Differential Equations Test

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Differential Equations Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

Solution of differential equation $$\left( { 2y+xy }^{ 3 } \right) dx+\left( x{ +x }^{ 2 }{ y }^{ 2 } \right) dy=0$$

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

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

C
$${ x }^{ 2 }y+\dfrac { { x }^{ 4 }{ y }^{ 4 } }{ 3 } =c$$

D
None of these

Solution
Question 2
1 / -0

The differential equation of all circles in the first quadrant which touch the coordinate axes is of order -

A
1

B
2

C
3

D
None of these

Solution

Centre of circle $$=a,a$$
Radius of circle $$=a$$
$$\therefore$$ Equation of circle -
$$\Rightarrow \left( x-a\right)^2+\left( y-a\right)^2=a^2$$
No. of arbitrary constants $$=1$$ ( i.e, a )
$$\therefore$$ Order of differential equations $$=1$$
Hence, the answer is $$1.$$
Question 3
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The population $$p(t)$$ at time $$t$$ of a certain mouse species satisfies the differential equation $$\dfrac { d p ( t ) } { d t } = 0.5 p ( t ) - 450.$$ If $$p ( 0 ) = 850 ,$$ then the time at which the population becomes zero is

A
$$\dfrac { 1 } { 2 } \ln 18$$

B
$$\ln 18$$

C
$$2 \ln 18$$

D
$$\ln 9$$

Solution

We have,
$$\begin{array}{l} \dfrac { { d\left( { p\left( t \right) } \right) } }{ { dt } } =\dfrac { 1 }{ 2 } p\left( t \right) -450 \\ \dfrac { { d\left( { p\left( t \right) } \right) } }{ { dt } } =\dfrac { { p\left( t \right) -900 } }{ 2 } \\ 2\int { \dfrac { { d\left( { p\left( t \right) } \right) } }{ { p\left( t \right) -900 } } } =\int { dt } \\ 2\, In\left| { \, p\left( t \right) -900 } \right| =t+c \\ t=0 \\ \Rightarrow 2\, In\, 50=0+c \\ \therefore \, 2\, In\, 900=t+2\, In\, 50 \\ t=2\, \, \left( { In\, 900-\, In\, 50 } \right) \\ =2\, In\, \left( { \dfrac { { 900 } }{ { 50 } } } \right) \\ =2\, In\, 18 \end{array}$$

Hence, option $$C$$ is correct answer.
Question 4
1 / -0

General solution of the differential equation $$\frac{{dy}}{{dx}} = 1 + xy$$ is

A
$$y = c \cdot \,{e^{ - x2/2}}$$

B
$$y = c \cdot \,{e^{ x2/2}}$$

C
$$y = (x + c) \cdot {e^{ - x2/2}}$$

D
None

Solution

$$\frac{dy}{dx}= 1+xy \Rightarrow \frac{dy}{dx}-ky=1$$
$$IF = e^{\int -xdx}= e^{-\frac{x^{2}}{2}}$$
$$\frac{dy}{dx}=1+xy\Rightarrow e^{-\frac{x^{2}}{2}}\frac{dy}{dx}-2yxe^{-\frac{x^{2}}{2}}=1.e^{-\frac{x^{2}}{2}}$$
$$\therefore \frac{d}{dx}(ye^{-\frac{x^{2}}{2}})=e^{-\frac{x^{2}}{2}}$$
$$ye^{-\frac{x^{2}}{2}}=e^{-\frac{x^{2}}{2}}+c$$
Question 5
1 / -0

The solution of the differential equation,
$$\dfrac{dy}{dx}=(x-y)^2$$, when $$y(1)=1$$, is :

