Differential Equations Test

Result

Differential Equations Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

The solution of the differential equation $$ \frac { d y } { d x } - k y = 0 , y ( 0 ) = 1 $$, approaches zero when $$ x \rightarrow \infty , $$ if

A
k = 0

B
k > 0

C
k < 0

D
k may be any real value
Question 2
1 / -0

Solve the differential equation:
$$(e^ {y}+1)\cos x\ dx+e^ {y}\sin x\ dy=0$$

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

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

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

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

Solution
Question 3
1 / -0

Solution of differential equation $$ \frac{d y}{d x}+x \sin ^{2} y=\sin y \cos y $$ is-

A
$$

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

$$

B
$$

\cot y=(x-1)+ ce^{-x}

$$

C

$$\tan y = (x-1) e^x +C$$

D
$$\cot y=(x-1) e^x+C$$

(where C is an arbitrary constant)
Question 4
1 / -0

The solution of $$y^5x+y-x\dfrac{dy}{dx}=0$$ is

A
$$\dfrac{x^4}{4}+\dfrac{1}{5}\left(\dfrac{x}{y}\right)^5=C$$

B
$$\dfrac{x^5}{4}+\dfrac{1}{5}\left(\dfrac{x}{y}\right)^4=C$$

C
$$\left(\dfrac{x}{y}\right)^5+\dfrac{x^4}{4}=C$$

D
$$(xy)^4+\dfrac{x^5}{5}=C$$

Solution
Question 5
1 / -0

Suppose y=y(x) satisfies the differential equation $$ydx+y^{2}dy=xdy.$$ If $$y(x) > 0 \forall x\epsilon R$$ and y(1)=1, then y(-3) equals

A
1

B
2

C
3

D
5

Solution
Question 6
1 / -0

$$y = (c_{1} + c_{2}x + x^{2}) e^{x}$$ is a solution of

A
$$\dfrac {d^{2}y}{dx^{2}} + y = 0$$

B
$$\dfrac {d^{2}y}{dx^{2}} + xy = 0$$

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

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

Solution
Question 7
1 / -0

Solution of the differential equation
$$x=1+\left(xy\dfrac{dy}{dx}\right)+\dfrac{(x^2y^2)}{2!}\left(\dfrac{dy}{dx}\right)^3+.........$$ is

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

B
$$y=(in\,x)^2+C$$

C
$$y=\pm\sqrt{(in\,x)^2+C}$$

D
$$x^y=xy+K$$
Question 8
1 / -0

$$y=3\sqrt{2x}$$ Find $$\frac{dy}{dx}$$

A
$$\frac{3}{\sqrt{2x}}$$

B
$$\frac{2}{\sqrt{2x}}$$

C
$$\frac{2}{\sqrt{3x}}$$

D
$$\frac{2}{\sqrt{4x}}$$

Solution
Question 9
1 / -0

Solution of differential equation $$2xy \dfrac {dy}{dx} = x^{2} + 3y^{2}$$.

A
$$x^{3} + y^{2} = px^{2}$$

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

C
$$x^{2} + y^{3} = px$$

D
$$x^{2} + y^{2} = px^{3}$$

Solution
Question 10
1 / -0

The solution of $$\dfrac{dy}{dx}+\dfrac{x}{1-x^{2}}y=x\sqrt{y}$$, is given by

A
$$3\sqrt{y}+(1-x^{2})=C(1-x^{2})^{1/4}$$

B
$$\dfrac{3}{2}\sqrt{y}+1(1-x^{2})=C(1-x^{2})^{3/2}$$

C
$$3\sqrt{y}-(1-x^{2})=C(1-x^{2})^{3/2}$$

D
$$None\ of\ these$$

Solution