Differential Equations Test

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Differential Equations Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

For $$x\epsilon R, x\neq 0$$, if $$y(x)$$ is a differentiable function such that $$x\int_{1}^{x}y (t) dt = (x + 1) \int_{1}^{x} t y (t) dt$$, then $$y (x)$$ equals:
(Where C is a constant)

A
$$Cx^{3} e^{\dfrac {1}{x}}$$

B
$$\dfrac {C}{x^{2}} e^{-\dfrac {1}{x}}$$

C
$$\dfrac {C}{x} e^{-\dfrac {1}{x}}$$

D
$$\dfrac {C}{x^{3}} e^{-\dfrac {1}{x}}$$

Solution

$$x\int_{1}^{x}y (t) dt = x \int_{1}^{x} ty (t) dt + \int_{1}^{x} ty (t) dt$$
differentiate w.r. to x.
$$\int_{1}^{x} y (t) dt + x [y (x) - y(1)] = \int_{1}^{x} ty (t) dt + x [xy (x) - y(1)] + xy (x) - y (1)$$
$$\int_{1}^{x} y (t) dt = \int_{1}^{x}t y (t) dt + x^{2} y (x) - y (1)$$
diff. again w.r to x
$$y(x) - y(1) = xy (x) - y(1) + 2xy (x) + x^{2}y^{'} (x)$$
$$(1 - 3x) y (x) = x^{2}y^{'} (x)$$
$$\dfrac {y^{'}(x)}{y(x)} = \dfrac {1 - 3x}{x^{2}}$$
$$\dfrac {1}{y}\dfrac {dy}{dx} = \dfrac {1 - 3x}{x^{2}} \Rightarrow ln\ y = -\dfrac {1}{x} - 3 ln\ x$$
$$ln (y x^{3}) = -\dfrac {1}{x}+lnc$$
$$yx^{3} = ce^{-\dfrac {1}{x}}$$
$$y = c\dfrac {e^{-\dfrac {1}{x}}}{x^{3}}$$ or $$y = \dfrac {ce^{-\dfrac {1}{x}}}{x^{3}}$$
Question 2
1 / -0

Let $$y = y(x)$$ be a solution of the differential equation, $$\sqrt{1-x^2}\dfrac{dy}{dx} + \sqrt{1-y^2} = 0, |x| < 1$$.
If $$y\left(\dfrac{1}{2}\right) = \dfrac{\sqrt{3}}{2}$$, then $$y \left(\dfrac{-1}{\sqrt{2}}\right)$$ is equal to:

A
$$\dfrac{\sqrt{3}}{2}$$

B
$$\dfrac{1}{\sqrt{2}}$$

C
$$-\dfrac{\sqrt{3}}{2}$$

D
$$-\dfrac{1}{\sqrt{2}}$$

Solution

$$\sqrt{1 - x^2} \dfrac{dy}{dx} + \sqrt{1-y^2} = 0$$

$$\Rightarrow \dfrac{dy}{\sqrt{1-y^2}} + \dfrac{dx}{\sqrt{1-x62}} = 0$$

Taking integration

$$\displaystyle\Rightarrow \int\dfrac{dy}{\sqrt{1-y^2}} + \int\dfrac{dx}{\sqrt{1-x^2}} = 0$$

$$\Rightarrow \sin^{-1}y + \sin^{-1} x = c$$ ....(1)

At $$x = \dfrac{1}{2}$$, $$y = \dfrac{\sqrt{3}}{2}$$

$$\Rightarrow\sin^{-1} \left(\dfrac{\sqrt{3}}{2} \right) + \sin^{-1}\left(\dfrac{1}{2}\right) = c$$

$$\Rightarrow\dfrac{\pi}3+\dfrac{\pi}6=c$$

$$\Rightarrow c = \dfrac{\pi}{2}$$

Put in equation (1)

$$\sin^{-1} y = \cos^{-1}x$$

$$y\left(\dfrac{-1}{\sqrt{2}}\right) = \sin\left(\cos^{-1}\left(\dfrac{-1}{\sqrt{2}}\right)\right) = \dfrac{1}{\sqrt{2}}$$

$$\therefore$$ $$\boxed{y\left(\dfrac{-1}{\sqrt{2}}\right) = \dfrac{1}{\sqrt{2}}}.....Answer$$

Hence option $$'B'$$ is the answer.
Question 3
1 / -0

The differential equation $$\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\sqrt{1-\mathrm{y}^{2}}}{\mathrm{y}}$$ determines a family of circles with:

A
variable radii and a fixed centre at $$(0,1)$$

B
variable radii and a fixed centre at $$(0, -1)$$

C
fixed radius 1 and variable centres along the x-axis

D
fixed radius 1 and variable centres along the y-axis

Solution

$$\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\sqrt{1-\mathrm{y}^{2}}}{\mathrm{y}}$$

$$\displaystyle \Rightarrow\int\frac{\mathrm{y}}{\sqrt{1-\mathrm{y}^{2}}}$$ dy $$=\displaystyle \int \mathrm{d}\mathrm{x}$$

$$\Rightarrow-\sqrt{1-\mathrm{y}^{2}}=\mathrm{x}+\mathrm{c}$$

$$\Rightarrow(\mathrm{x}+\mathrm{c})^{2}+\mathrm{y}^{2}=1
$$ centre $$(-\mathrm{c}, 0)$$ ; radius $$\sqrt{\mathrm{c}^{2}-\mathrm{c}^{2}+1}=1$$.
Question 4
1 / -0

Check whether the function is homogenous or not. If yes then find the degree of the function
$$g(x)=x^2-8x^3$$.