A
$$log_e \left| \dfrac{2-y}{2-x}\right|=2(y-1)$$

B
$$log_e \left| \dfrac{2-x}{2-y}\right|=x-y$$

C
$$-log_e \left| \dfrac{1+x-y}{1-x+y}\right|=x+y-2$$

D
$$-log_e \left| \dfrac{1-x+y}{1+x-y}\right|=2(x-1)$$

Solution

$$x-y =t \Rightarrow \dfrac{dy}{dx}=1-\dfrac{dt}{dx}$$
$$\Rightarrow 1- \dfrac{dt}{dx} \Rightarrow \int \dfrac{dt}{1-t^2} = \int 1dx$$
$$\Rightarrow\dfrac{1}{2}ln \left(\dfrac{1+t}{1-t}\right)=x+\lambda$$
$$\Rightarrow \dfrac{1}{2}ln \left(\dfrac{1+x-y}{1-x+y}\right)=x+\lambda$$

given $$y(1)=1$$
$$\Rightarrow \dfrac{1}{2}ln (1)=1+ \lambda \Rightarrow \lambda = -1$$
$$\Rightarrow ln \left(\dfrac{1+x-y}{1-x+y}\right)=2(x-1)$$
$$\therefore -log_e \left| \dfrac{1-x+y}{1+x-y}\right|=2(x-1)$$
Question 6
1 / -0

The solution of the differential equation $$\dfrac { dy }{ dx } =\dfrac { xy }{ { x }^{ 2 }+y^{ 2 } } $$ is -

A
$$ay^{ 2 }={ e }^{ { x }^{ 2 }/{ y }^{ 2 } }$$

B
$$ay={ e }^{ x/y }$$

C
$$y = c$$

D
$$y={ y }^{ 2 }+c$$

Solution
Question 7
1 / -0

The solution of differential equation $$\cos ^ { 2 } x \dfrac { d y } { d x } - ( \tan 2 x ) y = \cos ^ { 4 } x , | x | < \dfrac { \pi } { 4 } ,$$ where $$y \left( \dfrac { \pi } { 6 } \right) = \dfrac { 3 \sqrt { 3 } } { 8 }$$

A
$$y = \tan 2 x \cos ^ { 2 } x$$

B
$$y = \cot 2 x \cos ^ { 2 } x$$

C
$$2 y = \tan 2 x \cos ^ { 2 } x$$

D
$$2 y = \cot 2 x \cos ^ { 2 } x$$

Solution

Given that:-
$$\cos ^ { 2 } x \dfrac { d y } { d x } - ( \tan 2 x ) y = \cos ^ { 4 } x , | x | < \dfrac { \pi } { 4 } ,$$ where $$y \left( \dfrac { \pi } { 6 } \right) = \dfrac { 3 \sqrt { 3 } } { 8 }$$

To find:- Solution of the given equation.

Solution:- $$\because \cos^2x(\dfrac{dy}{dx})-(\tan 2x)y=\cos^4x$$

$$\Rightarrow \cos^2 x(\dfrac{dy}{dx})-\tan 2 xy =\cos^4x$$

$$\Rightarrow \dfrac{dy}{dx}-\dfrac{\tan 2 x}{\cos^2x}.y=\cos^4x$$

Now, we will compat the eqn i to

$$\dfrac{dy}{dx}+Py=B$$

$$\therefore P=f(x)$$

$$B=f(x)$$

So, Integration factor (I.F) $$=e^{\int Pdx}$$

Here we will find the value of $$-\dfrac{\tan 2x}{\cos^2x}$$

$$\displaystyle -\int \dfrac{\tan 2x}{\cos^2x}dx$$

$$\displaystyle =-\int \dfrac{2\sin 2x}{\cos 2x(2\cos^2x)}dx$$

$$\displaystyle =-\int \dfrac{2\sin 2x\,dx}{\cos 2x(1+\cos 2x)}$$

Let, $$t=\cos 2x$$

$$\Rightarrow dt=-2\sin x\,d\,x$$

$$\displaystyle \Rightarrow =\int \dfrac{dt}{t(1+t)}$$

$$\displaystyle =\int \dfrac{1}{t}-\dfrac{1}{1+t}dt$$

$$=ln\left|\dfrac{t}{1+t}\right|$$

$$=ln \left|\dfrac{\cos 2x}{1+\cos 2x}\right|$$

Now, putting $$ln\left|\dfrac{\cos 2x}{1+\cos 2x}\right|$$ in I.F

$$\therefore e^{Pdx}$$

$$=e^{ln}\left|\dfrac{\cos 2x}{1+\cos 2x}\right|$$

$$=\dfrac{\cos 2x}{1+\cos 2x}$$

$$=\dfrac{\cos 2x}{2\cos^2x}$$

Since, multiplying eqn i by I.F both side, we get.

