Differential Equations Test

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Differential Equations Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

$$\dfrac{dy}{dx}=\dfrac{4x+2y+1}{x-2y+3}$$ is a differential equation of the type:

A
variable separable

B
exact

C
Homogeneous

D
non homogeneous

Solution

Homogeneous differential equations involve only derivatives of y and terms involving y, and they’re set to 0, as in this equation:
$$\dfrac{d^4}{dx^4}+x\dfrac{d^2y}{dx^2}+y^2=02$$
Non homogeneous differential equations are the same as homogeneous differential equations, except they can have terms involving only x (and constants) on the right side, as in this equation
$$\dfrac{d^4}{dx^4}+x\dfrac{d^2y}{dx^2}+y^2=6x+3$$
Question 2
1 / -0

The solution of $$\dfrac{dy}{dx}=\dfrac{1+y^{2}}{\sec x}$$ is

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

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

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

D
$$\log\left ( 1+y^{2} \right )=\cos x+c$$

Solution

$$\dfrac{dy}{dx}=(1+y^{2})\cos\ x$$
$$\Rightarrow \int \dfrac{dy}{1+y^{2}}=\int \cos\ x\ dx$$
$$\Rightarrow \tan^{-1}y=\sin\ x+c$$
Question 3
1 / -0

The solution of $$\dfrac{dy}{dx}+2x=e^{3x}$$ is:

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

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

C
$$y=e^{3x}-x^{2x}+c$$

D
$$y=e^{3x}-x^{2}+c$$

Solution

$$\dfrac{dy}{dx}+2x=e^{3x}$$

$$\Rightarrow dy=(e^{3x}-2x)dx$$

$$\int dy =\int e^{3x}dx-2 \int x dx$$

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

$$y=\dfrac{e^{3x}}{3}-x^{2}+c$$
Question 4
1 / -0

The solution of $$x^{3}dy-y^{3}dx=0$$ is:

A
$$x^{4}-y^{4}=c$$

B
$$x^{2}-y^{2}=c$$

C
$$\dfrac{1}{x}-\dfrac{1}{y}=c$$

D
$$\dfrac{1}{x^{2}}-\dfrac{1}{y^{2}}=c$$

Solution

$$x^{3}dy-y^{3}dx=0$$

$$\Rightarrow \int \dfrac{dy}{y^{3}}=\int \dfrac{dx}{x^{3}}+c$$

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

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

The solution of $$\displaystyle \frac{dy}{dx}=\displaystyle \frac{3(y+1)}{x-2}$$ is:

A
$$y -1= c (x - 2)^{3}$$

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

C
$$y +1= c (x - 2)^{2}$$

D
$$y -1= c (x - 2)^{2}$$

Solution

$$\dfrac{dy}{dx}=\dfrac{3(y+1)}{x-2}$$

$$\Rightarrow \int \dfrac{dy}{3(1+y)}=\int \dfrac{dx}{x-2}+c$$

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

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

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

$$\Rightarrow (1+y)=(x-2)^{3}c$$
Question 6
1 / -0

The solution of $$\dfrac{d^{2}y}{dx^{2}}=xe^{x}+1$$ is:

A
$$\displaystyle y=(x-1)e^{x}+\frac{1}{2}x^{2}+C_{1}x+C_{2}$$

B
$$\displaystyle y=(x-2)e^{x}+\frac{1}{2}x^{2}+C_{1}x+C_{2}$$

C
$$\displaystyle y=(x+2)e^{x}+\frac{1}{2}x^{2}+C_{1}x+C_{2}$$

D
$$\displaystyle y=(x+2)e^{x}+C_{1}$$

Solution

$$\dfrac{d}{dx}\left(\dfrac{dy}{dx} \right)=xe^{x}+1$$
$$\displaystyle \Rightarrow \int d\left(\dfrac{dy}{dx} \right)=\int (xe^{x}+1)dx+c_{1}$$
$$\displaystyle \Rightarrow \frac{dy}{dx}=xe^{x}-e^{x}+x+c_{1}$$
$$\displaystyle \int dy =\int ( xe^{x}-e^{x}+x+c_{1})dx+c_{2}$$
$$\displaystyle \Rightarrow \int dy =xe^{x}-e^{x}-e^{x}+\frac{x^{2}}{2}+c_{1}x+c_{2}$$
$$\displaystyle \Rightarrow y= xe^{x}-2e^{x}+\frac{x^{2}}{2}+c_{1}x+c_{2}$$
$$\displaystyle y=e^{x}(x-2)+\frac{x^{2}}{2}+c_{1}x+c_{2}$$
Question 7
1 / -0

The solution of the differential equation
$$\displaystyle \frac{dy}{dx}=\displaystyle \frac{xy+y}{xy+x}$$ is

