$y = cos^{-1} x$ then

$(1 - x^2)y'' -xy'=?$ where

$y'= \dfrac{dy}{dx}$

0%
$1$
0%
$0$
0%
$2x$
0%
$2y$

Explanation

$\begin{array}{l} y={ \cos ^{ -1 } }x \\ \frac { { dy } }{ { dx } } =-\frac { 1 }{ { \sqrt { 1-{ x^{ 2 } } } } } \\ \left( { 1-{ x^{ 2 } } } \right) y_{ 1 }^{ 2 }-1=0 \\ { y_{ 1 } }=\frac { { dy } }{ { dx } } \, \, \, \, \, ,{ y_{ 2 } }=\frac { { { d^{ 2 } }y } }{ { d{ x^{ 2 } } } } \\ -2xy_{ 1 }^{ 2 }+\left( { 1-{ x^{ 2 } } } \right) 2{ y_{ 1 } }{ y_{ 2 } }=0 \\ -x{ y_{ 1 } }+\left( { 1-{ x^{ 2 } } } \right) { y_{ 2 } }=0 \\ \left( { 1-{ x^{ 2 } } } \right) { y_{ 2 } }-x{ y_{ 1 } }=0 \\ \left( { 1-{ x^{ 2 } } } \right) y''-xy'=0 \\ Option\, \, B\, \, \, is\, \, correct\, \, answer. \end{array}$

Solution of

${y^2}dx + ({x^2} - xy + {y^2})dy = 0$

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$\tan{^{ - 1}}\left( {\frac{x}{y}} \right) + \log y + c = 0$
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$2\tan{^{ - 1}}\left( {\frac{x}{y}} \right) + \log y + c = 0$
0%
$\log y\left( {y + \sqrt {{x^2} + {y^2}} } \right) + \log y + c = 0$
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$\log y\left( {y - \sqrt {{x^2} + {y^2}} } \right) + \log y + c = 0$

Explanation

${y^2}dx + \left( {{x^2} - xy + 42} \right)dy = 0$

$\displaystyle {{dx} \over {dy}} = {{ - \left( {{x^2} - xy + {y^2}} \right)} \over {{y^2}}}$

$\displaystyle {{dx} \over {dy}} = - {{{x^2}} \over {{y^2}}} + {x \over y} - 1$

Put $v = x/y\implies x=vy$

$dx/dy = v + ydv/dy$

$\Rightarrow v + ydv/dy = - {v^2} + v - 1$

$\Rightarrow ydv/dy = - \left( {{v^2} + 1} \right)$

$\displaystyle {{dv} \over {{v^2} + 1}} = - {1 \over y}dy$

Integrating both side

$\Rightarrow {\tan ^{ - 1}}v = - \log y + c$

$\Rightarrow {\tan ^{ - 1}}x/y = - \log y + c\,\,\,\,c = - c$

$\Rightarrow {\tan ^{ - 1}}\left( {x/y} \right) + \log y + c = 0$

$y = \sqrt{\dfrac{1-x}{1+x}}$ then find

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

0%
$1$
0%
$-1$
0%
$2$
0%
$0$

Explanation

$y= \sqrt{ \dfrac{1-x}{1+x} }$

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

$y.y = \dfrac{ (1+x)(-1)-(1-x)(1) }{ (1+x)^{2} }$

$y.y = \dfrac{-1-x-1+x}{(1+x)^{2}}$

$yy = \dfrac{-2}{ (1+x)^{2} }$

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

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

Given

$(1-x^{2}) \left( \dfrac{dy}{dx} \right)+y$

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

$= \dfrac{-(1-x)}{(1+x)}+ y$

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

$=y \left(1- \dfrac{(1-x)}{(1+x)} \times \dfrac{1+x}{1-x} \right)$

$=0$

If tangent at point p, with parameter t, on the curve $x=4t^2+3$ , $y=8t^3-1,t \epsilon R,$ meets the curve again at point Q, then the coordinates of Q are:-

0%
$t^2$ +3,- $t^3$ −1
0%
4 $t^2$ +3,−8 $t^2$ −1
0%
$t^2$ +3, $t^3$ −1
0%
16 $t^2$ +3,−64 $t^3$ −1

Explanation

Given,

$P(x=4t^2+3,y=8t^3-1)$

$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dy}{dt}}$

$=\dfrac{24t^2}{8t}=3t$

Tangent at P is given by,

$(y-8t^3+1)=3t(x-4t^2-3)$

Let,

$Q(4t_1^2+3,8t_1^3-1)$

Therefore Q satisfies the equation of tangent at P,

$8t_1^3-1-8t^3+1=3t(4t_1^2+3-4t^2-3)$

$8t_1^3-8t^3=3t(4t_1^2-4t^2)$

$8(t_1-t)(t^2+tt_1+t_1^2)=3t \times 4(t_1-t)(t_1+t)$

$2(t^2+tt_1+t_1^2)=3t(t_1+t)$

$2t_1^2-tt_1+t^2=0$

$(t_1-t)(2t_1+t)=0$

$t_1=-\dfrac{t}{2}$

$\therefore Q=(4t_1^2+3,8t_1^3-1)=(t^2+3,-t^3-1)$

The solution of

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

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$logy = (x - 1)e^x + c$
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$(x -1) logy = xe^x + c$
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$x logy = (x + 1)e^x + c$
0%
$x logy = (x - 1)e^x + c$

