Differential Equations Test

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Differential Equations Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Solution of differential equation $$\sin y.\dfrac {dy}{dx}+\dfrac {1}{x}\cos y=x^{4}\cos^{2}y$$ is

A
$$x\sec y=x^{6}+C$$

B
$$6x\sec y=x+C$$

C
$$6x\sec y=x^{6}+C$$

D
$$6x\sec y=6x^{6}+C$$
Question 2
1 / -0

The general solution of the differential equation $$\dfrac{dy}{dx}=e^{x+y}$$ is :

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

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

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

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

Solution

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

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

$$\dfrac{dy}{e^y}=e^xdx$$

On integrating both side we get

$$\displaystyle\int \dfrac{dy}{e^y}=\displaystyle\int e^xdx$$

$$\Rightarrow -e^{-y}=e^x+c$$

$$e^{-y}+e^x=c$$.
Question 3
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The value of the constant $$'m'$$ and $$'c'$$ for which $$y=mx+c$$ is a solution of the differential equation $$D^2y-3Dy-4y=-4x$$.

A
Is $$m=-1; c=3/4$$

B
Is $$m=1; c=-3/4$$

C
No such real $$m, c$$

D
Is $$m=1; c=3/4$$

Solution

We have,
$$y=mx+c, \implies \dfrac{dy}{dx}=m, \implies \dfrac{d^{2}y}{dx^{2}}=0$$
Substituting in
$$D^2y-3Dy-4y=-4x$$.
i.e. $$\dfrac{d^{2}y}{dx^{2}}-3\dfrac{dy}{dx}-4y=-4x$$
$$0-3m-4(mx+c)=-4x$$
$$-3m-4c-4mx=-4x$$
$$4mx+(3m+4c)=4x+0$$
Comparing coefiiecient of $$x$$ we get,
$$4m=4\implies m=1$$
Comparing constant terms we get
$$3m+4c=0\implies 3(1)=-4c\implies c=-3/4$$
Question 4
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The solution of the differential equation $${ 2x }^{ 2 }y\dfrac { dy }{ dx } =tan\left( { x }^{ 2 }{ y }^{ 2 } \right) -{ 2xy }^{ 2 }$$ given $$y(1)=\sqrt { \dfrac { \pi }{ 2 } } $$ is

A
sin $$\left( { x }^{ 2 }{ y }^{ 2 } \right) -1=0$$

B
$$cos\left( \dfrac { \pi }{ 2 } +{ x }^{ 2 }{ y }^{ 2 } \right) +x=0$$

C
$$sin\left( { x }^{ 2 }{ y }^{ 2 } \right) ={ e }^{ x-1 }$$

D
$$sin\left( { x }^{ 2 }{ y }^{ 2 } \right) ={ e }^{ 2(x-1) }$$

Solution
Question 5
1 / -0

The solution of the equation $$(x^2 +xy)dy=(x^2+y^2)dx $$is

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

B
$$logx=2log(x-y)+\frac{y}{x}+c$$

C
$$logx=log(x-y)+\frac{x}{y}+c$$

D
none of these above

Solution

Given: $$(x^2+xy)dy=(x^2+y^2)dx$$
To find: Solution of the eq
Now,
$$(x^2+xy)dy=(x^2+y^2)dx$$
divide $$x^2$$ bo both side.
$$\therefore (1+\dfrac{y}{x})dy=(1+\dfrac{y^2}{x^2})dx$$

or , $$(1+\dfrac{y}{x})\dfrac{dy}{dx}=(1+\dfrac{y^2}{x^2})$$

Let, $$\dfrac{y}{x}=v\Rightarrow y=xv$$

or $$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$

$$\therefore (1+v)(v+\dfrac{xdv}{dx})=(1+v^2)$$

or, $$v+x\dfrac{dv}{dx}=\dfrac{1+v^2}{1+v}-v$$

or, $$x\dfrac{dv}{dx}=\dfrac{1+v^2-v-v^2}{1+v}$$

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

Integrating both side

$$\therefore \int (\dfrac{1+v}{1-v})dv=\int \dfrac{dx}{x}$$

or, $$\int (\dfrac{2}{1-v}-1)dv=\int \dfrac{dx}{x}$$

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

or, $$-2 \log (1-v)-v=\log +c$$

or $$2\log (1-v)+\log x+v=c$$

or, $$\log (1-v)^2+\log x+v=c$$

or $$\log \{x(1-\dfrac{y}{x})^2\}+\dfrac{y}{x}=c$$

or $$\log \{x(\dfrac{x-y}{x})^2\}=c-\dfrac{y}{x}$$

or $$(x-y)^2.\dfrac{1}{x}=c.e^{-y/x}$$

or $$(x-y)^2=xe^{-y/x}.c$$
Question 6
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Solve the given differential equation $$\dfrac{dy}{dx}=(cosx-sinx),$$

