Differential Equations Test

Result

Differential Equations Test - 38

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The solution of $$\dfrac{dy}{dx}=\dfrac{y}{x}+\tan\left(\dfrac{y}{x}\right)$$.

A
$$\sin\left(\dfrac{y}{x}\right)=cx$$

B
$$\sin\left(\dfrac{y}{x}\right)=cy$$

C
$$\cos\left(\dfrac{y}{x}\right)=cx$$

D
$${ \sec \left( \dfrac { y }{ x }\right) +\tan \left( \dfrac { y }{ x }\right) }=xc$$

Solution

$$\begin{matrix} \dfrac { { dy } }{ { dx } } =\dfrac { g }{ n } +\tan \left( { \dfrac { y }{ x } } \right) \\ It\, \, is\, \, { { homogenouous\ diffenrential\ equation } } \\ { { Lety= } }\sqrt { x } \\ \Rightarrow \dfrac { { dy } }{ { dx } } =\sqrt { } +\dfrac { { xdv } }{ { dx } } \\ and\, \, put\, \, in\, \, differential\, \, equation\, \, \\ \Rightarrow \sqrt { } +x\dfrac { { dv } }{ { dx } } =\sqrt { } +\tan v \\ \Rightarrow \dfrac { { xdv } }{ { dx } } =\tan v \\ \Rightarrow \int { \dfrac { { dv } }{ { \tan v } } =\int { \dfrac { { dx } }{ x } } } \\ \Rightarrow \int { \sec v=\log x+c } \\ \Rightarrow \log \left| { \sec v=\tan v } \right| \log xc \\ \Rightarrow \log \left| { \sec \dfrac { y }{ x } +\tan \dfrac { y }{ x } } \right| =\log xc \\ \, \, \, \left| { \sec \dfrac { y }{ x } +\tan \dfrac { y }{ x } } \right| =xc\, \, \, Ans. \\ \end{matrix}$$
Question 2
1 / -0

Solve :

$$\displaystyle\int (x^x)^x(2xlog_ex+x)dx$$

A
$$x^{(x^x)}+C$$

B
$$(x^x)^x+C$$

C
$$x^x\cdot log_ex+C$$

D
$$(x^x)+C$$

Solution

Consider the given integral.

$$I=\int{{{\left( {{x}^{x}} \right)}^{x}}\left( 2x\log x+x \right)}dx$$

$$I=\int{\left( {{x}^{{{x}^{2}}}} \right)\left( 2x\log x+x \right)}dx$$

Let $$t={{x}^{{{x}^{2}}}}$$

$$ \log t={{x}^{2}}\log x $$

$$ \dfrac{1}{t}\dfrac{dt}{dx}=\left( 2x\log x+x \right) $$

$$ dt={{x}^{{{x}^{2}}}}\left( 2x\log x+x \right)dx $$

Therefore,

$$ I=\int{1dt} $$

$$ I=t+C $$

On putting the value of $$t$$, we get

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

Hence, this is the answer.
Question 3
1 / -0

The differential equation of all parabolas having their axis of symmetry with the axis of x is?

A
$$y\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^2=0$$

B
$$y\dfrac{d^2x}{dx^2}+\left(\dfrac{dy}{dx}\right)^2=0$$

C
$$y=2x\frac { { dy } }{ { dx } }$$

D
$$y\dfrac{d^2y}{dx^2}-\left(\dfrac{dy}{dx}\right)=0$$

Solution

$$Let\, \, a\, \, is\, \, { { constant } } \\ { { { y } }^{ 2 } }=4ax\to \left( i \right) \\ differentiate\, \, with\, \, respection \\ \Rightarrow 2y\, \, \frac { { dy } }{ { dx } } =4a\times 1 \\ \Rightarrow 2y\, \, \frac { { dy } }{ { dx } } =4a\to \left( { ii } \right) \\ Put\, \, 4a=2y\frac { { dy } }{ { dx } } \, \, in\, \, e{ q^{ n } }\, \left( i \right) \\ \Rightarrow { y^{ 2 } }=2y\frac { { dy } }{ { dx } } \times x \\ \Rightarrow y=2\frac { { dy } }{ { dx } } \times x \\ \therefore y=2x\frac { { dy } }{ { dx } } \\ It\, \, is\, \, required\, \, differentiate\, \, e{ q^{ n } }\, \, of\, \, parabola\, \, \, \\ \, \, \, \, \, \, \, \, $$
Question 4
1 / -0

If $$c$$ is any arbitrary constant, then the general solution of differential equation $$ydx-xdy=xydx$$ is given by -