Explanation

$x\dfrac{dy}{dx}+ylny = xye^{x}$

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

Let

$xlny = t$

$lny +\dfrac{x}{y}\dfrac{dy}{dx} = \dfrac{dt}{dx}$

$\dfrac{dt}{dx} = x.e^{x}$

$\displaystyle \int dt = \int x-e^{x}$

$\displaystyle t = xe^{x}-\int e^{x}dx$

$t = e^{x} (x-1)+c$

prove value of 1

$xlny = e^{x}(x-1)+c$

1066898_1070889_ans_26ece50c02494947b9de8e679606c67c.jpg

The real value of n for which substitution

$y = {u^n}$ will transform differential equation

$2{x^4}y\frac{{dy}}{{dx}} + {y^4} = 4{x^6}$ into homogeneous equation

0%
$\frac{1}{2}$
0%
$1$
0%
$\frac{3}{2}$
0%
$4$

Explanation

$2{x^4}y\frac{{dy}}{{dx}} + {y^4} = 4{x^6}$

$2y\frac{{dy}}{{dx}} + \frac{{{y^4}}}{{{x^4}}} = 4{x^2}$

$y = {u^n}$

$\frac{{dy}}{{dx}} = n{x^{n - 1}}$

$2n{u^{n - 1}}{u^n}\frac{{du}}{{dx}} + \frac{{{u^{4x}}}}{{{x^4}}} = 4{x^2}$

$2x - 1 = 2$

$2x = 3$

$x = \frac{3}{2}$

The solution of the differential equation

$x\dfrac{dy}{dx}=y+x\tan \dfrac{y}{x}$ is :

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$\sin \dfrac{x}{y}=cx$
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$\sin \dfrac{y}{x}=cx$
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$\sin \dfrac{x}{y}=cy$
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$\sin \dfrac{y}{x}=cy$

Explanation

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

Let

$y=vx$

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

$\therefore v+x\dfrac{dv}{dx}v+\tan(v)$

$\Rightarrow \dfrac{dv}{\tan(v)}=\dfrac{dx}{x}$

$\displaystyle \Rightarrow \int \cot(v)dv=\int\dfrac{1}{x}dx$

$\Rightarrow ln|\sin(v)|=|n|x|+k=|n|x|+|nle^k|$

$\Rightarrow ln|\sin(v)|+ln |e^kx|$

$\Rightarrow \sin(v)=cx$

$\{e^k=c\}$

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

1053509_1094142_ans_12bdb9399d9843c5a49256be30867955.png

Soluation of D.E.

$\dfrac{{dy}}{{dx}} = \dfrac{{3x + 4y + 3}}{{12x + 16y - 4}}$ is

0%
$y = 4x + \ell n\left| {3x + 4y} \right| + C$
0%
$4y = x + \ell n\left| {3x + 4y} \right| + C$
0%
$y = \ell n\left| {3x + 4y} \right| + C$
0%
$x + y = \ell n\left| {3x + 4y} \right| + C$

Explanation

$\dfrac {dy}{dx}=\dfrac {3x+4y+3}{12x+16y-4}$

Let

$2=3x+4y\ \Rightarrow \dfrac {dz}{dx}=3+4\dfrac {dy}{dx}=\dfrac {1}{4}\left (\dfrac {dz}{dx} -3 \right)$

$\therefore \ \dfrac {1}{4}\left (\dfrac {dz}{dx} -3 \right)=\dfrac {z+3}{4z-4}\Rightarrow \dfrac {dz}{dx}-3=\dfrac {4z+12}{4z-4}$

$\Rightarrow \ \dfrac {dz}{dx}=\dfrac {4z+12}{4z -4}+3=\dfrac {16z}{4z-4}=\dfrac {4z}{z-1}$

$\Rightarrow \ \dfrac {z-1}{z}dz=4dx$

$\Rightarrow \ \displaystyle \int \dfrac {z-1}{z}dz =4\displaystyle \int dx$

$\Rightarrow \ z-\ln |z|=4x+C$

$\Rightarrow \ 3x+4y-\ln |3x+4y|=4x+C$

$\Rightarrow \ \boxed {4y=x+\in |3x+4y|+C}$

Solution of the differential equation

$\tan{y}\sec ^{ 2 }{ x } dx+\tan { x } \sec ^{ 2 }{ y } dy=0$ is:

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$\tan{x}+\tan{y}=k$
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$\tan{x}-\tan{y}=k$
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$\cfrac{\tan{x}}{\tan{y}}=k$
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$\tan{x}.\tan{y}=k$

Explanation

$\tan { y } \sec ^{ 2 }{ x } { dx }+\tan { x } \sec ^{ 2 }{ y } { dy }={ 0 }\\ \Rightarrow \tan { y } \sec ^{ 2 }{ x } { dx }=-\tan { x } \sec ^{ 2 }{ y } { dy }\\ \Rightarrow \frac { \sec ^{ 2 }{ x } }{ \tan { x } } { dx }=-\frac { \sec ^{ 2 }{ y } }{ \tan { y } } { dy }\\ \Rightarrow \int { \frac { \sec ^{ 2 }{ x } }{ \tan { x } } { dx } } =\int { -\frac { \sec ^{ 2 }{ y } }{ \tan { y } } { dy } }$

let tanx be equal to b and tany equal to a

$\\ \Rightarrow \int { \frac { dx }{ b } } =-\int { \frac { dy }{ a } } \\ \Rightarrow \ln { b } =-\ln { a } +k\\ \Rightarrow \ln { ab } =k\\ \Rightarrow \ln { \tan { x } \tan { y } } =k\\ \\ \Rightarrow \tan { x } \tan { y } =k\\$

The solution of the differential equation

$ydx + (x+x^2y)dy = 0$ is-

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$\dfrac{1}{xy} + \log y = c$
0%
$\log y = cx$
0%
$-\dfrac{1}{xy} = c$
0%
$-\dfrac{1}{xy} + \log y = c$

Explanation

Given,

$ydx+(x+x^2y)dy=0$

divide above equation by

$x^2ydy$

$\dfrac{1}{x^2}.\dfrac{dx}{dy}+\frac{1}{xy}=1=0$

$~~~~~~~-(1)$

Let

$\dfrac{1}{x}=t$

$\to -\dfrac{1}{x}\dfrac{dx}{dy}=\dfrac{dt}{dy}$

$1\to -\dfrac{dt}{dy}+\dfrac{t}{y}+1=0$

$\dfrac{dt}{dy}-\dfrac{t}{y}=1$

$-y^{-1}$ ,q=1

so I.F.=

$e^{\int p.dy}$

$e^{\int \dfrac{-1}{y}dy}$

$e^{-logy}$

$\dfrac{1}{y}$

So complete solution

$t.\dfrac{1}{y}=\int 1.\dfrac{1}{y}dy$

$t.\dfrac{1}{y}=logy+c$

$\to \dfrac{1}{xy}=logy+c$

$\to -\dfrac{1}{xy}+logy=c$

The general solution of the differential equation

$\dfrac { dy }{ dx } + y\cot { x } = \csc { x }$ , is

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$x + y sin { x } = C$
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$x + y cos { x } = C$
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$y+x(\sin { x + \cos { x } } )$
0%
$y\sin { x = x + C }$

Explanation

$\dfrac{dy}{dx}+y\cot x=cosec n$

$I.F=e^{\displaystyle\int \omega t xdx}=e^{en(\sin n)}$

$=\sin n$

$y\cdot \sin x=\displaystyle\int cosec x. mn-dx+c$

$y\sin n=x+c$

$\neq y\sin x=x+c$ .

