Differential Equations Test

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Differential Equations Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The solution of the differential equation $$\sec^{2} x \cdot \tan y \,dx + \sec^{2} y \cdot \tan x\ dy = 0$$ is

A
$$\tan x \cdot \cot y = C$$

B
$$\cot x \cdot \tan y = C $$

C
$$\tan x \cdot \tan y = C$$

D
$$\sin x \cdot \cos y = C$$

Solution

Given, $$\sec^{2} x \cdot \tan y dx+\sec^{2} y \cdot \tan x dy =0$$

On separating the varaibales, we get

$$\Rightarrow \sec^{2} x\cdot \tan y dx = - \sec^{2} y\cdot \tan \, x dy$$

$$\Rightarrow \dfrac{\sec^2x}{\tan \, x}dx=- \dfrac{\sec^{2}y}{\tan y}dy$$

On integration both the sides, we get

$$\displaystyle \int \frac{\sec^{2}x}{\tan x}dx= -\int \frac{\sec^{2}y}{\tan y}dy$$

Let $$\tan x=u \Rightarrow \sec^{2}x=\dfrac{du}{dx}$$

$$\Rightarrow dx =\dfrac{du}{\sec^{2}x}$$ and $$\tan \, y=v$$

$$\Rightarrow \sec^{2}y=\dfrac{dv}{dy}$$

$$\Rightarrow dy =\dfrac{dv}{\sec^2 y}$$

$$\displaystyle \therefore \int \frac{\sec^{2} x}{u}\cdot \frac{du}{\sec^{2} x}=-\int \frac{\sec^2y}{v}\cdot \frac{dv}{\sec^2y}$$

$$\Rightarrow \displaystyle \int \dfrac{du}{u}=\int \dfrac{dv}{v}$$

$$\Rightarrow \log|u|=-\log |v|+\log|C|$$

$$\Rightarrow \log|\tan \, x|=-\log |\tan y| + \log |C|$$

$$\Rightarrow \log|\tan x\cdot \tan y| = \log |C|$$

$$(\because \log m + \log n=\log mn)$$

$$ (\because \log m =\log n\Rightarrow m =n)$$

$$\Rightarrow \tan x\cdot \tan y = C$$
Question 2
1 / -0

The general solution of the differential equation $$xdy-ydx={y}^{2}dx$$ is

A
$$y=\cfrac { x }{ C-x } $$

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

C
$$(C+x)(2x)$$

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

E
$$x=\cfrac { y }{ C-x } $$

Solution

We have $$xdy-ydx={ y }^{ 2 }dx$$
$$\Rightarrow x\cfrac { dy }{ dx } -y={ y }^{ 2 }$$
$$\Rightarrow \cfrac { x\cfrac { dy }{ dx } -y }{ { y }^{ 2 } } =1\quad $$
$$\Rightarrow -\left( \cfrac { y-x\cfrac { dy }{ dx } }{ { y }^{ 2 } } \right) =1$$
$$\Rightarrow -\cfrac { d }{ dx } \cfrac { x }{ y } =1$$
$$\Rightarrow \cfrac { x }{ y } =-x+C$$
$$\Rightarrow y=\cfrac { x }{ C-x } $$
Question 3
1 / -0

Solution of $$\cfrac { dx }{ dy } +mx=0$$, where $$m< 0$$ is:

A
$$x=C{ e }^{ my }$$

B
$$x=C{ e }^{ -my }$$

C
$$x=my+C$$

D
$$x=C$$

Solution

Given $$\dfrac{dx}{dy}+mx=0, m<0$$

$$\Rightarrow \dfrac{dx}{dy}=-mx$$

$$\Rightarrow \dfrac{dx}{x}=-m\:dy$$

Integrating both sides we get

$$\Rightarrow ln\: x=-my+c$$

Raising to power $$e$$ on both sides we get,

$$x=e^{-my+c}$$

$$\Rightarrow x=e^{-my} \cdot e^c$$

$$\Rightarrow x=Ce^{-my}$$
Question 4
1 / -0

The general solution of the differential equation $$\left( x+y \right) dx+xdy=0$$ is

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

B
$$2{ x }^{ 2 }-{ y }^{ 2 }=C$$

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

D
$${ y }^{ 2 }+2xy=C$$

Solution

Given differential equation is

$$\left( x+y \right) dx+xdy=0$$

$$\Rightarrow xdx=-\left( x+y \right) dy$$

$$\Rightarrow \dfrac { dy }{ dx } =\dfrac { -\left( x+y \right) }{ x }$$

It is a homogeneous differential equation.

