Differential Equations Test

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Differential Equations Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The general solution of the differential equation $$\log _{ e }{ \left( \cfrac { dy }{ dx } \right) } =x+y$$ is:

A
$${ e }^{ x }+{ e }^{ -y }=C$$

B
$${ e }^{ x }+{ e }^{ y }=C$$

C
$${ e }^{ y }+{ e }^{ x }=C$$

D
$${ e }^{ -x }+{ e }^{ -y }=C$$

Solution

Given $$\log _{ e }{ \left( \cfrac { dy }{ dx } \right) } =x+y$$

$$\Rightarrow \cfrac { dy }{ dx } ={ e }^{ x+y }={ e }^{ x }{ e }^{ y }$$
$$\Rightarrow { e }^{ -y }dy={ e }^{ x }dx$$

On integrating, we get

$$-{ e }^{ -y }={ e }^{ x }-C$$
$$\Rightarrow C={ e }^{ x }+{ e }^{ -y }$$
Question 2
1 / -0

Find the general solution of $$dy=y \sec x dx$$.

A
$$y=C(\sec x -2\tan x)$$

B
$$y=C(\sec x +\tan x)$$

C
$$y=C(2\sec x +\tan x)$$

D
None of these

Solution

Given Differential Equation is $$dy=y\sec { x } dx$$

Integrating with respect to $$x$$ gives,

$$\int { \cfrac { 1 }{ y } dy } =\int { \sec { x } } dx$$

$$\ln { y } =\ln { \left( \sec { x } +\tan { x } \right) } +\ln { c } $$

Where $$c$$ is the integration constant,

$$\therefore y=c\left( \sec { x } +\tan { x } \right) $$.
Question 3
1 / -0

The solution of the differential equation $$\dfrac{dx}{x}+\dfrac{dy}{y}=0$$ is

A
$$xy=c$$

B
$$x+y=c$$

C
$$log\, x\, log\, y = c$$

D
$$x^2+y^2=c$$

Solution

$$ \dfrac{dx}{x}+\dfrac{dy}{y} = 0 $$
$$ \Rightarrow \dfrac{dx}{x} = \dfrac{-dy}{y} $$
integrating,
$$\displaystyle = \int \dfrac{dy}{y} = -\int \dfrac{dx}{x} $$
$$ = lny = -lnx+c $$
$$ \Rightarrow lnx.y = c \Rightarrow \boxed {xy = constant} $$
Question 4
1 / -0

The order of the differential equation $$\left [1 + \left (\dfrac {dy}{dx}\right )^{5}\right ]^{\dfrac {2}{3}} = \dfrac {d^{3}y}{dx^{3}}$$ is:

A
$$2$$

B
$$1$$

C
$$3$$

D
$$\dfrac {2}{3}$$

Solution

$$\left [1 + \left (\dfrac {dy}{dx}\right )^{5}\right ]^{\dfrac {2}{3}} = \dfrac {d^{3}y}{dx^{3}}$$
$$\Rightarrow \left [1 + \left (\dfrac {dy}{dx}\right )^{5}\right ]^ {2} = \left(\dfrac {d^{3}y}{dx^{3}}\right)^3$$
Hence order of above differential equation is $$3$$
Question 5
1 / -0