1372829_1126684_ans_fbaea842c7044b93b5e881caeac10c84.jpg

The solution of

$\cfrac { dy }{ dx } =\left( \cfrac { ax+b }{ cy+d } \right)$ represents a parabola if:-

0%
a=0,c=0
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a=1,c=2
0%
a=0, $c\neq0$
0%
a=1,c=1

Explanation

We hyave,

$\dfrac{dy}{dx}=\dfrac{ax+b}{cy+d}$

For above differential equation to represent equation of parabola, one variable must have highest degree

$2$ and other must have degree

$1$ .

When we take

$a=0$ and

$c≠0$ , then we get

$\dfrac{dy}{dx}=\dfrac{0+b}{cy+d}$

$(cy+d)\ dy=b\ dx$

$\int (cy+d)\ dy=\int b\ dx$

$\dfrac{cy^2}{2}+dy=bx+C$ which is equation of parabola.

Hence, this is the answer.

The solution of the differential equation

$x \, dx + y\, dy = x^2y\, dy - y^2\, x\, dx$ is

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$x^2-1 = C(1+y^2)$
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$x^2+1 = C(1-y^2)$
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$x^3-1 = C(1+y^3)$
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$x^3+1 = C(1-y^3)$

Explanation

$x\, dx + y \, dy = x^2 y \, dy - y^2x \, dx$

$y\, dy (x^2-1) = x\, dx (1+y^2)$

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

$\Rightarrow x^2-1 = C (1+y^2)$

Solve differential equation

$dx=\dfrac{xdv}{v^2-a^2}$ .

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$x^{2a}=\dfrac{v-a}{v+a}\times c$
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$x^{2a}=\dfrac{v+a}{v-a}\times c$
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$x=\dfrac{v-a}{v+a}\times c$
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None of these

Solution for

$\dfrac{x + y \dfrac{dy}{dx}}{y - x \dfrac{dy}{dx}} = x^2 + 2y^2 + \dfrac{y^4}{x^2}$ is

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

The particular solution of the differential equation

$y(1+\log x)\dfrac{dx}{dy} -x =0$ when

$x=e, y=e^2$ is

0%
$y =ex \log x$
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$ey = x \log x$
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$xy = e \log x$
0%
$y \log x = ex$

Explanation

Correct equation is,

$y(1+logx)\dfrac{dx}{dy}-xlogx=0$

$\to \dfrac{dy}{y}=\dfrac{1+logx}{logx}dx$

$~~~~~-(1)$

Let

$xlogx=t$

$(1+logx)dx=dt$

$(1) \to \dfrac{dy}{y}=\dfrac{dt}{t}$

integrating Both the sides

$\to \int \dfrac{dy}{y}=\int \dfrac{dt}{t}$

$logy=logt+c$

$at x=e and y=e^2$

$loge^2=eloge+c$

$2-1=c$

c=1

$logy=logt+1$

$logy=logt+loge$

$logy=log(xlogx)+loge$

$y=xelogx$

Solution of differential equation

$xdy-yx=0$ represents:

0%
rectangular hyperbola
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straight line passing through origin
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parabola whose vertex is at origin
0%
circle whose center is at origin

Explanation

$xdy-ydx=0\Rightarrow \dfrac{dy}{y}=\dfrac{dx}{x}$

Integrating both sides

$\ln y=\ln x\Rightarrow y=x$

Solution of differential equation

$xdy-yx=0$ reperesnts straight line passing through origin

The solution of

$\cos \, y \, \log (\sec \, x + \tan \, x) dx = \cos \, x \, \log (\sec \, y + \tan \, y) dy$ is

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$[\log ( \sec \, x + \tan \, x)]^2 - [\log (\sec \, y + \tan \, y)]^2 = c$
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$[\log ( sec \, x + \tan \, x)]^2 + [\log (\sec \, y + \tan \, y)]^2 = c$
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$[\log ( \sec \, x - \tan \, x)]^2 - [\log (\sec \, y - \tan \, y)]^2 = c$
0%
$[\log ( \sec \, x - \tan \, x)]^2 + [\log (\sec \, y - \tan \, y)]^2 = c$

Explanation

Given,

$cosylog(secx+tanx)dx=cosxlog(secy+tany)dy$

$\dfrac{log(secy+tany)dy}{cosy}=\dfrac{log(secx+tanx)dx}{cosx}$

Integrate Both the sides

$\int \dfrac{log(secy+tany)dy}{cosy}=\int \dfrac{log(secx+tanx)dx}{cosx}$

$~~~~~~-(1)$

Let u=

$log(sey+tany)$

$du=\dfrac{1}{cosy}dy$

and

$log(secx+tanx)=v$

$\dfrac{1}{cosx}=dv$

$(1)\to \int udu=\int vdv$

$\to \dfrac{u^2}{2}=\dfrac{v^2}{2}+c$

$u^2=v^2+2c$

$[log(secy+tany)]^2-[log(secx+tanx)]^2=c$

The solution of

$\dfrac{{dy}}{{dx}} = \dfrac{{x{{{\mathop{\rm log x}\nolimits} }^2} + x}}{{\sin y + y\cos y}}$

0%
$y\sin y = {x^2}\log x + c$
0%
$y\sin y = {x^2} + c$
0%
$y\sin y = {x^2} + \log x + c$
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$y\sin y = x \log x + c$

