Differential Equations Test

Result

Differential Equations Test - 41

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Solution of the differential equation: $$\left( 2xcosy+{ y }^{ 2 }cosx \right) dx+\left( 2ysinx-{ x }^{ 2 }siny \right) dy=0$$ is :

A
$${ x }^{ 2 }sinx+y^{ 2 }cosx=c$$

B
$${ x }^{ 2 }siny+y^{ 2 }cosx=c$$

C
$${ x }^{ 2 }cosy+y^{ 2 }sinx=c$$

D
None of these

Solution
Question 2
1 / -0

Solution of the equation $$\dfrac{{dy}}{{dx}} = 1 + xy + x + y$$ is

A
$$\left( {1 + y} \right)\left( {1 + x} \right) = c$$

B
$$\log \left( {1 + y} \right) = 1 + x + c$$

C
$$\ln |1 + y| = x + \dfrac{{{x^2}}}{2} + c$$

D
$$\ln |1 + y| + x + \dfrac{{{x^2}}}{2} + c = 0$$

Solution

$$\dfrac { dy }{ dx } =1+xy+x+y$$
$$=1+y(x+1)+x$$
$$\dfrac { dy }{ dx } =\left( 1+y \right) \left( 1+x \right) $$
$$\Rightarrow \dfrac { dy }{ \left( 1+y \right) } =\left( 1+x \right) dx$$
$$\Rightarrow log\left| 1+y \right| =x+\dfrac { { x }^{ 2 } }{ 2 } +c$$ [C]
Question 3
1 / -0

The general solution of the differential equation $$\dfrac { dy }{ dx } +y={ x }^{ 3 }$$ is ______.

A
$${ ye }^{ x }={ e }^{ x }\left( { x }^{ 3 }+{ 3x }^{ 2 }-6x-6 \right) +c$$

B
$${ ye }^{ x }={ e }^{ x }\left( { x }^{ 3 }-{ 3x }^{ 2 }-6x+6 \right) +c$$

C
$$y=\left( { x }^{ 3 }-{ 3x }^{ 2 }+6x-6 \right) +c{ e }^{ -x }$$

D
$$y={ e }^{ x }$$

Solution
Question 4
1 / -0

Solution of differential equation $$ \cfrac {dy} {dx} + x{sin}^{2} y $$= $$sin y \quad cos y \quad $$is

A
$$ tany=(x-1)+ce-x $$

B
$$ coty=(x-1)+ce-x $$

C
$$ tany=(x-1)ex+c $$

D
$$ coty=(x-1)ex+c $$
Question 5
1 / -0

If $$\dfrac{dy}{dx}=y+3>0$$ and $$y(0)=2$$ then $$y(\ln{2})$$ is equal to :

A
$$7$$

B
$$5$$

C
$$13$$

D
$$-2$$

Solution

$$ \frac{dy}{dx} = y + 3 $$ and y(0) = 2
then y (ln 2) = (?)
$$ \therefore \frac{dy}{dx} = y+3$$
$$ \therefore \frac{1}{y+3} dy = dx$$
Integrating both side
$$ \int \frac{1}{y+3}dy =\int dx$$
$$ \therefore $$ ln(y + 3) = x + c
$$ \rightarrow $$ Initial condition y(0)=2
$$ \therefore ln(2+3)=0+c$$
$$ \therefore c=ln 5$$
$$ \rightarrow $$ ln(y + 3) = x + ln5
$$ \therefore $$ ln(y + 3) - ln5 = x
$$ \therefore ln\left ( \frac{y+3}{5} \right )=x$$
$$ \therefore \frac{y+3}{5} = e^{x}$$
$$ \therefore y=5e^{x} - 3$$
for y (ln2)
$$ y = 5e^{ln2}-3$$
$$ = 5 \times 2 - 3 $$
= 10 - 3
= 7
Question 6
1 / -0