A
Not homogenous

B
Homogenous with degree 4

C
Homogenous with degree 2

D
None of these

Solution

Given function is $$g\left( x \right) ={ x }^{ 2 }-8{ x }^{ 3 }$$
To check its Homogenitiy, we find $$g(kx)$$ which gives,
$$g\left( kx \right) ={ { k }^{ 2 }x }^{ 2 }-8{ k }^{ 3 }{ x }^{ 3 }$$
Now,
$$\cfrac { g\left( kx \right) }{ g\left( x \right) } =\cfrac { { k }^{ 2 }{ x }^{ 2 }-8{ k }^{ 3 }{ x }^{ 3 } }{ { x }^{ 2 }-8{ x }^{ 3 } } \neq { k }^{ n }$$ where $$k>0$$ and $$n$$ is an integer.
$$\therefore$$ this function is not homogenous.
Question 5
1 / -0

Solution of differential equation $$\displaystyle \frac{dy}{dx} = sin x + 2x$$, is

A
$$y = x^2 - cos x + c$$

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

C
$$y = cos x + 2$$

D
$$y = cos x + 2 + C$$

Solution

$$\displaystyle \frac{dy}{dx} = sin x + 2x$$
$$y = - cos x + \displaystyle \frac{2x^2}{2} = x^2 - cos x + c$$
Question 6
1 / -0

The order of the differential equation
$$2x^2\dfrac{d^2y}{dx^2} - 3\dfrac{dy}{dx} + y = 0$$ is

A
2

B
1

C
0

D
Not defined

Solution

Order is the highest derivative in the equation. Here it is $$2$$
Question 7
1 / -0

The solution of $$\dfrac{dy}{dx}=e^{logx}$$ is:

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

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

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

D
$$xy=x^{2}+c$$

Solution

$$\dfrac{dy}{dx}=e^{log\ x}=x$$
$$\Rightarrow \int dy=\int xdx+c$$
$$\Rightarrow y=\frac{x^{2}}{2}+c$$
$$\Rightarrow 2y=x^{2}+c$$
Question 8
1 / -0

The solution of $$\dfrac{dy}{dx}=\dfrac{2x}{3y^{2}}$$ is:

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

B
$$y^{2}+x^{3}=c$$

C
$$y^{3}-x^{2}=c$$

D
$$y^{3}-x=c$$

Solution

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

$$\Rightarrow \int 3y^{2}dy=\int2xdx+c$$

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

$$\rightarrow y^{3}-x^{2}=c$$
Question 9
1 / -0

The solution of $$\dfrac{dy}{dx}-\dfrac{2xy}{1+x^{2}}=0$$ is

A
$$y=c\left ( 1+x^{2} \right )$$

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

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

D
$$y=\dfrac{c}{\sqrt{1+x^{2}}}$$

Solution

$$\dfrac{dy}{dx}=\dfrac{2xy}{1+x^{2}}$$

$$\Rightarrow\displaystyle \int \dfrac{dy}{y}=\int \dfrac{2x}{1+x^{2}}dx$$

$$\Rightarrow \log\ y=\log(1+x^{2})c$$
$$y=c(1+x^{2})$$
Question 10
1 / -0

The solution of $$x^{2} \cfrac{dy}{dx}=2$$ is

A
$$y=-\cfrac{2}{x}+c$$

B
$$y=\cfrac{x}{2}+c$$

C
$$y=\cfrac{2}{x}+c$$

D
$$y = 2x + c$$

Solution

$$x^{2}\cfrac{dy}{dx}=2$$
$$\int dy=\int \cfrac{2}{x^{2}}dx+c$$
$$\Rightarrow y=\cfrac{-2}{x}+c$$
$$\Rightarrow y+\cfrac{2}{x}=c$$

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Differential Equations Test - 11

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Differential Equations Test - 11

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Question 1

$$Cx^{3} e^{\dfrac {1}{x}}$$

$$\dfrac {C}{x^{2}} e^{-\dfrac {1}{x}}$$

$$\dfrac {C}{x} e^{-\dfrac {1}{x}}$$

$$\dfrac {C}{x^{3}} e^{-\dfrac {1}{x}}$$

Solution

Question 2

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

$$\dfrac{1}{\sqrt{2}}$$

$$-\dfrac{\sqrt{3}}{2}$$

$$-\dfrac{1}{\sqrt{2}}$$

Solution

Question 3

variable radii and a fixed centre at $$(0,1)$$

variable radii and a fixed centre at $$(0, -1)$$

fixed radius 1 and variable centres along the x-axis

fixed radius 1 and variable centres along the y-axis

Solution

Question 4

Not homogenous

Homogenous with degree 4

Homogenous with degree 2

None of these

Solution

Question 5

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

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

$$y = cos x + 2$$

$$y = cos x + 2 + C$$

Solution

Question 6

2

1

0

Not defined

Solution

Question 7

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

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

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

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

Solution

Question 8

$$y^{3}+x^{2}=c$$

$$y^{2}+x^{3}=c$$

$$y^{3}-x^{2}=c$$

$$y^{3}-x=c$$

Solution

Question 9

$$y=c\left ( 1+x^{2} \right )$$

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

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

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

Solution

Question 10

$$y=-\cfrac{2}{x}+c$$

$$y=\cfrac{x}{2}+c$$

$$y=\cfrac{2}{x}+c$$

$$y = 2x + c$$

Solution

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