Differential Equations Test

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Differential Equations Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The solution of $$\frac{dy}{dx}=|x|$$ is :

A
$$y=\frac{x|x|}{2}+c$$

B
$$y=\frac{|x|}{2}+c$$

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

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

Solution

$$\cfrac{dy}{dx}=x$$ for $$x>0$$
$$\cfrac{dy}{dx}=-x$$ for $$x\le 0$$
For $$x>0,$$
$$\cfrac{dy}{dx}=x$$
$$\Rightarrow y=\cfrac{{x}^{2}}{2}+c$$ ......... [By integrating]
For $$x\le 0$$
$$\cfrac{dy}{dx}=-x$$
$$y=\cfrac{{-x}^{2}}{2}+c$$ ........ [By integrating]
$$\Rightarrow$$$$y=\cfrac { \left| x \right| x }{ 2 } +c$$
Question 2
1 / -0

Directions For Questions

For the next three (03) items that follow :
The general solution of the differential equation $$(x^2 + x + 1)dy + (y^2 + y + 1)dx =0$$ is
$$(x + y + 1) = A(1 + Bx + Cy + Dxy)$$ where B, C and D are constants and A is parameter.

...view full instructions

What is C equal to ?

A
1

B
-1

C
2

D
None of the above

Solution

$${\textbf{Step-1: Finding the value of D. }}$$
$${\text{Given,}}$$
$$\Rightarrow(x^2+x+1)dy+(y^2+y+1)dx=0$$
$$\Rightarrow(x^2+x+1)dy=-(y^2+y+1)dx$$
$$\Rightarrow\dfrac{dy}{y^2+y+1}=-\dfrac{dx}{x^2+x+1}$$
$${\text{Integrating both sides}}$$
   $$\Rightarrow \int \dfrac{dy}{y^2+y+1}=-\int\dfrac{dx}{x^2+x+1}$$
$${\text{By using perfect square method}}$$
$$\Rightarrow\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}=-\int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$$

$$\Rightarrow \int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}+\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$$

$${\text{use formula}}$$ $$\int\dfrac{dx}{a^2+x^2}=\dfrac{1}{a}tan^{-1}\dfrac{x}{a}$$

$$\Rightarrow \dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2x+1}{2}]+\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2y+1}{2}]+\dfrac{2}{\sqrt3} tan^{-1}C=0$$

$${\text{[.... Where C is the integral constant]}}$$

  $${\text{use formula}}$$ $$tan^{-1}A+tan^{-1}B$$

$$\Rightarrow\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{\dfrac{2x+1}{3}+\dfrac{2y+1}{\sqrt3}}{1-(\dfrac {2x+1}{\sqrt3})(\dfrac{2y+1}{\sqrt3})}]=-\dfrac{2}{\sqrt3}tan^{-1} C$$

$$\Rightarrow \dfrac{2x+1+2y+1}{\sqrt 3}\times \dfrac{\sqrt3 \sqrt3}{3-4xy-2x-2y-1}=C$$

$$\Rightarrow \dfrac{\sqrt3(2x+2y+2)}{2-4xy-2x-2y}=C$$

$$\therefore \sqrt3\times2(x+y+1)=c\times 2(1-2xy-x-y)$$

$$\therefore (x+y+1)=\dfrac{C}{\sqrt3}(1-x-y-2xy)$$

$${\text{ Compare this equation with the given equation gives,}}$$

$$\therefore C=-1$$

$${\textbf{Hence option B is correct}}$$
Question 3
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The solution of the differential equation $$\dfrac {dy}{dx} = \dfrac {yf'(x) - y^{2}}{f(x)}$$ is:

