Differential Equations Test

Result

Differential Equations Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

Let $$y=y(x)$$ be the solution of the differential equation $$\sin x\dfrac {dy}{dx}+y\cos x=4x,x\ \in\ (0,\pi)$$, if $$y\left(\dfrac {\pi}{2}\right)=0$$, then $$y\left(\dfrac {\pi}{6}\right)$$ is equal to

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

B
$$\dfrac {4}{9\sqrt {2}}\pi^{2}$$

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

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

Solution
Question 2
1 / -0

The solution of equation $$\dfrac { dy }{ dx } =\dfrac { 1 }{ x+y+1 } $$

A
$$x={ ce }^{ y }-y-2$$

B
$$y=x+{ ce }^{ y }-2$$

C
$$x+{ ce }^{ y }-2-2=0$$

D
$$y=x$$
Question 3
1 / -0

The solutions of the differential equation $$\frac{dy}{dx}= \frac{siny+x}{sin2y-x cos y }$$ is

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

B
$$sin^{2}y = xsin y -\frac{x^{2}}{2}+c$$

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

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

Solution
Question 4
1 / -0

Solution of differential equation $$xdy=(y-{x}^{2}-{y}^{2})dx$$ is (where $$c$$ is arbitrary constant)

A
$$y=x\cos{(c+x)}$$

B
$$y=x\tan{(x-c)}$$

C
$$y=x\csc{(c+x)}$$

D
none of these

Solution
Question 5
1 / -0

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

A
$$\log \left(\tan \dfrac{y}{2}\right) = c-2\sin x$$

B
$$\log\left( \tan \dfrac{y}{4}\right) = c-2\sin \left(\dfrac{x}{2}\right)$$

C
$$\log \left(\tan \left(\dfrac{y}{2} +\dfrac{\pi}{4}\right) \right)= c-2 \sin x$$

D
None of the above

Solution
Question 6
1 / -0

The solution of differential equation,$$\left(x+\tan y\right)dy=\sin 2y\,dx$$ is

A
$$xcoty=\dfrac{1}{2}\log |cosec2y-cot2y|+c$$

B
$$x=\tan y+c\sqrt{\tan y}$$

C
$$x=\cot y+c\sqrt{\cot y}$$

D
$$None\ of\ these$$

Solution
Question 7
1 / -0

If the general solution of the differential equation if $$y'=\frac { y }{ x } +\Phi \left( \frac { x }{ y } \right) $$, for some function $$\Phi $$, is given by $$yin|cx|=x$$, where c is an arbitrary constant, then $$\Phi $$ (2) is equal to :

A
4

B
$$\frac { 1 }{ 4 } $$

C
-4

D
$$-\frac { 1 }{ 4 } $$

Solution
Question 8
1 / -0

If a curve $$y=f(x)$$ passes through the point $$(1,-1)$$ and satisfies the differential equation, $$y(1+xy)dx=xdy$$, then $$f\left(-\dfrac {1}{2}\right)$$ is equal to

A
$$-\dfrac {4}{5}$$

B
$$\dfrac {2}{5}$$

C
$$\dfrac {4}{5}$$

D
$$-\dfrac {2}{5}$$

Solution
Question 9
1 / -0

General solution of differential equation $$x^{2}(x+y\frac{dy}{dx})+(x\frac{dy}{dx}-y)\sqrt{x^{2}+y^{2}}=0$$ is

A
$$\frac{1}{\sqrt{x^{2}+y^{2}}}+\frac{y}{x}=c$$

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

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

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

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

Solution
Question 10
1 / -0

Solution of differential equation $${ x }^{ 6 }dy+3{ x }^{ 5 }ydx=xdy-2y\ dx$$ is

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

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

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

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

Solution