A
$$y=sinx+cosx+c$$

B
$$y=sinx-cosx+c$$

C
$$y=tanx+secx$$

D
$$\text{None of these}$$

Solution

Given

$$\dfrac {dy}{dx}=\cos x-\sin x$$

$$ dy=(\cos x-\sin x) dx$$

Integrating on Both sides

$$\displaystyle \int dy=\int (\cos x-\sin x) dx$$

$$y=\displaystyle \int \cos xdx-\int \sin x dx$$

$$y=\sin x+\cos x +c$$
Question 7
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Which of the following functions is differentiable at x=0?

A
$${ e }^{ -\left| x \right| }-\left| x \right| $$

B
$${ e }^{ \left| x \right| }+\left| x \right| $$

C
$$\left| x \right| -{ e }^{ \left| x \right| }$$

D
$$\left| x \right| -{ e }^{ -\left| x \right| }$$

Solution
Question 8
1 / -0

The solution of the differential equation $$x\dfrac{dy}{dx}=y(log y-log x+1)$$ is

A
$$y=xe^{cx}$$

B
$$y+xe^{cx}=0$$

C
$$y+e^{x}$$

D
none of these

Solution

$$x\dfrac{dy}{dx}=y\left(\log{y}-\log{x}+1\right)$$

$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{y}{x}\left(\log{\dfrac{y}{x}}+1\right)$$

Put $$y=vx\Rightarrow\,\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$

$$\Rightarrow\,v+x\dfrac{dv}{dx}=v\left(\log{v}+1\right)$$
$$\Rightarrow\,v+x\dfrac{dv}{dx}=v\log{v}+v$$
$$\Rightarrow\,x\dfrac{dv}{dx}=v\log{v}$$
$$\Rightarrow\,\dfrac{dv}{v\log{v}}=\dfrac{dx}{x}$$

Integrating both sides,

$$\Rightarrow\,\log{\left(\log{v}\right)}=\log{x}+\log{c}=\log{cx}$$
$$\Rightarrow\,\log{v}=cx$$
$$\Rightarrow\,\log{\dfrac{y}{x}}=cx$$
$$\Rightarrow\,y=x{e}^{cx}$$
Question 9
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If $$\cos { x } \cfrac { dy }{ dx } -y\sin { x } =6x,(0<x<\cfrac { \pi }{ 2 } )$$ and $$\quad y\left( \cfrac { \pi }{ 3 } \right) =0\quad $$ then $$y\left( \cfrac { \pi }{ 6 } \right) $$ is equal to:

A
$$-\cfrac { { \pi }^{ 2 } }{ 4\sqrt { 3 } } $$

B
$$-\cfrac { { \pi }^{ 2 } }{ 2 } $$

C
$$-\cfrac { { \pi }^{ 2 } }{ 2\sqrt { 3 } } $$

D
$$\cfrac { { \pi }^{ 2 } }{ 2\sqrt { 3 } } $$

Solution

$$\cfrac { dy }{ dx } -y\tan { x } =6x\sec { x } $$
$$\quad y\left( \cfrac { \pi }{ 3 } \right) =0\quad ;y\left( \cfrac { \pi }{ 6 } \right) =7$$
$${ e }^{ \int { pdx } }={ e }^{ -\int { \tan { x } dx } }={ e }^{ \ln { \cos { x } } }=\cos { x } $$
$$y.\cos { x } =\int { 6x\sec { x } \cos { x } } dx$$
$$y.\cos { x } =\cfrac { 6{ x }^{ 2 } }{ 2 } +C\quad \quad $$
$$y=3{ x }^{ 2 }\sec { x } +C\sec { x } $$
$$0=3.\cfrac { { \pi }^{ 2 } }{ 9 } .(2)+C(2)$$
$$2C=\cfrac { -2{ \pi }^{ 2 } }{ 3 } \Rightarrow C=-\cfrac { 2{ \pi }^{ 2 } }{ 3 } $$
$$y(\pi /6)=3.\cfrac { { \pi }^{ 2 } }{ 36 } .\left( \cfrac { 2 }{ \sqrt { 3 } } \right) +\left( \cfrac { 2 }{ \sqrt { 3 } } \right) .\left( -\cfrac { { \pi }^{ 2 } }{ 3 } \right) \Rightarrow y=-\cfrac { { \pi }^{ 2 } }{ 2\sqrt { 3 } } $$
Question 10
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Solution of the differential equation of $${ (y }^{ 2 }-{ x }^{ 3 })dx-xydy=0\quad $$ is