So, putting $$y=vx\Rightarrow \dfrac { dy }{ dx } =v+x\dfrac { dv }{ dx }$$, we get

$$v+x\dfrac { dv }{ dx } =-\dfrac { x+vx }{ x } =-\dfrac { 1+v }{ 1 } $$

$$\Rightarrow x\dfrac { dv }{ dx } =-1-2v$$

$$\Rightarrow \displaystyle\int { \dfrac { dv }{ 1+2v } } =-\displaystyle\int { \dfrac { dx }{ x } } $$

$$\Rightarrow \log { \left( 1+2v \right) } =-\log { x } +\log { { C }_{ 1 } }$$

$$\Rightarrow \log { \left( 1+\dfrac { 2y }{ x } \right) } =2\log { \dfrac { { C }_{ 1 } }{ x } } $$

$$\Rightarrow \dfrac { x+2y }{ x } ={ \left( \dfrac { { C }_{ 1 } }{ x } \right) }^{ 2 }$$

$$\Rightarrow { x }^{ 2 }+2xy=C$$ (where, $$C={ C }_{ 1 }^{ 2 }$$)
Question 5
1 / -0

The solution of the differential equation $$\left( { x }^{ 2 }-y{ x }^{ 2 } \right) \dfrac { dy }{ dx } +{ y }^{ 2 }+x{ y }^{ 2 }=0$$ is

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

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

C
$$\log { \left( xy \right) } =\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +C$$

D
$$\log { \left( xy \right) } +\dfrac { 1 }{ x } +\dfrac { 1 }{ y } =C$$

Solution

Given differential equation is
$$\left( { x }^{ 2 }-y{ x }^{ 2 } \right) \dfrac { dy }{ dx } +{ y }^{ 2 }+x{ y }^{ 2 }=0$$
$$\Rightarrow \dfrac { 1-y }{ { y }^{ 2 } } dy+\dfrac { 1+x }{ { x }^{ 2 } } dx=0$$
$$\Rightarrow \left( \dfrac { 1 }{ { y }^{ 2 } } -\dfrac { 1 }{ y } \right) dy+\left( \dfrac { 1 }{ { x }^{ 2 } } +\dfrac { 1 }{ x } \right) dx=0$$
On integrating both sides, we get the required solution
$$\dfrac { -2 }{ y } -\log { y } -\dfrac { 2 }{ x } +\log { x } =C$$
$$\Rightarrow \log { \left( \dfrac { x }{ y } \right) } =\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +C$$
Question 6
1 / -0

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

A
$$\log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =-2\cos { \cfrac { x }{ 2 } } +C$$

B
$$\log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =2\cos { \cfrac { x }{ 2 } } +C$$

C
$$\log _{ e }\left|{ \tan { \cfrac { y }{ 2 } } }\right| =2 \sin { \cfrac { x }{ 2 } } +C$$

D
$$\log _{ e }\left|{ \tan { \cfrac { y }{ 2 } } }\right| =-2\sin { \cfrac { x }{ 2 } } +C\quad $$

Solution

We have
$$\cfrac { dy }{ dx } +\sin { \cfrac { x+y }{ 2 } } =\sin { \cfrac { x-y }{ 2 } } $$
$$\Rightarrow \cfrac { dy }{ dx } =\sin { \cfrac { x-y }{ 2 } } -\sin { \cfrac { \\ x+y }{ 2 } } $$
$$\Rightarrow \cfrac { dy }{ dx } =2\cos { \cfrac { \left( \cfrac { x-y }{ 2 } +\cfrac { x+y }{ 2 } \right) }{ 2 } } \sin { \cfrac { \left( \cfrac { x-y }{ 2 } -\cfrac { x+y }{ 2 } \right) }{ 2 } }$$ ..... $$\left[ \because \sin { C } -\sin { D } =2\cos { \cfrac { C+D }{ 2 } } .\sin { \cfrac { C-D }{ 2 } } \right] $$
$$\Rightarrow \cfrac { dy }{ dx } =2\cos { \cfrac { x }{ 2 } } .\sin { \cfrac { -y }{ 2 } } $$
$$\Rightarrow \cfrac { dy }{ dx } =-2\cos { \left( \cfrac { x }{ 2 } \right) } .\sin { \left( \cfrac { y }{ 2 } \right) } $$
$$\Rightarrow \cfrac { dy }{ \sin { \left( \cfrac { y }{ 2 } \right) } } =-2\cos { \left( \cfrac { x }{ 2 } \right) } $$
On integrating both sides, we get
$$\Rightarrow \displaystyle \int \text{cosec} { \left( \cfrac { y }{ 2 } \right) } dy=-2\int _{ }^{ }{ \cos { \left( \cfrac { x }{ 2 } \right) } dx } $$
$$\Rightarrow 2\log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =-4\sin { \left( \cfrac { x }{ 2 } \right) } +C$$
$$\Rightarrow \log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =-2\sin { \left( \cfrac { x }{ 2 } \right) } +C$$
Question 7
1 / -0

The solution of $$x \log x \displaystyle\frac{dy}{dx}+y=1$$ is?

