$$ is

A
$$ logy = (x - 1)e^x + c$$

B
$$(x -1) logy = xe^x + c$$

C
$$x logy = (x + 1)e^x + c$$

D
$$x logy = (x - 1)e^x + c$$

Solution

$$ x\dfrac{dy}{dx}+ylny = xye^{x} $$
$$ \dfrac{x}{y}\dfrac{dy}{dx}+lny = xe^{x} $$
Let $$ xlny = t $$
$$ lny +\dfrac{x}{y}\dfrac{dy}{dx} = \dfrac{dt}{dx} $$
$$ \dfrac{dt}{dx} = x.e^{x} $$
$$\displaystyle \int dt = \int x-e^{x} $$
$$\displaystyle t = xe^{x}-\int e^{x}dx $$
$$ t = e^{x} (x-1)+c $$
prove value of 1
$$ xlny = e^{x}(x-1)+c $$
Question 3
1 / -0

The real value of n for which substitution $$y = {u^n}$$ will transform differential equation $$2{x^4}y\frac{{dy}}{{dx}} + {y^4} = 4{x^6}$$ into homogeneous equation

A
$$\frac{1}{2}$$

B
$$1$$

C
$$\frac{3}{2}$$

D
$$4$$

Solution

$$2{x^4}y\frac{{dy}}{{dx}} + {y^4} = 4{x^6}$$

$$2y\frac{{dy}}{{dx}} + \frac{{{y^4}}}{{{x^4}}} = 4{x^2}$$

$$y = {u^n}$$

$$\frac{{dy}}{{dx}} = n{x^{n - 1}}$$

$$2n{u^{n - 1}}{u^n}\frac{{du}}{{dx}} + \frac{{{u^{4x}}}}{{{x^4}}} = 4{x^2}$$

$$2x - 1 = 2$$

$$2x = 3$$

$$x = \frac{3}{2}$$
Question 4
1 / -0

The solution of the differential equation $$x\dfrac{dy}{dx}=y+x\tan \dfrac{y}{x}$$ is :

A
$$\sin \dfrac{x}{y}=cx$$

B
$$\sin \dfrac{y}{x}=cx$$

C
$$\sin \dfrac{x}{y}=cy$$

D
$$\sin \dfrac{y}{x}=cy$$

Solution

$$\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}+\tan\left(\dfrac{y}{x}\right)$$
Let $$y=vx$$
$$\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$
$$\therefore v+x\dfrac{dv}{dx}v+\tan(v)$$
$$\Rightarrow \dfrac{dv}{\tan(v)}=\dfrac{dx}{x}$$
$$\displaystyle \Rightarrow \int \cot(v)dv=\int\dfrac{1}{x}dx$$
$$\Rightarrow ln|\sin(v)|=|n|x|+k=|n|x|+|nle^k|$$
$$\Rightarrow ln|\sin(v)|+ln |e^kx|$$
$$\Rightarrow \sin(v)=cx$$ $$\{e^k=c\}$$
$$\Rightarrow \boxed{\sin\left(\dfrac{y}{x}\right)=cx}$$
Question 5
1 / -0

If $$y(x)$$ is the solution of the differential equation $$(x+2)\dfrac{dy}{dx}=x^{2}+4x-9, x \neq -2$$ and $$y(0)=0$$, then $$y(-4)$$ is equal to

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

Consider the given equation.

$$ \int{dy}=\int{\dfrac{\left( {{x}^{2}}+4x-9\, \right)}{(x+2)}}dx $$

Integrate on both side w.r.t $$ dx & dy $$

$$ \int{dy}=\int{\dfrac{\left( {{x}^{2}}+4x-9\, \right)}{(x+2)}}dx $$

$$ \int{dy=\int{\dfrac{{{\left( x+2 \right)}^{2}}-13}{\left( x+2 \right)}}}dx $$

$$ \int{dy}=\int{\left( x+2 \right)}dx+\int{\dfrac{13}{\left( x+2 \right)}}dx $$

$$ y=\dfrac{{{x}^{2}}}{2}+2x+13\ln \left( x+2 \right)+C $$

$$ y\left( 0 \right)=0\,\,\,\,\,\,\,\,y(-4) $$

$$ 0=0+0+13\ln 2+C $$

$$ C=-13\ln 2\,\,\,\,........\left( 3 \right) $$

$$ y=\dfrac{{{x}^{2}}}{2}+2x+13\ln \left( x+2 \right)-13\ln 2\,\,\,\,\,..........\left( 4 \right) $$