A
$$f(x) = y + C$$

B
$$f(x) = y(x + C)$$

C
$$f(x) = x + C$$

D
None of the above

Solution

$$\dfrac {dy}{dx} = \dfrac {yf'(x) - y^{2}}{f(x)}$$
$$\Rightarrow yf'(x) dx - f(x) dy = y^{2}dx$$
$$\Rightarrow \dfrac {yf'(x) dx - f(x) dy}{y^{2}} = dx$$
$$\Rightarrow d\left \{\dfrac {f(x)}{y}\right \} = dx$$
On integration, we get
$$\dfrac {f(x)}{y} = x + C$$
$$\Rightarrow f(x) = y(x + C)$$
Question 4
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The solution of $$\dfrac{dy}{dx}=\sqrt{1-x^2-y^2+x^2y^2}$$ is:
where c is an arbitrary constant

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

B
$$2{\sin}^{-1}y=\sqrt{1-x^2}+{\sin}^{-1}x+c$$

C
$$2{\sin}^{-1}y=x \sqrt{1-x^2}+{\sin}^{-1}x+c$$

D
$$2{\sin}^{-1}y=x \sqrt{1-x^2}+{\cos}^{-1}x+c$$

Solution

Solution:
Given, $$\cfrac{dy}{dx}=\sqrt{1-x^2-y^2+x^2y^2}$$
$$=\sqrt{(1-x^2)-y^2(1-x^2)}$$
$$=\sqrt{(1-x^2)(1-y^2)}$$
$$=(\sqrt{1-x^2})(\sqrt{1-y^2})$$
or, $$\cfrac{dy}{\sqrt{1-y^2}}=(\sqrt{1-x^2})dx$$
On integrating both sides, we get
$$\int$$$$\cfrac{dy}{\sqrt{1-y^2}}=\int(\sqrt{1-x^2})dx$$
or, $$\sin^{-1}y=\cfrac12[x\sqrt{1-x^2}+\sin^{-1}x]+A$$
or, $$2\sin^{-1}y=x\sqrt{1-x^2}+\sin^{-1}x+C$$
Hence, C is the correct option.
Question 5
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The solution of the differential equation $$\dfrac {dy}{dx} = \tan \left (\dfrac {y}{x}\right ) + \dfrac {y}{x}$$ is:

A
$$\cos \left (\dfrac {y}{x}\right ) = cx$$

B
$$\sin \left (\dfrac {y}{x}\right ) = cx$$

C
$$\cos \left (\dfrac {y}{x}\right ) = cy$$

D
$$\sin \left (\dfrac {y}{x}\right ) = cy$$

Solution

$$\dfrac {dy}{dx} = \tan \left (\dfrac {y}{x}\right ) + \left (\dfrac {y}{x}\right )$$ ..... $$(i)$$
Take, $$\dfrac {y}{x} = v$$
$$\implies y = vx$$
$$\implies \dfrac {dy}{dx} = v + x\dfrac {dv}{dx}$$
$$\therefore$$ The given equation $$(i)$$ becomes
$$v + x\dfrac {dv}{dx} = \tan v + v$$
$$\implies \dfrac {1}{\tan v}dv = \dfrac {1}{x}dx$$
$$\implies \displaystyle \int \cot v\ dv = \int \dfrac {1}{x}dx$$
$$\implies \log |\sin v| = \log x + \log c=\log|xc|$$
$$\implies \sin v = xc$$
$$\therefore \sin \left (\dfrac {y}{x}\right ) = xc$$
Question 6
1 / -0

Directions For Questions

For the next three (03) items that follow :
The general solution of the differential equation $$(x^2 + x + 1)dy + (y^2 + y + 1)dx =0$$ is
$$(x + y + 1) = A(1 + Bx + Cy + Dxy)$$ where B, C and D are constants and A is parameter.

...view full instructions

What is D equal to ?