A
$$y=cxe^{x}$$

B
$$y=cx e^{-x}$$

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

D
$$ye^{x}=cx$$

Solution

$$ydx-xdy=xydx$$
$$ydx-xydx=xdy$$
$$y(1-x)dx=xdy$$
$$\dfrac{1-x}{x}dx=\dfrac{dy}{y}$$
Integrating both side
$$\int{\left(\dfrac{1-x}{x}\right)dx}=\int{\dfrac{dy}{y}}$$
$$\log_e x-x=\log_e y+c$$
$$\log_e {\dfrac{x}{y}}=x+c$$
$$\dfrac{x}{y}=e^c\cdot e^x$$
$$y=e^cxe^{-x}$$
suppose $$c=e^c$$
then,
$$y=cxe^{-x}$$
Question 5
1 / -0

The solution of the equation $$\dfrac {dy}{dx}+\sqrt {\dfrac { {1-y}^{2} }{ {1-x}^{2} }}=0$$ is

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

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

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

D
None of these

Solution

Let $$ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}} = c $$ be the solution of
the given equation $$ \dfrac{dy}{dx}+\sqrt{\dfrac{1-y^2}{1-x^2}} = 0$$
So, lets verify the solution,
differentiating equation w.r.t, x, we get,
$$ x(\dfrac{-2yy'}{2\sqrt{1-y^2}})+\sqrt{1-y^2}+y(\dfrac{-2x}{2\sqrt{1-x^2}})+y'\sqrt{1-x^2} = 0$$
$$ \dfrac{-xyy'}{\sqrt{1-y^{2}}}+\sqrt{1+y^2}-\dfrac{yx}{\sqrt{1-x^2}}+y'\sqrt{1+x^2} = 0$$
$$ y'(\dfrac{-xy}{\sqrt{1-y^2}}+\sqrt{1-x^2}) = \dfrac{yx}{\sqrt{1-x^2}} - \sqrt{1-y^2}$$
$$ y' (\dfrac{-xy+\sqrt{1-x^{2}}\sqrt{1-y^2}}{1-y^2}) = \dfrac{xy-\sqrt{1-y^2}\sqrt{1-x^2}}{\sqrt{1-x^2}}$$
$$ y' = \dfrac{(xy-\sqrt{1-y^2}\sqrt{1-x^2})(1-y^2)}{(1-x^2)(-(xy-\sqrt{1-x^2}\sqrt{1-y^2}))}$$
$$ \dfrac{dy}{dx} y' \Rightarrow -\dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$$
$$ \Rightarrow \dfrac{dy}{dx}+\sqrt{\dfrac{1-y^{2}}{1-x^{2}}} = 0$$ which is the given
differential equation.
So, option (C) is correct.
Question 6
1 / -0

The solution of $$\dfrac{dy}{dx}=\dfrac{\sqrt{x^2-y^2}+y}{x}$$.

A
$$\tan^{-1}\left(\dfrac{y}{x}\right)=log(cx)$$

B
$$\sin^{-1}\left(\dfrac{y}{x}\right)=log(cx)$$

C
$$\cos^{-1}\left(\dfrac{y}{x}\right)=log(cy)$$

D
$$\sec^{-1}\dfrac{y}{x}=log(cy)$$

Solution

Given,
$$\dfrac{dy}{dx}=\dfrac{\sqrt{x^2-y^2}+y}{x}$$
Let $$y=vx$$
So, $$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$
substituting,
$$=v+x\dfrac{dv}{dx}=\dfrac{\sqrt{x^2-(vx)^2}+vx}{x}$$
$$=v+x\dfrac{dv}{dx}=\dfrac{\sqrt{x^2-v^2x^2}+vx}{x}$$
$$=v+x\dfrac{dv}{dx}=\dfrac{\sqrt{x^2-(1-v^2)}+vx}{x}$$
$$=v+x\dfrac{dv}{dx}=\dfrac{x\sqrt{(1-v^2)}+vx}{x}$$
$$=v+x\dfrac{dv}{dx}=\sqrt{1-v^2}+v$$
$$=x\dfrac{dv}{dx}=\sqrt{1-v^2}$$
$$=\dfrac{dv}{\sqrt{1-v^2}}=\dfrac{dx}{x}$$
Integrating,
$$\displaystyle \int{\dfrac{dv}{\sqrt{1-v^2}}}=\displaystyle \int{\dfrac{dx}{x}}$$
as $$\displaystyle \int{\dfrac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}x$$
and $$\displaystyle \int{\dfrac{1}{x}dx}=\log x$$
$$=\sin^{-1}v=\log x+\log c$$
$$=\sin^{-1}v=\log(cx)$$
as $$\log a+\log b=\log ab$$
$$=\sin^{-1} \dfrac{y}{x}=\log (cx)$$
So, the solution of $$\dfrac{dy}{dx}=\dfrac{\sqrt{x^2-y^2}+y}{x}$$
is $$\sin^{-1}\dfrac{y}{x}=\log (cx)$$
Question 7
1 / -0

The solution of the differential equations $$\dfrac{dy}{dx}=\dfrac{x-2y+1}{2x-4y}$$ is?