$$ y\left( -4 \right)=\dfrac{{{\left( -4 \right)}^{2}}}{2}+2\left( -4 \right)+13\ln \left( \left( -4 \right)+2 \right)-13\ln 2 $$

$$ y\left( -4 \right)=0 $$
hence, this is the correct answer.
Question 6
1 / -0

The solution of the differential equation $$ydx + (x+x^2y)dy = 0$$ is-

A
$$\dfrac{1}{xy} + \log y = c$$

B
$$\log y = cx$$

C
$$-\dfrac{1}{xy} = c$$

D
$$-\dfrac{1}{xy} + \log y = c$$

Solution

Given,
$$ydx+(x+x^2y)dy=0$$

divide above equation by $$x^2ydy$$

$$\dfrac{1}{x^2}.\dfrac{dx}{dy}+\frac{1}{xy}=1=0$$$$~~~~~~~-(1)$$

Let$$\dfrac{1}{x}=t$$

$$\to -\dfrac{1}{x}\dfrac{dx}{dy}=\dfrac{dt}{dy}$$

$$1\to -\dfrac{dt}{dy}+\dfrac{t}{y}+1=0$$
or
$$\dfrac{dt}{dy}-\dfrac{t}{y}=1$$
p=$$-y^{-1}$$,q=1

so I.F.=$$e^{\int p.dy}$$

=$$e^{\int \dfrac{-1}{y}dy}$$

=$$e^{-logy}$$

=$$\dfrac{1}{y}$$
So complete solution

$$t.\dfrac{1}{y}=\int 1.\dfrac{1}{y}dy$$

$$t.\dfrac{1}{y}=logy+c$$

$$\to \dfrac{1}{xy}=logy+c$$

$$\to -\dfrac{1}{xy}+logy=c$$
Question 7
1 / -0

Soluation of D.E. $$\dfrac{{dy}}{{dx}} = \dfrac{{3x + 4y + 3}}{{12x + 16y - 4}}$$ is

A
$$y = 4x + \ell n\left| {3x + 4y} \right| + C$$

B
$$4y = x + \ell n\left| {3x + 4y} \right| + C$$

C
$$y = \ell n\left| {3x + 4y} \right| + C$$

D
$$x + y = \ell n\left| {3x + 4y} \right| + C$$

Solution

$$\dfrac {dy}{dx}=\dfrac {3x+4y+3}{12x+16y-4}$$
Let $$2=3x+4y\ \Rightarrow \dfrac {dz}{dx}=3+4\dfrac {dy}{dx}=\dfrac {1}{4}\left (\dfrac {dz}{dx} -3 \right)$$
$$\therefore \ \dfrac {1}{4}\left (\dfrac {dz}{dx} -3 \right)=\dfrac {z+3}{4z-4}\Rightarrow \dfrac {dz}{dx}-3=\dfrac {4z+12}{4z-4}$$
$$\Rightarrow \ \dfrac {dz}{dx}=\dfrac {4z+12}{4z -4}+3=\dfrac {16z}{4z-4}=\dfrac {4z}{z-1}$$
$$\Rightarrow \ \dfrac {z-1}{z}dz=4dx$$
$$\Rightarrow \ \displaystyle \int \dfrac {z-1}{z}dz =4\displaystyle \int dx$$
$$\Rightarrow \ z-\ln |z|=4x+C$$
$$\Rightarrow \ 3x+4y-\ln |3x+4y|=4x+C$$
$$\Rightarrow \ \boxed {4y=x+\in |3x+4y|+C}$$
Question 8
1 / -0

The general solution of the differential equation $$\dfrac { dy }{ dx } + y\cot { x } = \csc { x }$$, is

