Differential Equations Test

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Differential Equations Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

$$y = \sin kt$$ satisfies the differential equation $$y'' + 9y = 0$$. Then $$k=$$

A
$$\pm3$$

B
$$0$$

C
$$\pm2$$

D
$$\pm 4$$

Solution

Given, $$y = \sin kt$$ satisfies the differential equation $$y'' + 9y = 0$$.
Then, we have,
$$-k^2\sin kt+9\sin kt=0$$
or, $$k^2-9=0$$ [ Since $$\sin kt\ne 0$$]
or, $$k=\pm 3$$.
Question 2
1 / -0

The solution of the differential equation $$\dfrac{dy}{dx} = \dfrac{3y - 7x - 3}{3x - 7y + 7}$$ is

A
$$(y - x - 2)^5 (y + x - 5)^7 = c$$

B
$$(y - x - 5)^2 (y + x - 1)^7 = c$$

C
$$(y - x - 7)^2 (y + x - 5) = c$$

D
$$(y - x - 1)^2 (y + x - 1)^5 = c$$

Solution

$$\dfrac{dy}{dx}=\dfrac{3y-7x-3}{3x-7y+7}$$ put $$y=Y+1$$ and $$x=X$$

$$\dfrac{dy}{dx}=\dfrac{dY}{dX}$$

$$\dfrac{dY}{dX}=\dfrac{3Y-7X}{3X-7Y}$$ put $$Y=vX$$ $$\Rightarrow \dfrac{dY}{dX}=v+X\dfrac{dv}{dX}$$
$$\Rightarrow V+X\dfrac{dv}{dX}=\dfrac{3v-7}{3-7v}\Rightarrow \dfrac{3v-7v^2-3v+7}{(3-7v)}=-X\dfrac{dv}{dX}$$
$$X\dfrac{dv}{dX}=\dfrac{7(v+1)(v-1)}{(3-7v)}$$ $$\Rightarrow \boxed{\frac{dv(3-7v)}{(v+1)(v-1)}=7\frac{dx}{X}}$$
$$\dfrac{3-7v}{(v+1)(v-1)}=\dfrac{7-4-7v}{(v+1)(v-1)}=\dfrac{7(1-v)}{(v+1)(v-1)}-\dfrac{4}{(v+1)(v-1)}$$
$$=\dfrac{-7}{(v+1)}-\dfrac{4}{(v+1)(v-1)}$$
$$\Rightarrow \dfrac{3-7v}{(v+1)(v-1)}=\dfrac{-7}{(v+1)}-\dfrac{4}{(v+1)(v-1)}=\dfrac{-7}{(v+1}-\dfrac{4}{2}\left[\dfrac{1}{(v-1)}-\dfrac{1}{(v+1)}\right]=\dfrac{-7+2}{(v+1)}\left[\dfrac{1}{(v+1)}-\dfrac{1}{(v-1)}\right]$$
$$\boxed{\dfrac{3-7v}{(v+1)(v-1)}=\left[\dfrac{-5}{(v+1)}-\dfrac{2}{(v-1)}\right]}$$
$$\Rightarrow dv\left[\dfrac{-5}{(v+1)}-\dfrac{2}{(v-1)}\right]$$ $$7\dfrac{dx}{x}$$
$$\Rightarrow |(v+1)^5(v-1)^2|=7|n||x|$$
$$\Rightarrow X^7(v+1)^5(v-1)^2 = c$$
$$\Rightarrow x^7\dfrac{(Y+x)^5(Y-x)^2}{x^7}=c$$
$$\Rightarrow (y+x)^5(y-x)^2=c$$
$$\Rightarrow (y-1+x)^5(y-1-x)^2=c$$
$$\Rightarrow \boxed{(y+x-1)^5(y-x-1)^2=c}$$
Question 3
1 / -0

The differential equation for all the straight lines which are at a unit distance from the origin is

A
$${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1 - {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$$

B
$${\left( {y + x\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$$

C
$${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1+{\left( {\dfrac{{dy}}{{dx}}} \right)^2}$$

D
$${\left( {y + x\frac{{dy}}{{dx}}} \right)^2} = 1 - {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$$

Solution

Let the line be : $$y=mx+c$$

$$\Rightarrow 1=\dfrac {|c|} {\sqrt {m^{2}+1}}$$

$$\Rightarrow y=mx\pm \sqrt {m^{2}+1}$$

$$\Rightarrow \dfrac {dy} {dx} =m$$

$$\Rightarrow y=x\dfrac {dy} {dx} \pm \sqrt {\left (\dfrac {dy} {dx} \right) ^{2}+1} $$

$$\Rightarrow \left (y-x\dfrac {dy} {dx} \right) ^{2}=1+\left (\dfrac {dy} {dx} \right) ^{2}$$
Question 4
1 / -0

Which of the following differential equation is linear ?