Explanation

$\dfrac{dy}{dx}=\dfrac{2x\log{x}+x}{\sin{y}+y\cos{y}}$

$\Rightarrow\,\left(\sin{y}+y\cos{y}\right)dy=\left(2x\log{x}+x\right)dx$

$\Rightarrow\,\sin{y}dy+y\cos{y}dy=2x\log{x}dx+xdx$

Integrating both sides, we get

$\Rightarrow\,\displaystyle\int{\sin{y}dy}+\displaystyle\int{y\cos{y}dy}=2\displaystyle\int{x\log{x}dx}+\displaystyle\int{xdx}$

$\Rightarrow\,-\cos{y}+\displaystyle\int{y\cos{y}dy}=2\displaystyle\int{x\log{x}dx}+\dfrac{{x}^{2}}{2}+c$

Consider

$\displaystyle\int{y\cos{y}dy}$

Let

$u=y\Rightarrow\,du=dy$

$dv=\cos{y}dy\Rightarrow\,v=\sin{y}$

Consider

$\displaystyle\int{x\log{x}dx}$

Let

$u=\log{x}\Rightarrow\,du=\dfrac{1}{x}dx$

$dv=xdx\Rightarrow\,v=\dfrac{{x}^{2}}{2}$

$\Rightarrow\,-\cos{y}+y\sin{y}-\displaystyle\int{\sin{y}dy}=2\left[\dfrac{{x}^{2}\log{x}}{2}-\displaystyle\int{\dfrac{{x}^{2}}{2}\times \dfrac{1}{x}dx}\right]+\dfrac{{x}^{2}}{2}+c$

$\Rightarrow\,-\cos{y}+y\sin{y}+\cos{y}={x}^{2}\log{x}-\displaystyle\int{xdx}+\dfrac{{x}^{2}}{2}+c$

$\Rightarrow\,y\sin{y}={x}^{2}\log{x}+\dfrac{{x}^{2}}{2}-\dfrac{{x}^{2}}{2}+c$

$\Rightarrow\,y\sin{y}={x}^{2}\log{x}+c$

The solution to the differential equation

$\cos\ xdy=y(\sin x-y)dx$ ,

$0<x<\dfrac{\pi}{2}$ , is

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$\tan x=(\sec x+c)y$
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$\sec x=(\tan x+c)y$
0%
$y\sec x=\tan x+c$
0%
$y\tan x=\sec x+c$

Explanation

$cosx\, dy=y(sinx-y)dx$ ___(1)

$\dfrac{dy}{dx}=\dfrac{y(sinx-y)}{cos x}= ytanx-y^{2}secx$

$\dfrac{dy}{dx}-ytanx=-y^{2}secx$ ___(2)

[Divide by

$y^2$ ]

$y^{-2}\dfrac{dy}{dx}-y^{-1}tanx= -secx$ ___ (3)

Let

$+y^{-1}=z\Rightarrow -y^{-2}\dfrac{dy}{dx}=\dfrac{dz}{dx}$

$-\dfrac{dz}{dx}-ztanx= -secx$ ___ (4)

$=\dfrac{dz}{dx}+ztanx=secx$

$I.F=e^{\displaystyle + \int tanx}= e^{+log\,secx}= secx$

$\displaystyle z\times I.F= \int secx*I.F\, dx$

$\displaystyle zsecx=\int sec^{2}x\, dx$

$\displaystyle \frac{1}{y}secx= tan x+c$

$secx=y(tan x+c)$

1120585_1144328_ans_ca19011e28414bccbf882bb1f3a2a1bd.jpg

The solution of

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

0%
$\sin\left(\dfrac{y}{x}\right)=cx$
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$\sin\left(\dfrac{y}{x}\right)=cy$
0%
$\cos\left(\dfrac{y}{x}\right)=cx$
0%
${ \sec \left( \dfrac { y }{ x }\right) +\tan \left( \dfrac { y }{ x }\right) }=xc$

Explanation

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

$c$ is any arbitrary constant, then the general solution of differential equation

$ydx-xdy=xydx$ is given by -

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$y=cxe^{x}$
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$y=cx e^{-x}$
0%
$y+e^{x}=cx$
0%
$ye^{x}=cx$

Explanation

$ydx-xdy=xydx$

$ydx-xydx=xdy$

$y(1-x)dx=xdy$

$\dfrac{1-x}{x}dx=\dfrac{dy}{y}$

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,

$y=cxe^{-x}$

Solve :

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

0%
$x^{(x^x)}+C$
0%
$(x^x)^x+C$
0%
$x^x\cdot log_ex+C$
0%
$(x^x)+C$

Explanation

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.

The solution of the equation

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

0%
$x\sqrt { {1-y}^{2} }-y\sqrt { {1-x}^{2} }=c$
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$x\sqrt { {1-y}^{2} }+y\sqrt { {1-x}^{2} }=c$
0%
$x\sqrt { {1+y}^{2} }+y\sqrt { {1+x}^{2} }=c$
0%
None of these

Explanation

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.

1151408_1144581_ans_8d8dc50229224c1c8f9e6fc779a85296.jpeg

The solution of

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

0%
$\tan^{-1}\left(\dfrac{y}{x}\right)=log(cx)$
0%
$\sin^{-1}\left(\dfrac{y}{x}\right)=log(cx)$
0%
$\cos^{-1}\left(\dfrac{y}{x}\right)=log(cy)$
0%
$\sec^{-1}\dfrac{y}{x}=log(cy)$

Explanation

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,

$\displaystyle \int{\dfrac{dv}{\sqrt{1-v^2}}}=\displaystyle \int{\dfrac{dx}{x}}$

$\displaystyle \int{\dfrac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}x$

and

$\displaystyle \int{\dfrac{1}{x}dx}=\log x$

$=\sin^{-1}v=\log x+\log c$

$=\sin^{-1}v=\log(cx)$

$\log a+\log b=\log ab$

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

So, the solution of

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

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

The solution of the differential equations

$\dfrac{dy}{dx}=\dfrac{x-2y+1}{2x-4y}$ is?