A
$$x + y sin { x } = C$$

B
$$x + y cos { x } = C$$

C
$$y+x(\sin { x + \cos { x } } )$$

D
$$y\sin { x = x + C }$$

Solution

$$\dfrac{dy}{dx}+y\cot x=cosec n$$
$$I.F=e^{\displaystyle\int \omega t xdx}=e^{en(\sin n)}$$
$$=\sin n$$
$$y\cdot \sin x=\displaystyle\int cosec x. mn-dx+c$$
$$y\sin n=x+c$$
$$\neq y\sin x=x+c$$.
Question 9
1 / -0

The solution of $$\cfrac { dy }{ dx } =\left( \cfrac { ax+b }{ cy+d } \right) $$ represents a parabola if:-

A
a=0,c=0

B
a=1,c=2

C
a=0,$$c\neq0$$

D
a=1,c=1

Solution

We hyave,
$$\dfrac{dy}{dx}=\dfrac{ax+b}{cy+d}$$

For above differential equation to represent equation of parabola, one variable must have highest degree $$2$$ and other must have degree $$1$$.

When we take $$a=0$$ and $$c≠0$$, then we get
$$\dfrac{dy}{dx}=\dfrac{0+b}{cy+d}$$

$$(cy+d)\ dy=b\ dx$$

$$\int (cy+d)\ dy=\int b\ dx$$

$$\dfrac{cy^2}{2}+dy=bx+C$$ which is equation of parabola.

Hence, this is the answer.
Question 10
1 / -0

The integrating factor of the differential equation $$\cfrac { dy }{ dx } -y\tan { x } =\cos { x } $$ is,

A
$$\cos x$$

B
$$\sin x$$

C
$$\sec x$$

D
$$\tan x$$

Solution

We have,
$$\cfrac { dy }{ dx } -y\tan { x } =\cos { x } $$

We know that the general equation
$$\dfrac{dy}{dx}+Py=Q$$

So,
$$P=\ -tan x, Q=\cos x$$

We know that
$$I.F=e^{\int P\ dx}$$

Therefore,
$$I.F=e^{\int \tan x\ dx}$$

$$I.F=e^{-\ln |\cos x|}$$

$$I.F=e^{\ln \left ( \dfrac{1}{\sec x} \right )}$$

$$I.F= \dfrac{1}{\sec x}$$

$$I.F=\cos x$$

Hence, this is the answer.

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 36

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Differential Equations Test - 36

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SHARING IS CARING

Question 1

$$1$$

$$-1$$

$$2$$

$$0$$

Solution

Question 2

$$ logy = (x - 1)e^x + c$$

$$(x -1) logy = xe^x + c$$

$$x logy = (x + 1)e^x + c$$

$$x logy = (x - 1)e^x + c$$

Solution

Question 3

$$\frac{1}{2}$$

$$1$$

$$\frac{3}{2}$$

$$4$$

Solution

Question 4

$$\sin \dfrac{x}{y}=cx$$

$$\sin \dfrac{y}{x}=cx$$

$$\sin \dfrac{x}{y}=cy$$

$$\sin \dfrac{y}{x}=cy$$

Solution

Question 5

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 6

$$\dfrac{1}{xy} + \log y = c$$

$$\log y = cx$$

$$-\dfrac{1}{xy} = c$$

$$-\dfrac{1}{xy} + \log y = c$$

Solution

Question 7

$$y = 4x + \ell n\left| {3x + 4y} \right| + C$$

$$4y = x + \ell n\left| {3x + 4y} \right| + C$$

$$y = \ell n\left| {3x + 4y} \right| + C$$

$$x + y = \ell n\left| {3x + 4y} \right| + C$$

Solution

Question 8

$$x + y sin { x } = C$$

$$x + y cos { x } = C$$

$$y+x(\sin { x + \cos { x } } )$$

$$y\sin { x = x + C }$$

Solution

Question 9

a=0,c=0

a=1,c=2

a=0,$$c\neq0$$

a=1,c=1

Solution

Question 10

$$\cos x$$

$$\sin x$$

$$\sec x$$

$$\tan x$$

Solution

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