The general solution of the differential equation $$e^xdy+(ye^x+2x)dx=0$$ is

A
$$xe^x+x^2=C$$

B
$$xe^x-y^2=C$$

C
$$ye^x+x^2=C$$

D
$$ye^x-x^2=C$$

Solution

$$e^x dy +(ye^x+2x)dx=0$$
$$e^x dy=-(ye^x+2x)dx$$
$$\dfrac{dy}{dx}=\dfrac{-(ye^x+2x)}{e^x}$$

$$\dfrac{dy}{dx}=-\dfrac{ye^x}{e^x}-\dfrac{2x}{e^x}$$

$$\dfrac{dy}{dx}=-y-\dfrac{2x}{e^x}$$

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

This is of the form $$\dfrac{dy}{dx}+Py=Q$$

where $$P=1$$ and $$Q=-\dfrac{2x}{e^x}$$

$$I.F=e^{\int Pdx}$$
$$=e^{\int 1 dx}=e^x$$

Solution is :
$$ye^x=\int \dfrac{-2x}{e^x} e^x \ dx+C$$

$$ye^x=-\int 2x \ dx +C$$
$$ye^x=-x^2+C$$
$$ye^x+x^2=C$$
Question 7
1 / -0

The solution of differential equation $$\cos x.\sin y dx+\sin x. \cos ydy=0$$ is

A
$$\dfrac{\sin x}{\sin y}=c$$

B
$$\sin x. \sin y=c$$

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

D
$$\cos x.\cos y=c$$

Solution

Given,
$$\left( \cos { x } .\sin { y } \right) dx+\left( \sin { x } .\cos { y } \right) dy=0$$
$$\Rightarrow \left( \cos { x } .\sin { y } \right) dx=-\sin x.\cos y dy$$
$$\Rightarrow \dfrac{\cos x}{\sin x}dx=\dfrac{-\cos x}{\sin y}dy$$
( All we have done is separated the variables )
$$\Rightarrow \cot x\;dx=-\cot y\;dy$$
( Now integrating both the sides we get )
$$\Rightarrow \int \cot x\;dx =-\int \cot y\;dy$$
$$\Rightarrow \int \cot y\;dy =-\int \cot x\;dx$$
$$ln\;\sin y=-ln \sin x +ln C$$ ( $$ln$$ C is interpretation const. )
$$ln\sin y =ln \dfrac{1}{\sin x}+ln C$$
$$ln \sin y=ln \dfrac{C}{\sin x}$$ $$\left( \because ln\;a+ln\;b=ln \;ab\right)$$
$$\sin y=\dfrac{C}{\sin x}$$
$$\Rightarrow \sin x.\sin y=C$$
Hence, the answer is $$\sin x.\sin y=C.$$

'
Question 8
1 / -0

For the given differential equation find the general solution:
$$\dfrac { dy }{ dx } +2y=sin x$$

A
$$\dfrac 1 5 [2sin x -cosx ]+C e ^{-2x}$$

B
$$\dfrac 1 5 [2sin x -cosx ]+C $$

C
$$\dfrac 1 2 [5sin x -cosx ]+C e ^{-2x}$$

D
None of these

Solution
Question 9
1 / -0

The differential equation of the system of circles touching the $$x$$-axis at origin is

A
$$\left( { x }^{ 2 }-{ y }^{ 2 } \right) \dfrac { dy }{ dx } -2xy=0$$

B
$$\left( { x }^{ 2 }+{ y }^{ 2 } \right) \dfrac { dy }{ dx } +2xy=0$$

C
$$\left( { x }^{ 2 }-{ y }^{ 2 } \right) \dfrac { dy }{ dx } +2xy=0$$

D
$$\left( { x }^{ 2 }+{ y }^{ 2 } \right) \dfrac { dy }{ dx } -2xy=0$$

Solution
Question 10
1 / -0

Let y=y(x) be the solution of the differential equation $$sinx\dfrac { dy }{ dx } +ycosx=4x,x\in (0,\pi )$$. If $$y=\left( \dfrac { \pi }{ 2 } \right) =0,then\quad y\left( \dfrac { \pi }{ 6 } \right) $$ is equal to :

A
$$\dfrac { 4 }{ 9\sqrt { 3 } } { \pi }^{ 2 }$$

B
$$\dfrac { 8 }{ 9\sqrt { 3 } } { \pi }^{ 2 }$$

C
$$-\dfrac { 8 }{ 9 } { \pi }^{ 2 }$$

D
$$-\dfrac { 4 }{ 9 } { \pi }^{ 2 }$$

Solution