A
$$\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}+2y=0$$

B
$$\dfrac{d^2y}{dx^2}+y\dfrac{dy}{dx}+x=0$$

C
$$\dfrac{d^{2}y}{dx^2}+\dfrac{y}{x}+\sin y=x^2$$

D
$$(1+x)\dfrac{dy}{dx}-xy=1$$

Solution

Only in the equation $$(1+x)\dfrac{dy}{dx}-xy=1$$ , both the terms $$\dfrac{dy}{dx}$$ and $$y$$ are linear. Hence this equation is a linear differential equation.
Question 5
1 / -0

Solution of the different equation, $$ydx - xdy + x{y^2}dx = 0$$ can be.

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

B
$$2y + {y^2}x = \lambda y$$

C
$$2y - {y^2}x = \lambda y$$

D
none of these

Solution

$$\cfrac{ydx-xdy}{y^2} = d(\cfrac{x}{y})$$
Hence,
$$\cfrac{ydx-xdy}{y^2} + xdx = 0$$
$$d(\cfrac{x}{y}) + xdx = 0$$
Integrating both sides we get,
$$\cfrac{x}{y} + \cfrac{x^2}{2} = C$$
$$2x+x^2 y = \lambda y$$
Question 6
1 / -0

The solution of the differential equaton $$3{e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}ydy = 0$$ is

A
$$\tan y = c{\left( {1 - {e^x}} \right)^3}$$

B
$${\left( {1 - {e^x}} \right)^3}\tan y = c$$

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

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

Solution

$$3{e}^{x}\tan y dx+(1-{e}^{x})\sec ^{ 2 }{ y } dy=0$$
$$\cfrac { 3{ e }^{ x } }{ 1-{ e }^{ x } } dx+\cfrac { \sec ^{ 2 }{ y } dy\quad }{ \tan { y } } =0$$
Integrating both sides
$$-3\log{1-{ e }^{ x }}+\log{\tan y)}=\log{e}$$
$$\log{(\tan y)}=\log C+3\log{(1-{ e }^{ x })}$$
$$\log{(\tan y)}=\log { \left( C{ \left( 1-{ e }^{ x } \right) }^{ 3 } \right) } $$
$$\tan y=C{(1-{e}^{x})}^{3}$$
Question 7
1 / -0

Solve $$(1+\cos x)dy=(1-\cos x)dx$$.

A
$$y = \displaystyle \cot \cfrac{x}{2} - x +C$$

B
$$y = \displaystyle \tan \cfrac{x}{2} - x +C$$

C
$$y = \displaystyle \sin \cfrac{x}{2} - x +C$$

D
None of these

Solution

$$dy = \cfrac{1-\cos x}{1+\cos x} dx$$
Integrating both sides,we get
$$y = \displaystyle \int -\cfrac{\cos x+1-2}{\cos x+1}dx$$
$$y = \displaystyle \int -dx+ \int 2\cfrac{1}{\cos x+1}dx$$
$$y = \displaystyle -x+\int \cfrac{\sec^2 \cfrac{x}{2}}{2} dx$$
$$y = \displaystyle \tan \cfrac{x}{2} - x +C$$
Question 8
1 / -0

If $$\dfrac{dy}{dx}+y\tan x=\sin 2x$$ and $$y(0)=1$$, then $$y(\pi)=?$$

A
$$1$$

B
$$-1$$

C
$$-5$$

D
$$5$$

Solution

$$I.F. = e^{\int tanxdx}$$
$$\implies I.F. = e^{-lncosx}$$
$$I.F. = \cfrac{1}{cosx}$$
Multiplying with $$I.F.$$ in the given equation,we get-
$$\cfrac{dy}{cosx dx} + \cfrac{ysinx}{cos^2x} = 2sinx$$
$$d(\cfrac{y}{cosx}) = 2sinxdx$$
Integrating both sides, we get-
$$\cfrac{y}{cosx} = -2cosx + C$$
$$y(0) = 1$$
$$1 = -2+C$$
$$C = 3$$
For $$x=\pi$$ we get
$$\cfrac{y}{-1} = 2+3$$
$$y = -5$$
Question 9
1 / -0