0%
$(x-2y)^2+2x=c$
0%
$(x-2y)^2+x=c$
0%
$(x-2y)+2x^2=c$
0%
$(x-2y)+x^2=c$

Explanation

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

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

The solution of

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

0%
$\sin h^{-1}x+\sin h^{-1}y=c$
0%
$\sqrt{1+x^2}+\sqrt{1+y^2}=c$
0%
$(1+x^2)(1+y^2)=c$
0%
$\sqrt{\dfrac{1+x^2}{1+y^2}}=c$

Explanation

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,

$\dfrac{1}{2}\displaystyle \int{\dfrac{dt}{\sqrt t}}=\dfrac{-1}{2}\displaystyle \int{\dfrac{du}{\sqrt{ t^2}}}$

$\dfrac{1}{2}\displaystyle \int{t-\dfrac{1}{2}dt}=\dfrac{-1}{2}\displaystyle \int{-\dfrac{1}{2}\displaystyle \int{u-\dfrac{1}{2}du}}$

$\displaystyle \int{x^n dx}=\dfrac{x^{n+1}}{n+1}$

$=\dfrac{1}{2}\dfrac{t^{-1/2+1}}{\dfrac{-1}{2}+1}=\dfrac{-1}{2}\dfrac{u^{-1/2+1}}{\dfrac{-1}{2}+1}+c$

$=\dfrac{t^{1/2}}{1/2}=\dfrac{-u^{1/2}}{1/2}+c$

$=t^{1/2}+u^{1/2}=c$

$=\sqrt t+\sqrt u=c$

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

So, the solution of

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

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

The solution of

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

0%
$y-x=c(1+xy)$
0%
$y+x=c(1+xy)$
0%
$y=(1+2x)c$
0%
$xy=x^2+x+x$

Explanation

$\cfrac{dy}{dx}=(1+y^2)(1+x^2)^{-1}$

$\implies \cfrac{dy}{(1+y^2)}=\cfrac{dx}{(1+x^2)}$

$\implies \int \cfrac{dy}{(1+y^2)}=\int \cfrac{dx}{(1+x^2)}$

$\implies \tan^{-1}y=\tan^{-1}x+\tan^{-1}c$

$\tan^{-1}(\cfrac{y-x}{1+xy}=\tan^{-1}c$

$\implies \cfrac{y-x}{1+xy}=c$

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

The solution of

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

0%
$y=ce^{xy}$
0%
$y=\dfrac{e^{\frac{y}{x}}}{c}$
0%
$log y=xy+c$
0%
$log x=xy+c$

Explanation

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

$f:R \rightarrow R$ be a continuous function such that

$f(x)=\displaystyle \int^{x}_{1}2tf(t)dt$ , then which of the following does not hold(s) good?

0%
$f(\pi)=e^{\pi^{2}}$
0%
$f(1)=e$
0%
$f(0)=1$
0%
$f(2)=-2$

The solution of the differential equation

${x}^{2}dy=-2xydx$ is

0%
$x{y}^{2}=c$
0%
${x}^{2}{y}^{2}=c$
0%
${x}^{2}y=c$
0%
$xy=c$

Explanation

$x^{2}dy=-2xydx$

$xdy=-2ydx$

$xdy+2ydx=0$

$\displaystyle\int d(xy^{2}) =\int 0$

$xy^{2}=C$

Solution of the differential equation

$\dfrac{dy}{dx}+y\sec x=\tan x\left(\le x< \dfrac{\pi}{2}\right)$ is

0%
$y\left( \sec { x } -\tan { x } \right) =\left( \sec { x } +\tan { x } \right) -x+C$
0%
$y\left( \sec { x } +\tan { x } \right) =\left( \sec { x } -\tan { x } \right) -x+C$
0%
$y\left( \sec { x } +\tan { x } \right) =\left( \sec { x } +\tan { x } \right) -x+C$
0%
$none of these$

Explanation

$\dfrac{dy}{dx}+y\sec x=\tan x$

$\dfrac{dy}{dx}+Py=Q$

$=\displaystyle\int Pdx=e^{\log|\sec x+\tan x|}$

$=\sec x+\tan x$

$Y(IF)=\displaystyle\int(Q\times IF) dx+C$

$Y(\sec x+\tan x)=\displaystyle\int \tan x(\sec x+\tan x)+C$

$Y(\sec x+\tan x)=\displaystyle\int \tan x\sec x+\int \tan^{2}x dx$

$Y(\sec x+\tan x)=\sec x+\displaystyle\int\sec^{2}-1dx+C$

$Y(\sec x+\tan x)=\sec x+\tan x-x+C$

Solution of the differential equation

$xdy-ydx=\sqrt {x^{2}+y^{2}}dx$ is

0%
$y+\sqrt {x^{2}+y^{2}}=cx$
0%
$y+\sqrt {x^{2}+y^{2}}=cx^{2}$
0%
$y+\sqrt {x^{2}+y^{2}}=C$
0%
$none\ of\ these$

Explanation

$xdy-ydx=\sqrt{x^2+y^2}+x$

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

$\Rightarrow F(x,y)=\dfrac{\sqrt{x^2+y^2}+y}{x}$

$F(kx\cdot ky)=\dfrac{\sqrt{k^2x^2+k^2y^2+ky}}{kx}=k^o\cdot F(x, y)$

$y=vx$ and

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

$v+x\dfrac{dv}{dx}=\dfrac{\sqrt{x^2+v^2x^2}+vx}{x}$

$v+x\dfrac{dv}{dx}=\sqrt{1+v^2}+v$

$\dfrac{xdv}{dx}=\sqrt{1+v^2}$

$\dfrac{dv}{\sqrt{1+v^2}}=\dfrac{dx}{k}$

$log|v+\sqrt{1-v^2})l=log x+log e$

$v=\dfrac{y}{x}$

$log\left(\dfrac{y}{x}\right)+\sqrt{1+y^2/x^2}=log cx$

$log \dfrac{y+\sqrt{x^2+y^2}}{x}=log cx$

$log(y+\sqrt{x^2+y^2})=log cx^2$

$y+\sqrt{x^2+y^2}=cx^2$ .