A
$$x + y = log \left (\displaystyle \frac{cy}{cx} \right )$$

B
$$x + y = \log (cxy)$$

C
$$x - y = \log \left ( \displaystyle \frac{cx}{y} \right )$$

D
$$y - x = \log \left ( \displaystyle \frac{cx}{y} \right )$$

Solution

$$\displaystyle \dfrac{dy}{dx}=\displaystyle \dfrac{y(x+1)}{x(y+1)}$$
$$\displaystyle \dfrac{(y+1)}{y}dy=\displaystyle \dfrac{(x+1)}{x}dx$$
$$\displaystyle \int (1+{1}/{y})dy=\int (1+{1}/{x})dx+c_{1}$$
$$\displaystyle \Rightarrow y+log\ y=x+log\ x+c_{1}$$
$$\displaystyle (y-x)+log{y}/{x}=c_{1}$$
$$\displaystyle \Rightarrow (y-x)=-log\ c_{2}-log{y}/_{x}$$ , $$c_{1}=-log\ c_{2}$$
$$\displaystyle =-log(\ {c_{2}y}/{x})$$
$$\displaystyle (y-x)=log(\ {c_{3}x}/{y})$$ , $$\displaystyle c_{3}={1}/{c_{2}}$$
Question 8
1 / -0

The solution of $$\dfrac{dy}{dx}=\text{cosech }y$$ is:

A
$$x=\cosh y+c$$

B
$$x=\text{sech}\ y+c$$

C
$$x=\sinh y+c$$

D
$$x=\coth y+c$$

Solution

$$\dfrac{dy}{dx}=\text{cosech }y$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac1{\text{sinh }y}$$
$$\Rightarrow \sinh y\ dy =dx$$
Integrating both sides, we get
$$\displaystyle \int \sinh y\ dy=x+c$$
$$\Rightarrow \cosh y =x+c$$
Question 9
1 / -0

The solution of $$(1+e^{x})ydy=e^{x}dx$$ is:

A
$$y^{2}=\log c(e^{x}+1)$$

B
$$\dfrac{y^{2}}{2}=\log ce^{x}$$

C
$$\dfrac{y^{2}}{2}=\log c(e^{x}+1)$$

D
$$2y = \log c (e^{x}+1)$$

Solution

$$(1+e^{x})ydy=e^{x}dx$$

$$\int ydy =\int (\dfrac{e^{x}}{1+e^{x}})dx+c$$

$$\Rightarrow \dfrac{y^{2}}{2}=\log(1+e^{x})c$$
Question 10
1 / -0

The solution of the differential equation $$(1+x^{2})\displaystyle \frac{dy}{dx}=2 x \cot y$$, is:

A
$$\sec y = c (1+x^{2})$$

B
$$\cos y = c (1+x^{2})$$

C
$$\tan y = c (1+x^{2})$$

D
$$\cot y = c (1+x^{2})$$

Solution

Given:
$$(1+x^{2})\displaystyle \frac{dy}{dx}=2x\ \cot\ y$$
Rearranging this equation
$$\displaystyle\dfrac{dy}{\cot y}=\displaystyle\frac{2x}{1+x^{2}}dx$$
Integrating both sides, we get
$$\displaystyle\int \tan\ y\ dy=\displaystyle \int \frac{2x}{1+x^{2}}dx+c$$
$$\Rightarrow \log\left |\sec\ y \right |=\log(1+x^{2})+\log\ c$$
$$\Rightarrow\log\ \sec\ y=\log\ c(1+x^{2})$$
$$\Rightarrow \sec\ y=c(1+x^{2})$$
where, $$c$$ is the constant of integration.
Hence, option A.

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 16

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Differential Equations Test - 16

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SHARING IS CARING

Question 1

variable separable

exact

Homogeneous

non homogeneous

Solution

Question 2

$$\tan^{-1}y=\cos x+c$$

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

$$y^{3}=\csc x +c$$

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

Solution

Question 3

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

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

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

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

Solution

Question 4

$$x^{4}-y^{4}=c$$

$$x^{2}-y^{2}=c$$

$$\dfrac{1}{x}-\dfrac{1}{y}=c$$

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

Solution

Question 5

$$y -1= c (x - 2)^{3}$$

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

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

$$y -1= c (x - 2)^{2}$$

Solution

Question 6

$$\displaystyle y=(x-1)e^{x}+\frac{1}{2}x^{2}+C_{1}x+C_{2}$$

$$\displaystyle y=(x-2)e^{x}+\frac{1}{2}x^{2}+C_{1}x+C_{2}$$

$$\displaystyle y=(x+2)e^{x}+\frac{1}{2}x^{2}+C_{1}x+C_{2}$$

$$\displaystyle y=(x+2)e^{x}+C_{1}$$

Solution

Question 7

$$x + y = log \left (\displaystyle \frac{cy}{cx} \right )$$

$$x + y = \log (cxy)$$

$$x - y = \log \left ( \displaystyle \frac{cx}{y} \right )$$

$$y - x = \log \left ( \displaystyle \frac{cx}{y} \right )$$

Solution

Question 8

$$x=\cosh y+c$$

$$x=\text{sech}\ y+c$$

$$x=\sinh y+c$$

$$x=\coth y+c$$

Solution

Question 9

$$y^{2}=\log c(e^{x}+1)$$

$$\dfrac{y^{2}}{2}=\log ce^{x}$$

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

$$2y = \log c (e^{x}+1)$$

Solution

Question 10

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

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

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

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

Solution

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