A
$$(x-2y)^2+2x=c$$

B
$$(x-2y)^2+x=c$$

C
$$(x-2y)+2x^2=c$$

D
$$(x-2y)+x^2=c$$

Solution

Given,
$$\dfrac{dy}{dx}=\dfrac{x-2y+1}{2x-4y}$$
Let $$x-2y t$$
So, $$1-\dfrac{2dy}{dx}=\dfrac{dt}{dx}$$
$$=1-\dfrac{dt}{dx}=2\dfrac{dy}{dx}$$
$$=\dfrac{1-\dfrac{dt}{dx}}{2}=\dfrac{dy}{dx}$$
substituting,
$$\dfrac{1-\dfrac{dt}{dx}}{2}=\dfrac{t+1}{2t}$$
$$=1-\dfrac{dt}{dx}=\dfrac{t+1}{t}$$
$$=1-\dfrac{dt}{dx}=1+\dfrac{1}{t}$$
$$=-\dfrac{dt}{dx}=\dfrac{1}{t}$$
$$=-tdt=dx$$
Integrating,
$$-\displaystyle \int{t dt}=\displaystyle \int{dx}$$
$$=-\dfrac{t^2}{2}=x+C$$
$$=C=x+\dfrac{t^2}{2}$$
$$=C=x+\dfrac{(x-2y)^2}{2}$$
$$=C=2x+(x-2y)^2$$
So, the solution of
$$\dfrac{dy}{dx}=\dfrac{x-2y+1}{2x-4y}$$ is
$$(x-2y)^2+2x=C$$
Question 8
1 / -0

The solution of $$x\sqrt{1+y^2}dx+y\sqrt{1+x^2}dy=0$$.

A
$$\sin h^{-1}x+\sin h^{-1}y=c$$

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

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

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

Solution

Given,
$$x\sqrt{1+y^2}dx+y\sqrt{1+x^2}dy=0$$
separating the variable,
$$x\sqrt{1+y^2}dx=-\sqrt{1+x^2}dy$$
$$=\dfrac{x}{\sqrt{1+x^2}}dx=-\dfrac{y}{\sqrt{1+y^2}}dy$$
Let $$1+x^2=t$$
So, $$2x dx=dt$$
Let $$1+y^2=u$$
$$2y dy=du$$
substituting,
$$\dfrac{dt}{2\sqrt t}=-\dfrac{dt}{2\sqrt u}$$
Integrating,
$$\dfrac{1}{2}\displaystyle \int{\dfrac{dt}{\sqrt t}}=\dfrac{-1}{2}\displaystyle \int{\dfrac{du}{\sqrt{ t^2}}}$$
$$\dfrac{1}{2}\displaystyle \int{t-\dfrac{1}{2}dt}=\dfrac{-1}{2}\displaystyle \int{-\dfrac{1}{2}\displaystyle \int{u-\dfrac{1}{2}du}}$$
as $$\displaystyle \int{x^n dx}=\dfrac{x^{n+1}}{n+1}$$
$$=\dfrac{1}{2}\dfrac{t^{-1/2+1}}{\dfrac{-1}{2}+1}=\dfrac{-1}{2}\dfrac{u^{-1/2+1}}{\dfrac{-1}{2}+1}+c$$
$$=\dfrac{t^{1/2}}{1/2}=\dfrac{-u^{1/2}}{1/2}+c$$
$$=t^{1/2}+u^{1/2}=c$$
$$=\sqrt t+\sqrt u=c$$
$$=\sqrt{1+x^2}+\sqrt{1+y^2}=c$$
So, the solution of $$x\sqrt{1+y^2}dx+y\sqrt{1+x^2}dy=0$$ is $$\sqrt{1+x^2}+\sqrt{1+y^2}=c$$
Question 9
1 / -0

The solution of $$\dfrac{dy}{dx}=(1+y^2)(1+x^2)^{-1}$$ is?

A
$$y-x=c(1+xy)$$

B
$$y+x=c(1+xy)$$

C
$$y=(1+2x)c$$

D
$$xy=x^2+x+x$$

Solution

$$\cfrac{dy}{dx}=(1+y^2)(1+x^2)^{-1}$$
$$\implies \cfrac{dy}{(1+y^2)}=\cfrac{dx}{(1+x^2)}$$
$$\implies \int \cfrac{dy}{(1+y^2)}=\int \cfrac{dx}{(1+x^2)}$$
$$\implies \tan^{-1}y=\tan^{-1}x+\tan^{-1}c$$
$$\tan^{-1}(\cfrac{y-x}{1+xy}=\tan^{-1}c$$
$$\implies \cfrac{y-x}{1+xy}=c$$
$$\implies y-x=c(1+xy)$$
Question 10
1 / -0

The solution of $$\dfrac{dy}{dx}=\dfrac{y^2}{xy-x^2}$$.