1190795_1157908_ans_4d1e22cb44b54a43b670765dccc3a974.jpg

The solution of the differential equation

$\cfrac{dy}{dx}=1+x+y+xy$ is

0%
$\log{(1+y)}=x+\cfrac{{x}^{2}}{2}+c$
0%
${(1+y)}^{2}=x+\cfrac{{x}^{2}}{2}+c$
0%
$\log{(1+y)}=\log{(1+x)}+c$
0%
None of these

Explanation

$\dfrac{dy}{dx}= 1 + x + y + xy = 1 + x + y (1+x)$

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

$\dfrac{dy}{1+x}= \displaystyle\int (1+x)dx$

$ln(1+y)=x+\dfrac{x^{2}}{2}+C$

A continuously differentiable function

$\phi(x)$ in

$(0,\pi)$ satisfying

$y'=1+{y}^{2},y(0)=0=y(\pi)$ is

0%
$\tan{x}$
0%
$x(x-\pi)$
0%
$(x-\pi)(1-{e}^{x})$
0%
Not possible

Explanation

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

$\displaystyle \int \frac{dy}{1+y^{2}}=\int dx$

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

$x=0;y=0$

$c=0$

$y=\tan x$

Solution of the differential equation

$(1+x^{2})dy+2xy dx=\cot x dx$ is

0%
$y=\log|\sin x|(1+x^{2})+C(1+x^{2})^{-1}$
0%
$y=\log|\sin x|(1+x^{2})^{-1}+C(1+x^{2})$
0%
$y=\log|\sin x|(1+x^{2})+C(1+x)$
0%
$y=\log|\sin x|(1+x^{2})^{-1}+C(1+x^{2})^{-1}$

Explanation

$(1+x^2)dy+2xydx=\cot xdx$

Put in form

$\dfrac{dy}{dx}+PY=Q$

$(1+x)^2dy+2xydx=\cot xdx$

dividing both sides by

$(1+x^2)$

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

$\dfrac{dy}{dx}+\left(\dfrac{2x}{1+x^2}\right)y=\dfrac{\cot x}{1+x^2}$

Find P and Q

$\dfrac{dy}{dx}+PY=Q$

$P=\dfrac{2x}{1+x^2}$

$Q=\dfrac{\cot x}{1+x^2}$

$=e^{\displaystyle\int \dfrac{2x}{1+x^2}}$

$t=1+x^2$

$=e^{\displaystyle\int \dfrac{dt}{t}}=e^{log t}=t=1+x^2$

Solution:

$Y\times IF=\displaystyle\int Q\times IFdx+C$

$Y(1+x^2)=\displaystyle\int \dfrac{\cot x}{1+x^2}\times (1+x^2)dx+c$

$Y(1+x^2)=\displaystyle\int \cot xdx+c$

$Y(1+x^2)=log|\sin x|+c$

$Y=(1+x^2)^{-1}log|\sin x|+c(1+x^2)^{-1}$ .

1191520_1157095_ans_5d48a4f501754c9d86470142f015cd35.jpg

The solution of the differential equation

$x^3 \dfrac{dy}{dx} + 4x^2 \tan y = e^x \sec y$ satisfying

$y(1) = 0$ is

0%
$\tan \, y = (x - 2) e^x \log \, x$
0%
$\sin \, y = e^x (x - 1) x^{-4}$
0%
$\tan \, y = (x - 1) e^x x^{-3}$
0%
$\sin \, y = e^x (x - 1) x^{-3}$

Explanation

Given,

$x^3\dfrac{dy}{dx}+4x^2tany=e^xsecy$

Dividing whole equation by

$x^3secy$

$cosy\dfrac{dy}{dx}+\dfrac{4}{x}siny=\dfrac{e^x}{x^3}$

let siny=t

$cosydy=dt$

Put in above equation

$\dfrac{dt}{dx}+\dfrac{4}{x}t=\dfrac{e^x}{x^3}$

$\dfrac{4}{x},Q=\dfrac{e^x}{x^3}$

I.f.=

$e^{\int P.dx}$

$e^{\int \dfrac{4}{x}dx}$

$e^{4logx}$

$e^{logx^4}$

$x^4$

Complete solution,

$t\times x^4=\int x^4\times \dfrac{e^x}{x^3}dx+c$

$t\times x^4=\int xe^xdx+c$

$t\times x^4=xe^x-e^x+c$

$siny\times x^4=xe^x-e^x+c$

at x=1,y=0

$0=1.e-e+c$

c=0

$siny \times x^4=e^x(x-1)+0$

$siny=x^{-4}e^x(x-1)$

The solution of

$\cfrac{dy}{dx}=\cfrac{1+{y}^{2}}{\sec{x}}$ is

0%
$\tan ^{ -1 }{ y } =\cos { x } +c$
0%
$\tan ^{ -1 }{ y } =\sin { x } +c$
0%
${ y }^{ 3 }=co\sec { x } +c$
0%
$\log { \left( 1+{ y }^{ 2 } \right) } =\cos { x } +c$

Explanation

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

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

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

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

The solution of

$ydx-xdy=0$ is

0%
${y}^{2}=cx$
0%
$y=c{x}^{3}$
0%
$y=cx$
0%
${x}^{2}=cy$

Explanation

$y\ dx=x\ dy=0$

$y\ dx=x\ dy$

$\displaystyle \int \dfrac {dx}{x} = \displaystyle \int \dfrac {dy}{y}$

$\ln x=\ln y+\ln c$

$\ln \left (\dfrac {x}{y}\right)=\ln c\ \Rightarrow \dfrac {x}{y}=c$

$y=\dfrac {x}{c}=xc'$

$y=xc'$

Solution of differential equation

$\dfrac{dy}{dx}-2xy=x$ is

0%
$y=Ce^{x^2}-\dfrac {1}{2}$
0%
$y=Ce^{x^2}+\dfrac {1}{2}$
0%
$y=Cx^2-\dfrac {1}{2}$
0%
None

Explanation

Given

$\dfrac{dy}{dx}-2xy=x$ within in first order lines differential

equation of form

$\dfrac{dy}{dx}+ Py=Q$

$I.F= e^{\int Pdx} = e^{\int (-2x)dx}= C^{-x^{2}}$

Multiplying

$I.F$ on both sides

$e^{-x^{2}} \dfrac{dy}{dx} - 2xy. e^{-x^{2} }= xe^{-x^{2}}$

$d (e^{-x^{2} }.y) =xe^{-x^{2} }dx$

Integrating both sides

$e^{-x^{2} }y= \int xe^{-x^{2} } dx$

$=\dfrac{-e^{-x^{2}}}{2}+c$

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

$\therefore$ Option

$A$ is correct

$2x = {y^{\dfrac{1}{5}}} + {y^{\dfrac{{ - 1}}{5}}}{\text{and}}\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + \lambda x\dfrac{{dy}}{{dx}} + {\text{ky}} = 0,\;{\text{then}}\;\lambda + {\text{K}}$ is equal to.