The general solution of $$\cfrac{dy}{dx}=\cfrac{2x-y}{x+2y}$$ is

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

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

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

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

Solution

Given $$\cfrac{dy}{dx}=\cfrac{2x-y}{x+2y}$$
Put $$y=vx$$
$$\cfrac{dy}{dx}=v+x\cfrac{dv}{dx}$$
$$\therefore$$ $$v+x\cfrac { dv }{ dx } =\cfrac{2-v}{1+2v}$$
$$\Rightarrow$$ $$x\cfrac { dv }{ dx } =\cfrac{2-v-v(1+2v)}{1+2v}$$
$$\Rightarrow$$ $$x\cfrac { dv }{ dx } =\cfrac{2-2v-2{v}^{2}}{1+2v}$$
$$\Rightarrow$$ $$\int { \cfrac { 1+2v }{ 2(1-v-{ v }^{ 2 }) } } =\int { \cfrac { 1 }{ x } } dx$$
$$\Rightarrow$$ $$\log{a}-\cfrac{1}{2}\log{1-v-{v}^{2})}=\log{x}$$
$$\Rightarrow$$ $$2\log {a}-\log{(1-v-{v}^{2}}=2\log {x}$$
$$\Rightarrow$$ $$\log {c}=\log{[{x}^{2}(1-v-{v}^{2}]}$$ [Let $${a}^{2}=c]$$
$$\Rightarrow$$ $$c={x}^{2}(1-v-{v}^{2})$$
$$\Rightarrow$$ $$c={x}^{2}(1-\cfrac{y}{x}-\cfrac{{y}^{2}}{{x}^{2}})$$
$$\Rightarrow$$ $$c={x}^{2}(\cfrac{{x}^{2}-xy-{y}^{2}}{{x}^{2}})$$
$$\Rightarrow$$ $${x}^{2}-xy-{y}^{2}=c$$
Question 6
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If the general solutions of a differential equation is $$(y+c)^2=cx$$, where $$c$$ is an arbitrary constant, then the order and degree of differential equation are:

A
$$1,2$$

B
$$2,1$$

C
$$1,1$$

D
None of these

Solution

Given general solution is $${(y+c)}^{2}=cx$$
Now differentiate it with respect to $$x$$ on both sides
we get $$2(y+c)\frac{dy}{dx} = c$$ , Let $$\frac{dy}{dx}$$ be $$Y$$
$$\Rightarrow c=\frac { 2yY }{ 1-2Y } $$
By substituting $$c$$ in general solution , we get $${ (y+\frac { 2yY }{ 1-2Y } ) }^{ 2 }=\frac { 2yY }{ 1-2Y } x$$
$$\Rightarrow { y }^{ 2 }=2yxY(1-2Y)=2yxY-4yx{ Y }^{ 2 }$$
Therefore the order of equation is $$1$$ and the degree of equation is $$2$$
Therefore option $$A$$ is correct
Question 7
1 / -0

If $$f(x)$$ and $$g(x)$$ are twice differentiable functions on $$(0, 3)$$ satisfying $$f''(x) = g'', f'(1) = 4, g'(1) = 6, f(2) = 3, g(2) = 9$$, then $$f(1) - g(1)$$ is:

A
$$4$$

B
$$-4$$

C
$$0$$

D
$$-2$$

Solution

According to question,

$$f''(x) = g''(x)$$

Integrating w.r.t. $$x$$, we get

$$f'(x) = g'(x) + C_{1}$$

Put $$x = 1\Rightarrow f'(1) = g'(1) + C_{1}$$

$$\Rightarrow 4 = 6 + C_{1}$$

$$\therefore C_{1} = 2$$

$$\therefore f'(x) = g'(x) - 2$$

Again, integrating w.r.t. $$x$$, we get

$$f(x) = g(x) - 2x + C_{2}$$

$$\Rightarrow 3 = 9 - 4 + C_{2} \Rightarrow C_{2} = -2$$

$$\therefore f(x) = g(x) - 2x - 2$$

Put $$x = 1$$, we get

$$f(1) - g(1) = -2(1) - 2 = -4$$
Question 8
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The general solution of the differential equation $$\displaystyle \frac { dy }{ dx } +\frac { 1+\cos { 2y } }{ 1-\cos { 2x } } =0$$ is given by:

A
$$\tan { y } +\cot { x } =c$$

B
$$\displaystyle \tan { y } -\cot { x } =c$$

C
$$\displaystyle \tan { x } -\cot { y } =c$$

D
$$\displaystyle \tan { x } +\cot { x } =c$$

Solution

$$ \displaystyle \frac { dy }{ dx } +\frac { 1+\cos { 2y } }{ 1-\cos { 2x } } =0$$

$$\displaystyle \Rightarrow \frac { dy }{ dx } = - \frac { \cos ^{ 2 }{ y } }{\sin ^{ 2 }{ x } } $$

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

$$\int \sec^2y dy = - \int \csc^2x dx$$

$$\tan y +c = \cot x +C$$
$$\tan y - \cot x = c$$
Question 9
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Which of the following equation is a linear differential equation of order $$3$$ ?