Solve:
$$ \dfrac {dy}{dx} = y \cot x$$

A
$$y=k\tan x$$

B
$$y=k\csc x$$

C
$$y=k\sin x$$

D
$$y=k\cos x$$

Solution

$$\dfrac{dy}{dx}=y \cot x$$
$$\Rightarrow \dfrac{dy}{y}= \cot x dx$$
$$\Rightarrow \dfrac{dy}{y} = \dfrac{\cos x}{\sin x} dx$$
$$\Rightarrow ln |y|+ln |c|= ln |\sin x|$$
$$\Rightarrow ln |Cy| = ln |\sin \, x|$$
$$\Rightarrow cy= \sin x$$
$$\Rightarrow y= k \sin x$$ where k= constant
Question 10
1 / -0

The D.E. obtained from $$(y - a)^2 = 4 (x - b)$$ is given by

A
$$2y_2 + y_3 = 0$$

B
$$2y_2 = y_3$$

C
$$2y_1 + y^3_2 = 0$$

D
$$2y_2 + y^3_1 = 0$$

Solution

$$(y-a^2)=4(x-b)$$
$$y^2+a^2-2ay=4x-4b$$
Differentiating w.r.t to x, we get
$$2yy_1-2ay_1=4$$
$$yy_1-ay_1=2$$
$$yy_1-2=ay_1$$
$$y-\dfrac{2}{y_1}=a$$
Differentiating both sides wrt $$x_1$$ we get
$$y_1=\dfrac{2}{y_1^2}y_2=0$$
$$y^3_1+2y_2=0$$
$$\therefore y^3_1+2y_2=0$$
where $$y_1=\dfrac{dy}{dx}$$ & $$y_2=\dfrac{d^2y}{dx^2}$$.

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 14

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Differential Equations Test - 14

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SHARING IS CARING

Question 1

$$\pm3$$

$$0$$

$$\pm2$$

$$\pm 4$$

Solution

Question 2

$$(y - x - 2)^5 (y + x - 5)^7 = c$$

$$(y - x - 5)^2 (y + x - 1)^7 = c$$

$$(y - x - 7)^2 (y + x - 5) = c$$

$$(y - x - 1)^2 (y + x - 1)^5 = c$$

Solution

Question 3

$${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1 - {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$$

$${\left( {y + x\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$$

$${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1+{\left( {\dfrac{{dy}}{{dx}}} \right)^2}$$

$${\left( {y + x\frac{{dy}}{{dx}}} \right)^2} = 1 - {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$$

Solution

Question 4

$$\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}+2y=0$$

$$\dfrac{d^2y}{dx^2}+y\dfrac{dy}{dx}+x=0$$

$$\dfrac{d^{2}y}{dx^2}+\dfrac{y}{x}+\sin y=x^2$$

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

Solution

Question 5

$$2x + {x^2}y = \lambda y$$

$$2y + {y^2}x = \lambda y$$

$$2y - {y^2}x = \lambda y$$

none of these

Solution

Question 6

$$\tan y = c{\left( {1 - {e^x}} \right)^3}$$

$${\left( {1 - {e^x}} \right)^3}\tan y = c$$

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

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

Solution

Question 7

$$y = \displaystyle \cot \cfrac{x}{2} - x +C$$

$$y = \displaystyle \tan \cfrac{x}{2} - x +C$$

$$y = \displaystyle \sin \cfrac{x}{2} - x +C$$

None of these

Solution

Question 8

$$1$$

$$-1$$

$$-5$$

$$5$$

Solution

Question 9

$$y=k\tan x$$

$$y=k\csc x$$

$$y=k\sin x$$

$$y=k\cos x$$

Solution

Question 10

$$2y_2 + y_3 = 0$$

$$2y_2 = y_3$$

$$2y_1 + y^3_2 = 0$$

$$2y_2 + y^3_1 = 0$$

Solution

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