0%
26
0%
-24
0%
-23
0%
-26

Explanation

Given

$2x=y^{\dfrac{1}{5}}+y^{-\dfrac{1}{5}}$

$y^{\dfrac{1}{5}}+y^{-\dfrac{1}{5}}=2x$

Differentiate with respect to x on both the sides,

$\therefore \dfrac{1}{5}.y^{\dfrac{-4}{5}}\dfrac{dy}{dx}-\dfrac{1}{5}.y^{\dfrac{-6}{5}}.\dfrac{dy}{dx}=2$

$\dfrac{y^{-1}}{5}(y^{\dfrac{1}{5}}-y^{\dfrac{-1}{5}})\dfrac{dy}{dx}=2$

$(y^{\dfrac{1}{5}}-y^{-\dfrac{1}{5}})\dfrac{dy}{dx}=10y$

$(2\sqrt{x^2-1})\dfrac{dy}{dx}=10y$

$~~~~~~~-(1)$

Again Differentiating above equation with respect to x,$$

$2\sqrt{x^2-1}\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}.2.\dfrac{2x}{2\sqrt{x^2-1}}=10\dfrac{dy}{dx}$

$(x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=5\sqrt{x^2-1}\dfrac{dy}{dx}$

$(x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=5.5y$ ( from equation (1))

On comparing with the above equation,

we get ,

$\lambda=1 and k=-25$

$\lambda+k=1-25=-24$

The solution of

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

0%
$\frac { x } { y } + e ^ { x ^ { 3 } } = c$
0%
$\frac { x } { y } - e ^ { x ^ { 3 } } = c$
0%
$\frac { x } { y } = - e ^ { x ^ { 3 } } + c$
0%
none of these

Explanation

$\displaystyle\int{\dfrac{ydx-xdy}{{y}^{2}}}+3\displaystyle\int{{x}^{2}{e}^{{x}^{3}}dx}=0$

Consider

$\displaystyle\int{\dfrac{ydx-xdy}{{y}^{2}}}$

We know that

$\dfrac{d}{dx}\left(\dfrac{x}{y}\right)=\dfrac{ydx-xdy}{{y}^{2}}$

$\displaystyle\int{\dfrac{d}{dx}\left(\dfrac{x}{y}\right)dx}$

$=\dfrac{x}{y}+{c}_{1}$

Consider

$3\displaystyle\int{{x}^{2}{e}^{{x}^{3}}dx}$

$=\displaystyle\int{3{x}^{2}{e}^{{x}^{3}}dx}$

Let

$t={x}^{3}\Rightarrow\,dt=3{x}^{2}dx$

$=\displaystyle\int{{e}^{t}dt}$

$={e}^{t}+{c}_{2}$

$={e}^{{x}^{3}}+{c}_{2}$

$\therefore\,\displaystyle\int{\dfrac{ydx-xdy}{{y}^{2}}}+3\displaystyle\int{{x}^{2}{e}^{{x}^{3}}dx}=0$

$\dfrac{x}{y}+{c}_{1}+{e}^{{x}^{3}}+{c}_{2}=0$

$\Rightarrow\,\dfrac{x}{y}+{e}^{{x}^{3}}=-\left({c}_{1}+{c}_{2}\right)$

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

$c=-\left({c}_{1}+{c}_{2}\right)$

The solution of,

$\dfrac{xdy}{x^2 + y^2} = \left(\dfrac{y}{x^2 + y^2} - 1 \right)dx$ , is given by

0%
$\tan^{-1} \left(\dfrac{x}{y}\right) + x = C$
0%
$\tan^{-1} \left(\dfrac{y}{x}\right) + x = C$
0%
$\tan^{-1} \left(\dfrac{y}{x}\right) + xy = C$
0%
$\tan^{-1} \left(\dfrac{y}{x}\right) + x^2 = C$

Explanation

$\dfrac{xdy}{x^2+y^2}=\left(\dfrac{y}{x^2+y^2}-1\right)dx$

$\dfrac{xdy}{x^2+y^2}=\dfrac{ydx}{x^2+y^2}-dx$

$\dfrac{xdy-ydx}{x^2+y^2}=-dx$

$\dfrac{d}{dx}\tan^{-1}\dfrac{y}{x}=-dx$

Integrating both sides, we get

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

$\tan^{-1}\dfrac{y}{x}+x=c$ .

The solution of the differential equation

$\operatorname { xdy } \left( y ^ { 2 } e ^ { x y } + e ^ { \tfrac { x } { y } } \right) = y d x \left( e ^ { \frac { x } { y } } - y ^ { 2 } e ^ { x y } \right)$ is-

0%
$x y = \ln \left| e ^ { x / y } + C \right|$
0%
$e ^ { x y } = \ln ( x y + C )$
0%
$x y = e ^ { x / y } + C$
0%
$x y = e ^ { x / y } + \frac { x } { y }$

Explanation

$xdy(y^2e^{xy}+e^{x/y}=ydx(e^{x/y}-y^2e^{xy})$

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

$y'(xy^2e^{xy}+xe^{x/y})=ye^{x/y}-y^3e^{xy}$

$e^{xy}(xy^2y'+y^3)=(y-y'x)e^{x/y}$

$e^{xy}(xy'+y)=\left(\dfrac{1}{y}-\dfrac{y'x}{y^2}\right)e^{x/y}$

(Dividing by

$y^2$ )

$\dfrac{d}{dx}(e^{xy})=\dfrac{d}{dx}(e^{x/y})$

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

$\Rightarrow xy=ln|e^{x/y}+c|$ .