[Note: The original question asks for linear equation, but it should be linear differential equation]

A
$$\dfrac{d^3y}{dx^3}+\dfrac{d^2y}{dx^2}\dfrac{dy}{dx}+y=x$$

B
$$\dfrac{d^3y}{dx^3}+\dfrac{d^2y}{dx^2}+y^2=x^2$$

C
$$x\dfrac{d^3y}{dx^3}+\dfrac{d^2y}{dx^2}=e^x$$

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

Solution

Linear equation is an equation between two variables that gives a straight line and order of linear equation is highest order derivative in linear equation.
Since, in option A equation is linear equation in which highest order is $$3$$.
$$\therefore$$ option A is correct.
Question 10
1 / -0

Find a particular solution for the following differential equation.
$$y'-4y'-12y=te^{4t}$$

A
$$y(t)=\dfrac{1}{32}(3t+1)e^{4t}$$

B
$$y(t)=-\dfrac{1}{18}(3t+1)e^{4t}$$

C
$$y(t)=-\dfrac{1}{36}(3t+1)e^{4t}$$

D
None of these

Solution

Given, $${ y }'-4{ y '}-12y=t{ e }^{ 4t }$$
$$\Rightarrow -{ 3y }'-12y=t{ e }^{ 4t }\\ \Rightarrow { y }'+4y=-\dfrac{t{ e }^{ 4t }}{3}\\ \Rightarrow \dfrac { dy }{ dt } +4y=-\dfrac{1}{3}t{ e }^{ 4t }$$
This is a linear form of differential equation with
$$\displaystyle{\text{I.F.}={ e }^{ \int 4dt }={ e }^{ 4t }}$$
Solution of the differential equation is
$$\displaystyle{y(\text{I.F.})=\int { Q( } \text{I.F.})dt\\ y{ e }^{ 4t }=\int { \dfrac { -t{ e }^{ 4t } }{ 3 } { e }^{ 4t } } dt\\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \int { t{ e }^{ 8t } } dt\\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \left\{ t\int { { e }^{ 8t } } -\int { \dfrac { d }{ dt } (t)\int { { e }^{ 8t } } } \right\} \\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \left\{ \dfrac { t{ e }^{ 8t } }{ 8 } -\int { \dfrac { { e }^{ 8t } }{ 8 } } \right\} \\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \left\{ \dfrac { t{ e }^{ 8t } }{ 8 } -\dfrac { { e }^{ 8t } }{ 64 } \right\} \\ y{ e }^{ 4t }=\dfrac { -{ e }^{ 8t } }{ 3 } \left\{ \dfrac { 8t-1 }{ 64 } \right\} \\ y=\dfrac { -{ e }^{ 4t } }{ 192 } (8t-1)}$$
So, option D is correct.

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Differential Equations Test - 29

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Differential Equations Test - 29

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SHARING IS CARING

Question 1

$${ e }^{ x }+{ e }^{ -y }=C$$

$${ e }^{ x }+{ e }^{ y }=C$$

$${ e }^{ y }+{ e }^{ x }=C$$

$${ e }^{ -x }+{ e }^{ -y }=C$$

Solution

Question 2

$$y=C(\sec x -2\tan x)$$

$$y=C(\sec x +\tan x)$$

$$y=C(2\sec x +\tan x)$$

None of these

Solution

Question 3

$$xy=c$$

$$x+y=c$$

$$log\, x\, log\, y = c$$

$$x^2+y^2=c$$

Solution

Question 4

$$2$$

$$1$$

$$3$$

$$\dfrac {2}{3}$$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$1,2$$

$$2,1$$

$$1,1$$

None of these

Solution

Question 7

$$4$$

$$-4$$

$$0$$

$$-2$$

Solution

Question 8

$$\tan { y } +\cot { x } =c$$

$$\displaystyle \tan { y } -\cot { x } =c$$

$$\displaystyle \tan { x } -\cot { y } =c$$

$$\displaystyle \tan { x } +\cot { x } =c$$

Solution

Question 9

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

$$\dfrac{d^3y}{dx^3}+\dfrac{d^2y}{dx^2}+y^2=x^2$$

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

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

Solution

Question 10

$$y(t)=\dfrac{1}{32}(3t+1)e^{4t}$$

$$y(t)=-\dfrac{1}{18}(3t+1)e^{4t}$$

$$y(t)=-\dfrac{1}{36}(3t+1)e^{4t}$$

None of these

Solution

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