A
$$y=ce^{xy}$$

B
$$y=\dfrac{e^{\frac{y}{x}}}{c}$$

C
$$log y=xy+c$$

D
$$log x=xy+c$$

Solution

$$\frac { { dy } }{ { dx } } =\frac { { { y^{ 2 } } } }{ { xy-{ x^{ 2 } } } } \\ \frac { { dy } }{ { dx } } =\frac { { { y^{ 2 } } } }{ { { x^{ 2 } }\left( { \frac { y }{ x } -1 } \right) } } \\ Continue\, \, \\ \frac { { dy } }{ { dx } } =\frac { { { { \left( { \frac { y }{ x } } \right) }^{ 2 } } } }{ { \left( { \frac { y }{ x } -1 } \right) } } \to \left( i \right) \, \, \, \, \, \, \left[ { It\, \, is\, \, ogenous\, \, e{ q^{ n } } } \right] \\ Let\, \, y=vx\, \, ,\, \, \frac { y }{ x } =v \\ \frac { { dy } }{ { dx } } =v+\frac { { xdv } }{ { dx } } \, \, and\, \, put\, \, in\, \, e{ q^{ n } } \\ \Rightarrow v+\frac { { xdv } }{ { dx } } =\frac { { { v^{ 2 } } } }{ { \left( { v-1 } \right) } } \\ \Rightarrow \frac { { xdv } }{ { dx } } =\frac { { { v^{ 2 } } } }{ { v-1 } } -v \\ \Rightarrow \frac { { xdv } }{ { dx } } =\frac { { { v^{ 2 } }-{ v^{ 2 } }+v } }{ { v-1 } } \\ \Rightarrow \int { \frac { { \left( { v-1 } \right) } }{ v } dx=\int { \frac { { dx } }{ x } } } \\ \Rightarrow \int { \frac { v }{ v } dv-\int { \frac { 1 }{ v } dv=\log x+c } } \\ \Rightarrow \int { dv-\log v=\log x+c } \\ \Rightarrow v-\log v=\log x+c \\ \Rightarrow \frac { y }{ x } -\log \frac { y }{ x } =\log xc \\ \Rightarrow \frac { y }{ x } =\log xc+\log \frac { y }{ x } \\ \Rightarrow \frac { y }{ x } =\log cy \\ \therefore cy={ e^{ \frac { y }{ x } } } \\ \, \, \, \, \boxed { y=\frac { { { e^{ \frac { y }{ x } } } } }{ c } } \, $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Differential Equations Test - 38

Result

Differential Equations Test - 38

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\sin\left(\dfrac{y}{x}\right)=cx$$

$$\sin\left(\dfrac{y}{x}\right)=cy$$

$$\cos\left(\dfrac{y}{x}\right)=cx$$

$${ \sec \left( \dfrac { y }{ x }\right) +\tan \left( \dfrac { y }{ x }\right) }=xc$$

Solution

Question 2

$$x^{(x^x)}+C$$

$$(x^x)^x+C$$

$$x^x\cdot log_ex+C$$

$$(x^x)+C$$

Solution

Question 3

$$y\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^2=0$$

$$y\dfrac{d^2x}{dx^2}+\left(\dfrac{dy}{dx}\right)^2=0$$

$$y=2x\frac { { dy } }{ { dx } }$$

$$y\dfrac{d^2y}{dx^2}-\left(\dfrac{dy}{dx}\right)=0$$

Solution

Question 4

$$y=cxe^{x}$$

$$y=cx e^{-x}$$

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

$$ye^{x}=cx$$

Solution

Question 5

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

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

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

None of these

Solution

Question 6

$$\tan^{-1}\left(\dfrac{y}{x}\right)=log(cx)$$

$$\sin^{-1}\left(\dfrac{y}{x}\right)=log(cx)$$

$$\cos^{-1}\left(\dfrac{y}{x}\right)=log(cy)$$

$$\sec^{-1}\dfrac{y}{x}=log(cy)$$

Solution

Question 7

$$(x-2y)^2+2x=c$$

$$(x-2y)^2+x=c$$

$$(x-2y)+2x^2=c$$

$$(x-2y)+x^2=c$$

Solution

Question 8

$$\sin h^{-1}x+\sin h^{-1}y=c$$

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

$$(1+x^2)(1+y^2)=c$$

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

Solution

Question 9

$$y-x=c(1+xy)$$

$$y+x=c(1+xy)$$

$$y=(1+2x)c$$

$$xy=x^2+x+x$$

Solution

Question 10

$$y=ce^{xy}$$

$$y=\dfrac{e^{\frac{y}{x}}}{c}$$

$$log y=xy+c$$

$$log x=xy+c$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.