The solution of the differential equation

$( x \cot y + \log \cos x ) d y$

$+ ( \log \sin y - y \tan x ) d x = 0$ is:-

0%
$( \sin x ) ^ { y } ( \cos y ) ^ { x } = c$
0%
$( \sin y ) ^ { x } ( \cos x ) ^ { y } = c$
0%
$( \sin x ) ^ { x } ( \cos y ) ^ { y } = c$
0%
none of these

The solution of the differential equation

$y ^ { 2 } d y = x ( y d x - x d y ) \text { is } y = y ( x )$ . If

$y ( \sqrt { 3 } e ) = e$ and

$y \left( x _ { 0 } \right) = 1$ then

$x_0$ is

0%
$0$
0%
$1$
0%
$\frac { 1 } { e }$
0%
$-1$

Explanation

${ y }^{ 2 }dy=x(ydx-xdy)$

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

let

$x=vy$

then,

$\dfrac { dx }{ dy } =v+y\dfrac { dv }{ dy }$

now putting this in equation (1), we get

$1+{ v }^{ 2 }=v(v+vy\dfrac { dv }{ dy } )$

$1=vy \dfrac { dv }{ dy }$

integrating the above, we get

$\int { \dfrac { dy }{ y } } =\int { vdv } = log y=\dfrac { { v }^{ 2 } }{ 2 } +C$

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

Using the initial conditions, we get

$1=\dfrac { 3 }{ 2 } +C\Longrightarrow C=\dfrac { -1 }{ 2 }$

$log (y)=\dfrac { { x }^{ 2 } }{ { 2y }^{ 2 } } -\dfrac { 1 }{ 2 }$ when

$y=0,$ then

$log (1)=\dfrac { { x }^{ 2 } }{ { 2(1) }^{ 2 } } -\dfrac { 1 }{ 2 }$

$0=\dfrac { { x }^{ 2 } }{ { 2 } } -\dfrac { 1 }{ 2 } \Longrightarrow \dfrac { { x }^{ 2 } }{ { 2 } } =\dfrac { 1 }{ 2 }$

${ x }^{ 2 }=1\Longrightarrow { x }_{ 0 }=1$ (Answer)

So, option (B) is correct.

The differential equation whose solution is

$y = Ax^{5} + Bx^{4}$ is

0%
$x^{2} \dfrac {d^{2}y}{dx^{2}} + 8x \dfrac {dy}{dx} + 20y = 0$
0%
$x^{2} \dfrac {d^{2}y}{dx^{2}} + 8x \dfrac {dy}{dx} - 20y = 0$
0%
$x^{2} \dfrac {d^{2}y}{dx^{2}} - 8x \dfrac {dy}{dx} + 20y = 0$
0%
$x^{2} \dfrac {d^{2}y}{dx^{2}} - 8x \dfrac {dy}{dx} - 20y = 0$

Solution of the equation

$\dfrac{{dy}}{{dx}} = 1 + xy + x + y$ is

0%
$\left( {1 + y} \right)\left( {1 + x} \right) = c$
0%
$\log \left( {1 + y} \right) = 1 + x + c$
0%
$\ln |1 + y| = x + \dfrac{{{x^2}}}{2} + c$
0%
$\ln |1 + y| + x + \dfrac{{{x^2}}}{2} + c = 0$

Explanation

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

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

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

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

$\Rightarrow log\left| 1+y \right| =x+\dfrac { { x }^{ 2 } }{ 2 } +c$ [C]

The general solution of the differential equation

$\dfrac { dy }{ dx } +y={ x }^{ 3 }$ is ______.

0%
${ ye }^{ x }={ e }^{ x }\left( { x }^{ 3 }+{ 3x }^{ 2 }-6x-6 \right) +c$
0%
${ ye }^{ x }={ e }^{ x }\left( { x }^{ 3 }-{ 3x }^{ 2 }-6x+6 \right) +c$
0%
$y=\left( { x }^{ 3 }-{ 3x }^{ 2 }+6x-6 \right) +c{ e }^{ -x }$
0%
$y={ e }^{ x }$

$\dfrac{dy}{dx}=y+3>0$ and

$y(0)=2$ then

$y(\ln{2})$ is equal to :

0%
$7$
0%
$5$
0%
$13$
0%
$-2$

Explanation

$\frac{dy}{dx} = y + 3$ and y(0) = 2

then y (ln 2) = (?)

$\therefore \frac{dy}{dx} = y+3$

$\therefore \frac{1}{y+3} dy = dx$

Integrating both side

$\int \frac{1}{y+3}dy =\int dx$

$\therefore$ ln(y + 3) = x + c

$\rightarrow$ Initial condition y(0)=2

$\therefore ln(2+3)=0+c$

$\therefore c=ln 5$

$\rightarrow$ ln(y + 3) = x + ln5

$\therefore$ ln(y + 3) - ln5 = x

$\therefore ln\left ( \frac{y+3}{5} \right )=x$

$\therefore \frac{y+3}{5} = e^{x}$

$\therefore y=5e^{x} - 3$

for y (ln2)

$y = 5e^{ln2}-3$

$= 5 \times 2 - 3$

= 10 - 3

= 7

1120103_1247901_ans_a161d533fef34570b691bb6351a9abce.jpg

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 6 - MCQExams.com

If tangent at point p, with parameter t, on the curve $x=4t^2+3$ , $y=8t^3-1,t \epsilon R,$ meets the curve again at point Q, then the coordinates of Q are:-

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 6 - MCQExams.com

If tangent at point p, with parameter t, on the curve x=4t2+3x=4t^2+3,y=8t3−1,tϵR,y=8t^3-1,t \epsilon R, meets the curve again at point Q, then the coordinates of Q are:-

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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If tangent at point p, with parameter t, on the curve $x=4t^2+3$ , $y=8t^3-1,t \epsilon R,$ meets the curve again at point Q, then the coordinates of Q are:-