A
$$\log x=\displaystyle\frac{c}{(y-1)}$$

B
$$y\log x\displaystyle\frac{dy}{dx}+y=1$$

C
$$xy=\log (\log x)+c$$

D
$$\displaystyle\frac{x}{y}\log y=c$$

Solution

$$\left( x\log { x } \right) dy=\left( 1-y \right) dx$$
$$\cfrac { dy }{ 1-y } =\cfrac { dx }{ x\log { x } } $$
Integrating both sides
$$\int { \cfrac { dy }{ 1-y } } =\int { \cfrac { dx }{ x\log { x } } } $$
Let $$\log { x } =t,dt=\cfrac { 1 }{ x } dx$$
$$-\ln { \left( 1-y \right) } =\int { \cfrac { dt }{ t } } =\ln { t } +\ln { c }(constant) $$
$$\ln { \left( \cfrac { 1 }{ 1-y } \right) } =\ln { \left( tc \right) } $$
$$tc=\cfrac { 1 }{ 1-y } $$
$$\log { x } =\cfrac { 1 }{ c\left( 1-y \right) } =\cfrac { k }{ y-1 } $$
where $$k=-\cfrac { 1 }{ c } $$
Question 8
1 / -0

Solution of the differential equation $$(x^{2} + y^{3}) (2x^{2}dx + 3ydy) = 12x\ dx + 18y^{2}dy$$ is

A
$$\dfrac {2}{3}x^{3} + \dfrac {3}{2}y^{2} = 6ln (x^{2} + y^{3}) + c$$

B
$$x^{2} + y^{3} = 9ln (x^{2} + y^{3}) + c$$

C
$$\dfrac {2}{3}x^{3} + \dfrac {3}{2}y^{2} = 6ln (x^{3} + y^{2}) + c$$

D
$$x^{3} + y^{2} = 6ln (x^{2} + y^{3}) + c$$

Solution

$$LHS\longrightarrow (x^{ 2 }+y^{ 3 })(2{ x }^{ 2 }dx+3ydy)\\ =2{ x }^{ 4 }dx+3{ x }^{ 2 }ydy+2{ x }^{ 2 }y^{ 3 }dx+3{ y }^{ 4 }dy\\ \Rightarrow 2{ x }^{ 4 }dx+3{ x }^{ 2 }ydy+2{ x }^{ 2 }y^{ 3 }dx+3{ y }^{ 4 }dy=12xdx+18{ y }^{ 2 }dy\\ (or)\; the\; given\; question\; can\; be\; written\; as\\ (2{ x }^{ 2 }dx+3ydy)=\cfrac { 6(2{ x }^{ }dx+3y^2dy) }{ x^{ 2 }+y^{ 3 } } \\ integrating\; \Rightarrow \cfrac { 2 }{ 3 } { x }^{ 3 }+\cfrac { 3 }{ 2 } { y }^{ 2 }=6ln(x^{ 2}+y^{ 3 })+C\\ $$
Question 9
1 / -0

The solution of $$y' = e^{x - y} + x^{2} e^{-y}$$ is

A
$$3(e^{y} - e^{x}) - x^{3} = c$$

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

C
$$e^{y} - e^{x} + x^{3} = c$$

D
$$3(e^{y} - e^{x}) + x^{3} = c$$

Solution

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

$$e^y dy =(e^x +x^2)dx$$

integrating both side,

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

$$3(e^y -e^x) -x^3 = c$$
Question 10
1 / -0

If $$y=\sqrt{(a-x)(x-b)}-(a-b)\tan^{-1}\sqrt{\displaystyle\frac{a-x}{x-b}}(a > b)$$ then $$\displaystyle\frac{dy}{dx}=$$.