A
$${ y }^{ 2 }+2{ x }^{ 3 }+c{ x }^{ 2 }=0\quad $$

B
$${ y }^{ 2 }-2{ x }^{ 3 }+c{ x }^{ 2 }=0$$

C
$${ y }^{ 2 }+2{ x }^{ 3 }-c{ x }^{ 2 }=0$$

D
$${ y }^{ 2 }+3{ x }^{ 3 }+c{ x }^{ 2 }=0 $$

Solution

$${ (y }^{ 2 }-{ x }^{ 3 })dx-xydy=0$$
$$\quad \Rightarrow y(ydx-xdy)={ x }^{ 3 }dx\Rightarrow \cfrac { y }{ x } \left( \cfrac { ydx-xdy }{ { x }^{ 2 } } \right) =dx\Rightarrow \cfrac { y }{ x } d\left( \cfrac { y }{ x } \right) =dx\Rightarrow -\cfrac { 1 }{ 2 } { \left( \cfrac { y }{ x } \right) }^{ 2 }=x+k$$
$$\Rightarrow -{ y }^{ 2 }=2{ x }^{ 3 }+2{ x }^{ 2 }k\Rightarrow { y }^{ 2 }+2{ x }^{ 3 }+c{ x }^{ 2 }=0\quad $$

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Re-Attempt Weekly Quiz Competition

Differential Equations Test - 44

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Differential Equations Test - 44

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SHARING IS CARING

Question 1

$$x\sec y=x^{6}+C$$

$$6x\sec y=x+C$$

$$6x\sec y=x^{6}+C$$

$$6x\sec y=6x^{6}+C$$

Question 2

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

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

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

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

Solution

Question 3

Is $$m=-1; c=3/4$$

Is $$m=1; c=-3/4$$

No such real $$m, c$$

Is $$m=1; c=3/4$$

Solution

Question 4

sin $$\left( { x }^{ 2 }{ y }^{ 2 } \right) -1=0$$

$$cos\left( \dfrac { \pi }{ 2 } +{ x }^{ 2 }{ y }^{ 2 } \right) +x=0$$

$$sin\left( { x }^{ 2 }{ y }^{ 2 } \right) ={ e }^{ x-1 }$$

$$sin\left( { x }^{ 2 }{ y }^{ 2 } \right) ={ e }^{ 2(x-1) }$$

Solution

Question 5

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

$$logx=2log(x-y)+\frac{y}{x}+c$$

$$logx=log(x-y)+\frac{x}{y}+c$$

none of these above

Solution

Question 6

$$y=sinx+cosx+c$$

$$y=sinx-cosx+c$$

$$y=tanx+secx$$

$$\text{None of these}$$

Solution

Question 7

$${ e }^{ -\left| x \right| }-\left| x \right| $$

$${ e }^{ \left| x \right| }+\left| x \right| $$

$$\left| x \right| -{ e }^{ \left| x \right| }$$

$$\left| x \right| -{ e }^{ -\left| x \right| }$$

Solution

Question 8

$$y=xe^{cx}$$

$$y+xe^{cx}=0$$

$$y+e^{x}$$

none of these

Solution

Question 9

$$-\cfrac { { \pi }^{ 2 } }{ 4\sqrt { 3 } } $$

$$-\cfrac { { \pi }^{ 2 } }{ 2 } $$

$$-\cfrac { { \pi }^{ 2 } }{ 2\sqrt { 3 } } $$

$$\cfrac { { \pi }^{ 2 } }{ 2\sqrt { 3 } } $$

Solution

Question 10

$${ y }^{ 2 }+2{ x }^{ 3 }+c{ x }^{ 2 }=0\quad $$

$${ y }^{ 2 }-2{ x }^{ 3 }+c{ x }^{ 2 }=0$$

$${ y }^{ 2 }+2{ x }^{ 3 }-c{ x }^{ 2 }=0$$

$${ y }^{ 2 }+3{ x }^{ 3 }+c{ x }^{ 2 }=0 $$

Solution

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