A
-1

B
1

C
-2

D
None of the above

Solution

$${\textbf{Step-1: Finding the value of D. }}$$
$${\text{Given,}}$$
$$\Rightarrow(x^2+x+1)dy+(y^2+y+1)dx=0$$
$$\Rightarrow(x^2+x+1)dy=-(y^2+y+1)dx$$
$$\Rightarrow\dfrac{dy}{y^2+y+1}=-\dfrac{dx}{x^2+x+1}$$
$${\text{Integrating both sides}}$$
   $$\Rightarrow \int \dfrac{dy}{y^2+y+1}=-\int\dfrac{dx}{x^2+x+1}$$
$${\text{By using perfect square method}}$$
$$\Rightarrow\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}=-\int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$$

$$\Rightarrow \int\dfrac{dx}{(x+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}+\int\dfrac{dy}{(y+\dfrac{1}{2})^2+(\dfrac{\sqrt3}{2})^2}$$

$${\text{use formula}}$$ $$\int\dfrac{dx}{a^2+x^2}=\dfrac{1}{a}tan^{-1}\dfrac{x}{a}$$

$$\Rightarrow \dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2x+1}{2}]+\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{2}{\sqrt3}\times\dfrac{2y+1}{2}]+\dfrac{2}{\sqrt3} tan^{-1}C=0$$

$${\text{[.... Where C is the integral constant]}}$$

  $${\text{use formula}}$$ $$tan^{-1}A+tan^{-1}B$$

$$\Rightarrow\dfrac{2}{\sqrt3}tan^{-1}[\dfrac{\dfrac{2x+1}{3}+\dfrac{2y+1}{\sqrt3}}{1-(\dfrac {2x+1}{\sqrt3})(\dfrac{2y+1}{\sqrt3})}]=-\dfrac{2}{\sqrt3}tan^{-1} C$$

$$\Rightarrow \dfrac{2x+1+2y+1}{\sqrt 3}\times \dfrac{\sqrt3 \sqrt3}{3-4xy-2x-2y-1}=C$$

$$\Rightarrow \dfrac{\sqrt3(2x+2y+2)}{2-4xy-2x-2y}=C$$

$$\therefore \sqrt3\times2(x+y+1)=c\times 2(1-2xy-x-y)$$

$$\therefore (x+y+1)=\dfrac{C}{\sqrt3}(1-x-y-2xy)$$

$${\text{ Compare this equation with the given equation gives,}}$$

$$\therefore D=-2$$

$${\textbf{Hence option C is correct }}$$
Question 7
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What is the number of arbitrary constants in the particular solution of differential equation of third order ?

A
0

B
1

C
2

D
3

Solution

$$\textbf{Step 1: Defining}$$
$$\text{We know that a particular solution of a differential equation contains no constants}$$
$$\text{This is applicable for all order of differential equations}$$
$$\textbf{Hence, Number of arbitrary constants in the particular solution of a differential equation }$$
$$\textbf{of third order is 0}$$
Question 8
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What is the equation of a curve passing through (0, 1) and whose differential equation is given by dy = y tan x dx ?

A
y = cos x

B
y = sin x

C
y = sec x

D
y = cosec x

Solution

Given : $$dy=y \ \tan x \ dx$$
$$\cfrac{dy}{y}=\tan x \ dx$$

On integrating both sides, we get
$$\Rightarrow \ln y \ = \ -\ln \cos x \ +c$$

The equation satisfies $$(0,1)$$
$$\Rightarrow 0=0+c$$
$$\Rightarrow c=0$$
We get
$$y=\cfrac{1}{\cos x}$$
$$y=\sec x$$
Question 9
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The solution of the differential equation $$\dfrac {dy}{dx} = (x + y)^{2}$$ is:

A
$$\dfrac {1}{x + y} = c$$

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

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

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

Solution

Given DE is $$\dfrac {dy}{dx} = (x + y)^{2}$$
Let's substitute $$t=x+y$$
$$\Rightarrow 1+\cfrac{dy}{dx}=\cfrac{dt}{dx}$$
$$\Rightarrow \cfrac{dt}{dx}-1=\dfrac{dy}{dx}$$
$$\Rightarrow \cfrac{dt}{dx}-1=t^2$$
$$\Rightarrow \cfrac{dt}{dx}=t^2+1$$
$$\Rightarrow \cfrac{dt}{t^2+1}=dx$$
On integrating both sides we get
$$\displaystyle \int \cfrac{dt}{t^2+1}=\int dx$$
$$\Rightarrow \tan^{-1}{t}=x+c$$
$$\Rightarrow \tan^{-1}{(x+y)}=x+c$$
Question 10
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The solution of the differential equation $$(1+{ x }^{ 2 }{ y }^{ 2 })ydx+({ x }^{ 2 }{ y }^{ 2 }-1)xdy=0$$ is

A
$$xy=\log { \dfrac { x }{ y } } +C$$

B
$$xy=2\log { \dfrac { y }{ x } } +C$$

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

D
$${x}^{2}{y}^{2}=2\log { \dfrac { y }{ x } } +C$$

Solution

Given differential equation is $$(1+{ x }^{ 2 }{ y }^{ 2 })ydx+({ x }^{ 2 }{ y }^{ 2 }-1)xdy=0$$ .... $$(i)$$
$$\implies (1+{ x }^{ 2 }{ y }^{ 2 })ydx=-({ x }^{ 2 }{ y }^{ 2 }-1)xdy$$

$$\implies \dfrac{(1+{ x }^{ 2 }{ y }^{ 2 })}{x}dx=\dfrac{(1-{ x }^{ 2 }{ y }^{ 2 })}{y}dy$$
$$\implies \left(\dfrac{1}{x}+x y^{2}\right)\ dx=\left(\dfrac{1}{y}-{x}^{2} y\right)\ dy$$
Integrating both sides
$$\implies \displaystyle \int \left(\dfrac{1}{x}+x y^{2}\right)\ dx=\int \left(\dfrac{1}{y}-{x}^{2} y\right)\ dy$$
$$\Rightarrow \displaystyle \int\left({\dfrac{dx}x-\dfrac{dy}y}\right)+\displaystyle \int\left({xy^2dx+x^2ydy}\right)=0$$
$$\Rightarrow \displaystyle \int d\left(\log{\dfrac xy}\right)+\displaystyle\int\dfrac12d(x^2y^2)+C=0$$
$$\therefore \log\dfrac xy+\dfrac12 x^2y^2+C=0$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 31

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Differential Equations Test - 31

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SHARING IS CARING

Question 1

$$y=\frac{x|x|}{2}+c$$

$$y=\frac{|x|}{2}+c$$

$$y=\frac{x^2}{2}+c$$

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

Solution

Question 2

1

-1

2

None of the above

Solution

Question 3

$$f(x) = y + C$$

$$f(x) = y(x + C)$$

$$f(x) = x + C$$

None of the above

Solution

Question 4

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

$$2{\sin}^{-1}y=\sqrt{1-x^2}+{\sin}^{-1}x+c$$

$$2{\sin}^{-1}y=x \sqrt{1-x^2}+{\sin}^{-1}x+c$$

$$2{\sin}^{-1}y=x \sqrt{1-x^2}+{\cos}^{-1}x+c$$

Solution

Question 5

$$\cos \left (\dfrac {y}{x}\right ) = cx$$

$$\sin \left (\dfrac {y}{x}\right ) = cx$$

$$\cos \left (\dfrac {y}{x}\right ) = cy$$

$$\sin \left (\dfrac {y}{x}\right ) = cy$$

Solution

Question 6

-1

1

-2

None of the above

Solution

Question 7

0

1

2

3

Solution

Question 8

y = cos x

y = sin x

y = sec x

y = cosec x

Solution

Question 9

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

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

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

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

Solution

Question 10

$$xy=\log { \dfrac { x }{ y } } +C$$

$$xy=2\log { \dfrac { y }{ x } } +C$$

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

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

Solution

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