A
$$\sqrt{\displaystyle \frac{a-x}{x-b}}$$

B
$$\sqrt{(a-x)(x-b)}$$

C
$$0$$

D
$$1$$

Solution

Given, $$y=\sqrt {(a-x) (x-b)}-(a-b) \tan^{-1} \sqrt {\dfrac {a-x}{x-b}}$$Let $$u=(a-x) (x-b), \surd =\sqrt {\dfrac {a-x}{x-b}}$$$$\Rightarrow \ y=\sqrt u - (a-b)\tan^{-1}v$$Now, differentiating wrt $$x$$$$\Rightarrow \ \dfrac {dy}{dx} =\dfrac {d}{dx}(\sqrt u)- (a-b)\dfrac {d}{dx} \tan^{-1}v$$$$=\dfrac {d(\sqrt u)}{du}.\dfrac {du}{dx}-(a-b) \dfrac {d\tan^{-1}v}{dv}.\dfrac {dv}{dx}$$ [chain rule]$$=\dfrac {1}{2\sqrt u}. \dfrac {du}{dx} -(a-b) \dfrac {1}{1+v^2}.\dfrac {dv}{dx}......(1)$$$$\dfrac {du}{dx} =\dfrac {d}{dx} (a-x) (x-b) =\dfrac {d}{dx} (ax-av-x^2 +bx)=a-2x+b$$$$\dfrac {dv}{dx}=\dfrac {d}{dx} \left (\sqrt {\dfrac {a-x}{x-b}}\right)=\dfrac {1}{2\sqrt {\dfrac {a-x}{x-b}}}\times \dfrac {(x-b)(-1)-(a-x)(1)}{(x-b)^2}$$ [quotent rule]$$=\dfrac {\sqrt {x-b}}{2\sqrt {a-x}}\times \dfrac {(b-a)}{(x-b)^2}$$substituting in $$(1)$$$$\dfrac {dy}{dx}=\dfrac {1}{2\sqrt {(a-x)(x-b)}}\times (a-2x+b)-(a-b)\dfrac {1}{1+\dfrac {a-x}{x-b}}\times \dfrac {\sqrt {x-b}}{\sqrt {b-x}}\times \dfrac {(b-a)}{(x-b)^2}$$$$=\dfrac {a-2x+b}{2\sqrt {(a-x)(x-b)}}-\dfrac {(ab)(xb)}{(a-b)}\times \dfrac {\sqrt {x-b}}{2\sqrt {a-x}}\dfrac {(b-a)}{(x-b)^2}$$$$=\dfrac {a-2x+b+a-b}{2\sqrt {(a-x)(x-b)}}$$$$=\dfrac {2(a-x)}{2\sqrt {(a-x)(x-b)}}$$$$\Rightarrow \boxed {\dfrac {dy}{dx}=\sqrt {\dfrac {a-x}{x-b}}}$$option $$A$$ is corect

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 33

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Differential Equations Test - 33

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SHARING IS CARING

Question 1

$$\tan x \cdot \cot y = C$$

$$\cot x \cdot \tan y = C $$

$$\tan x \cdot \tan y = C$$

$$\sin x \cdot \cos y = C$$

Solution

Question 2

$$y=\cfrac { x }{ C-x } $$

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

$$(C+x)(2x)$$

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

$$x=\cfrac { y }{ C-x } $$

Solution

Question 3

$$x=C{ e }^{ my }$$

$$x=C{ e }^{ -my }$$

$$x=my+C$$

$$x=C$$

Solution

Question 4

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

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

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

$${ y }^{ 2 }+2xy=C$$

Solution

Question 5

$$\log { \left( \dfrac { x }{ y } \right) } =\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +C$$

$$\log { \left( \dfrac { y }{ x } \right) } =\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +C$$

$$\log { \left( xy \right) } =\dfrac { 1 }{ x } +\dfrac { 1 }{ y } +C$$

$$\log { \left( xy \right) } +\dfrac { 1 }{ x } +\dfrac { 1 }{ y } =C$$

Solution

Question 6

$$\log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =-2\cos { \cfrac { x }{ 2 } } +C$$

$$\log _{ e }{ \left| \tan { \cfrac { y }{ 2 } } \right| } =2\cos { \cfrac { x }{ 2 } } +C$$

$$\log _{ e }\left|{ \tan { \cfrac { y }{ 2 } } }\right| =2 \sin { \cfrac { x }{ 2 } } +C$$

$$\log _{ e }\left|{ \tan { \cfrac { y }{ 2 } } }\right| =-2\sin { \cfrac { x }{ 2 } } +C\quad $$

Solution

Question 7

$$\log x=\displaystyle\frac{c}{(y-1)}$$

$$y\log x\displaystyle\frac{dy}{dx}+y=1$$

$$xy=\log (\log x)+c$$

$$\displaystyle\frac{x}{y}\log y=c$$

Solution

Question 8

$$\dfrac {2}{3}x^{3} + \dfrac {3}{2}y^{2} = 6ln (x^{2} + y^{3}) + c$$

$$x^{2} + y^{3} = 9ln (x^{2} + y^{3}) + c$$

$$\dfrac {2}{3}x^{3} + \dfrac {3}{2}y^{2} = 6ln (x^{3} + y^{2}) + c$$

$$x^{3} + y^{2} = 6ln (x^{2} + y^{3}) + c$$

Solution

Question 9

$$3(e^{y} - e^{x}) - x^{3} = c$$

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

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

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

Solution

Question 10

$$\sqrt{\displaystyle \frac{a-x}{x-b}}$$

$$\sqrt{(a-x)(x-b)}$$

$$0$$

$$1$$

Solution

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