Differential Equations Test

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Differential Equations Test - 48

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Weekly Quiz Competition

Question 1
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lf $$f (x)$$ and $$g (x)$$ are two solutions of the differential equation $$a\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$$, then $$f (x) - g (x)$$ is the solution of

A
$$a^{2}\displaystyle \frac{d^{2}y}{dx^{2}}+\displaystyle \frac{dy}{dx}+y=e^{\displaystyle x}$$

B
$$\displaystyle {a}^{2}\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$$

C
$$a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=0$$

D
$$a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$$

Solution

Given, Differential equation as $$a\dfrac{\mathrm{d^2} y}{\mathrm{d} x^2}+x^2\dfrac{\mathrm{d} y}{\mathrm{d} x}+y=e^x$$ has $$f(x),g(x)$$ as two solutions.

On substituting $$f(x)$$ in the given Differential equation,

$$\Rightarrow af''(x)+x^2f'(x)+f(x)=e^x ................ A$$

On substituting $$g(x)$$ in the given differential equation,

$$\Rightarrow ag''(x)+x^2g'(x)+g(x)=e^x ................B$$

Subtracting B from A,

$$\Rightarrow a\times (f''(x)-g''(x))+x^2\times(f'(x)-g'(x))+(f(x)-g(x))=0$$

It is in the form of $$a\dfrac{\mathrm{d^2} y}{\mathrm{d} x^2}+x^2\dfrac{\mathrm{d} y}{\mathrm{d} x}+y=0$$, which have a solution of $$y=f(x)-g(x)$$
Question 2
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Solution of $$\dfrac{d^{2}y}{dx^{2}}$$= $$\log x$$ is:

A
$$y=\dfrac{1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$$

B
$$y=\dfrac{1}{2}x^{2}\log x+\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$$

C
$$y=\dfrac{-1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$$

D
$$y=3x^{2}+4x+c_{1}$$

Solution

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

$$\Rightarrow \dfrac{d}{dx}y_{1}=\log x$$

$$\Rightarrow \int dy_{1}=\int \log x dx+c_{1}$$

$$\int \log dx=x\log x-x+{ c }_{ 1 }$$

$$\Rightarrow y_{1}=x\log x-x+c_{1}$$

$$\int dy = \int (x \log x-x+c_{1})dx+c_{2}$$

$$y=\int x\log xdx-\dfrac{x^{2}}{2}+c_{1}x+c_{2}$$

$$\int x\log xdx=x(x\log x-x)-\int(x\log x-x)dx$$

$$\rightarrow 2\int x\log xdx=x^{2}\log x-x^{2}+x^{2}/_2=x^{2}\log x^-x^{2}/_2$$

$$\Rightarrow \int x\log xdx=\dfrac{x^{2}\log x}{2}-x^{2}/_4$$

$$\Rightarrow y=\dfrac{x^{2}\log x}{2}-\dfrac {3x^{2}}{4}+c_{1}x+c_{2}$$
Question 3
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lf the solution of the differential equation
$$\displaystyle \frac{1}{\sin^{-1}x}(\frac{dy}{dx})=1$$ is $$y=x \sin^{-1}x+f(x)+c$$ then f (x) is

A
$$1-x^{2}$$

B
$$\sqrt{1-x^{2}}$$

C
$$1+x^{2}$$

D
$$\sqrt{1+x^{2}}$$

Solution
Question 4
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The solution of $$\displaystyle \frac{dy}{dx}+\frac{x(1+y^{3})}{y^{2}(1+x^{2})}=0$$ is:

A
$$(1-x^{2})+(1+y^{3})=c$$

B
$$(1+y^{2})+(1+x^{3})=c$$

C
$$(1+x^{2})^{3}(1+y^{3})^{2}=c$$

D
$$(1+x^{2})+(1+y^{3})=cxy^{2}$$

Solution

$$\dfrac{dy}{dx}+\dfrac{x\left ( 1+y^{3} \right )}{y^{2}\left ( 1+x^{2} \right )}=0$$
$$\Rightarrow \dfrac{1}{3}\left ( \dfrac{3y^{2}}{1+y^{3}}dy \right )+\dfrac{1}{2}\left ( \dfrac{2x}{1+x^{2}}dx \right )=0$$
put$$1+x^{2}=t;$$
$$1+y^{3}=k$$
$$\Rightarrow 2xdx=dt$$
$$\Rightarrow 3y^{2}dy=dk$$
$$\displaystyle\Rightarrow \dfrac{1}{3}\int \dfrac{dk}{k}+\dfrac{1}{2}\int \dfrac{dt}{t}=c$$
$$\Rightarrow \log k^{\frac{1}{3}}t^{\frac{1}{2}}=c$$
$$\Rightarrow \log k^{2}+3=log c$$
$$\left ( 1+y^{3} \right )^{2}\left ( 1+x^{2} \right )^{3}=c$$
Question 5
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The solution of $$(1-x^{2})\displaystyle \frac{dy}{dx}+xy=xy^{2}$$ is:

A
$$ y(c+\sqrt{1-x^{2}})=1-x^{2}$$

B
$$y(c-\sqrt{1-x^{2}})=1-x^{2}$$

C
$$y(c+\sqrt{1-x^{2}})=\sqrt{1-x^{2}}$$

D
$$y(c-\sqrt{1-x^{2}})=\sqrt{1-x^{2}}$$

Solution

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

$$\dfrac{dy}{dx}=\dfrac{xy(y-1)}{1-x^{2}}$$

$$\Rightarrow \dfrac{dy}{y(y-1)}=\dfrac{x}{1-x^{2}}dx$$

$$\Rightarrow\displaystyle \int \left(\frac{1}{y-1}-\dfrac{1}{y}\right)dy=-\dfrac{1}{2}\int \dfrac{-2x}{1-x^{2}}dx+c_{1}$$

$$\Rightarrow \log\left(\dfrac{y-1}{y}\right)=-\dfrac{1}{2}\log(1-x^{2})+\log{c_{1}}$$

$$\Rightarrow \dfrac{y-1}{y}=\dfrac{c_{1}}{\sqrt{1-x^{2}}}$$

$$\Rightarrow y\sqrt{1-x^{2}}-\sqrt{1-x^{2}}=c_{1}y$$

$$\Rightarrow y(\sqrt{1-x^{2}}-c_{1})=\sqrt{1-x^{2}}$$

Put $$c_{1}=-c$$

$$\Rightarrow y(c+\sqrt{1-x^{2}})=\sqrt{1-x^{2}}$$
Question 6
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Solution of $$\>y-x\displaystyle \frac{dy}{dx}=5\left(y^{2}+\frac{dy}{dx}\right)$$, is:

A
$$(5+x)(1-5y)=cy$$

B
$$\dfrac { 5+x }{ 1-5y } =cy$$

C
$$(5-x)(1+5y)=cy$$

D
$$(5+x)(5-x)=c$$

Solution

$$5y^{2} - y = -\dfrac{dy}{dx} (5+x)$$

$$\Rightarrow -\dfrac{dx}{5+x} = \dfrac{dy}{y(5y-1)}$$

$$\Rightarrow -\dfrac{dx}{5+x} = \dfrac{5y - (5y-1)}{y(5y-1)}dy$$

Integrating both sides, we get
$$\Rightarrow\displaystyle\int\dfrac{dx}{5+x} = \int\left(\dfrac{5}{1-5y} + \frac{1}{y}\right)dy$$

$$\Rightarrow \log(5+x) = -\log(1-5y)+\log y + \log c; (\log c$$ is a constant$$)$$

$$\Rightarrow yc = (5+x)(1-5y)$$
Question 7
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If $$ \dfrac{dy}{dx}=xy+2x+3y+6$$, then find the value of $$y(-1)-e^2y(-3)$$.

A
$$e^2-1$$

B
$$e^2+1$$

C
$$2(e^2-1)$$

D
$$-2(e^2-1)$$

Solution

$$\dfrac{dy}{dx}=(x+3)(y+2)$$, If $$X=x+3, Y=y+2$$

$$\dfrac{dY}{Y}=XdX $$
$$ \Rightarrow Y=A. e^{X^2/2}, y=-2 +Ae^{(x+3)^2/2}$$
$$y(-1)=-2+Ae^2, y(-3)=-2+A$$
$$y(-1)-e^2 y(-3)=-2 +2e^2 =2(e^2-1)$$
Question 8
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A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k, then show that the differential equation describing such curves Is, $$\displaystyle y\frac{dy}{dx}=\pm \sqrt{k^{2}-y^{2}}$$. Find the equation of such a curve passing through (0.k), the curve is a:

A
line

B
circle

C
parabola

D
ellipse

Solution

Equation of normal at point $$P(x,y)$$ to the curve is
$$\displaystyle Y-y=-\frac { dx }{ dy } \left( X-x \right) $$
This straight line meets the x-axis in $$O(X,0)$$ where the number $$X$$ satisfies.
$$\displaystyle 0-y=-\frac { dx }{ dy } \left( X-x \right) \Rightarrow X=x+y\frac { dy }{ dx } $$
Thus $$\displaystyle { PO }^{ 2 }={ \left( X-x \right) }^{ 2 }+{ \left( 0-y \right) }^{ 2 }\Rightarrow { \left( y\frac { dy }{ dx } \right) }^{ 2 }$$
$$\displaystyle \Rightarrow y\frac { dy }{ dx } =\pm \sqrt { { k }^{ 2 }-{ y }^{ 2 } } ={ k }^{ 2 }-{ y }^{ 2 }$$
To find the equation we rewrite it as $$\displaystyle \frac { ydy }{ \sqrt { { k }^{ 2 }-{ y }^{ 2 } } } =\pm dx$$
Integrating we get
$$\displaystyle \int { \frac { ydy }{ \sqrt { { k }^{ 2 }-{ y }^{ 2 } } } } =\pm \int { dx } $$
$$\Rightarrow -\sqrt { { k }^{ 2 }-{ y }^{ 2 } } =\pm x+c$$ ...(1)
As the curve passes through $$(0,k)$$ we get
$$-\sqrt { { k }^{ 2 }-k^{ 2 } } =\pm \left( o \right) +c\Rightarrow c=o$$
Therefore equations (1) can be witten as $$-\sqrt { { k }^{ 2 }-{ y }^{ 2 } } =\pm x.$$
$$\Rightarrow { k }^{ 2 }-{ y }^{ 2 }={ x }^{ 2 }\Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ k }^{ 2 }$$
Question 9
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The solution of differential equation $$(e^x + 1) y dy = (y + 1) e^x dx$$ is :

A
$$(e^x + 1) (y + 1) = ce^y$$

B
$$(e^x + 1) (y + 1) = ce^{-y}$$

C
$$(e^x + 1) (y + 1) = ce^{2y}$$

D
none of the above

Solution

The given differential equation is $$(e^x + 1) y dy = (y + 1) e^x dx$$
$$\Rightarrow \displaystyle \frac{ydy}{(y+1)}=\displaystyle \frac{e^{x}}{(e^{x}+1)}dx;$$ Integrating both sides
$$\Rightarrow y- log |y + 1| = log (e^x + 1) + log k$$
$$\Rightarrow y = log |(y + 1)(e^x + 1)| + log k$$
$$\Rightarrow (y + 1)(e^x + 1) = ce^y$$
Question 10
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The solution of differential equation $$(e^x + 1)y dy = (y + 1) e^x dx$$ is:

A
$$(e^x + 1) (y + 1) = c . e^y$$

B
$$(e^x + 1) |(y + 1)| = ce^{-y}$$

C
$$(e^x + 1) (y + 1) = c$$

D
none of these

Solution

The given differential equation is $$(e^x + 1) y . dy = (y + 1) . e^x . dx$$
$$\Rightarrow \displaystyle \frac{y.dy}{(y+1)}=\displaystyle \frac{e^{x}}{(e^{x}+1)}.dx$$
$$\Rightarrow \int\left ( 1-\displaystyle \frac{1}{y+1} \right ).dy=\int \displaystyle \frac{e^{x}}{e^{x}+1}.dx$$
$$\Rightarrow y - log |y +1| = log (e^x + 1) + log k$$
$$\Rightarrow (y+1) (e^x+1)=e^y.c$$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 48

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Differential Equations Test - 48

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SHARING IS CARING

Question 1

$$a^{2}\displaystyle \frac{d^{2}y}{dx^{2}}+\displaystyle \frac{dy}{dx}+y=e^{\displaystyle x}$$

$$\displaystyle {a}^{2}\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$$

$$a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=0$$

$$a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$$

Solution

Question 2

$$y=\dfrac{1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$$

$$y=\dfrac{1}{2}x^{2}\log x+\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$$

$$y=\dfrac{-1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}$$

$$y=3x^{2}+4x+c_{1}$$

Solution

Question 3

$$1-x^{2}$$

$$\sqrt{1-x^{2}}$$

$$1+x^{2}$$

$$\sqrt{1+x^{2}}$$

Solution

Question 4

$$(1-x^{2})+(1+y^{3})=c$$

$$(1+y^{2})+(1+x^{3})=c$$

$$(1+x^{2})^{3}(1+y^{3})^{2}=c$$

$$(1+x^{2})+(1+y^{3})=cxy^{2}$$

Solution

Question 5

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

$$y(c-\sqrt{1-x^{2}})=1-x^{2}$$

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

$$y(c-\sqrt{1-x^{2}})=\sqrt{1-x^{2}}$$

Solution

Question 6

$$(5+x)(1-5y)=cy$$

$$\dfrac { 5+x }{ 1-5y } =cy$$

$$(5-x)(1+5y)=cy$$

$$(5+x)(5-x)=c$$

Solution

Question 7

$$e^2-1$$

$$e^2+1$$

$$2(e^2-1)$$

$$-2(e^2-1)$$

Solution

Question 8

line

circle

parabola

ellipse

Solution

Question 9

$$(e^x + 1) (y + 1) = ce^y$$

$$(e^x + 1) (y + 1) = ce^{-y}$$

$$(e^x + 1) (y + 1) = ce^{2y}$$

none of the above

Solution

Question 10

$$(e^x + 1) (y + 1) = c . e^y$$

$$(e^x + 1) |(y + 1)| = ce^{-y}$$

$$(e^x + 1) (y + 1) = c$$

none of these

Solution

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