$$\dfrac{\cos 2x}{2\cos^2x}.\dfrac{dy}{dx}-\dfrac{\sin 2x}{\cos 4 x}y=\dfrac{\cos 2x}{2}$$

$$\dfrac{\cos 2x}{2\cos^2x}.dy-(\dfrac{\sin 2x}{\cos 4 x})dx.y=\dfrac{\cos 2x}{2}dx$$

$$\displaystyle \Rightarrow \int d(y\dfrac{\cos 2x}{2\cos^2x})=\int \dfrac{\cos 2x}{2}dx$$

$$\Rightarrow y.\dfrac{\cos 2x}{2\cos^2x}=\dfrac{\sin 2x}{4}+C$$...........(II)

$$\Rightarrow \dfrac{3\sqrt{3}}{8}(\dfrac{1/2}{2.\dfrac{3}{4}})=\dfrac{\sqrt{3}}{8}+c$$

$$\Rightarrow x=\dfrac{\pi}{6}\,\,y=\dfrac{3\sqrt{3}}{8}$$

$$\Rightarrow \dfrac{\sqrt{3}}{8}=\dfrac{\sqrt{3}}{8}+c$$

$$\therefore c=0$$

putting the value of c in eq ii.

$$\therefore y.\dfrac{\cos 2x}{2\cos^2x}=\dfrac{\sin 2x}{4}+0$$

$$\Rightarrow y=\dfrac{\sin 2x.2\cos^2x}{\cos 2x.4}$$

$$\therefore y=\dfrac{\tan 2x.\cos^2x}{2}$$

$$\therefore 2y=\tan 2x.\cos^2x$$

hence, the required solution is

$$2y=\tan 2x.\cos^2x$$
Question 8
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A differential equation associated with the primitive $$y=a+b\ e^{5x}+c\ e^{7x}$$ is

A
$$y_{3}+2y_{2}-y_{1}=0$$

B
$$y_{3}+2y_{2}-35y_{1}=0$$

C
$$4y_{3}+5y_{2}-20y_{1}=0$$

D
$$none\ of\ these$$

Solution
Question 9
1 / -0

If $$\sqrt { \dfrac { \upsilon }{ \mu } } +\sqrt { \dfrac { \mu }{ \upsilon } } =6$$, then $$\dfrac { d\upsilon }{ d\mu } =$$

A
$$\dfrac { 17\mu -\upsilon }{ \mu -17\upsilon } $$

B
$$\dfrac { \mu -17\upsilon }{ 17\mu -\upsilon } $$

C
$$\dfrac { 17\mu +\upsilon }{ \mu -17\upsilon } $$

D
$$\dfrac { \mu +17\upsilon }{ 17\mu -\upsilon } $$

Solution

$$\sqrt{\dfrac{v}{\mu}}+\sqrt{\dfrac{\mu}{v}}=6$$
$$\Rightarrow \dfrac{v+\mu}{\sqrt{v\mu}}=6$$
$$\Rightarrow {\left(v+\mu\right)}^{2}=36v\mu$$
$$\Rightarrow {v}^{2}+{\mu}^{2}+2v\mu-36v\mu=0$$
$$\Rightarrow {v}^{2}+{\mu}^{2}-34v\mu=0$$
$$\Rightarrow 2v\dfrac{dv}{d\mu}+2\mu-34\left(v+\mu\dfrac{dv}{d\mu}\right)=0$$
$$\Rightarrow \left(2v-34\mu\right)\dfrac{dv}{d\mu}=34v-2\mu$$
$$\Rightarrow -2\left(17\mu-v\right)\dfrac{dv}{d\mu}=-2\left(\mu-17v\right)$$
$$\therefore \dfrac{dv}{d\mu}=\dfrac{\mu-17v}{17\mu-v}$$
Question 10
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The number of arbitrary constant in the particular solution of a differential equation is

A
$$3$$

B
$$4$$

C
infinite

D
zero

Solution

According to the definition of particular solution of a differential equation, it does not contain any arbitrary constants.
So number of arbitrary constants in a particular solution of a differential equation is $$0$$.