Math, asked by rohithrvs4, 19 hours ago

Find the general solution of the following differential equation ху у'=x2 + y2​

Answers

Answered by senboni123456
1

Answer:

Step-by-step explanation:

We have,

\tt{xy\,\dfrac{dy}{dx}=x^2+y^2}

\sf{\implies\,xy\,\dfrac{dy}{dx}-y^2=x^2}

\sf{\implies\,y\,\dfrac{dy}{dx}-\dfrac{y^2}{x}=x}

\sf{Put\,\,\,y^2=v}

\sf{\implies\,2\,y\,\dfrac{dy}{dx}=\dfrac{dv}{dx}}

\sf{\implies\,y\,\dfrac{dy}{dx}=\dfrac{1}{2}\cdot\dfrac{dv}{dx}}

So,

\sf{\implies\,\dfrac{1}{2}\cdot \dfrac{dv}{dx}-\dfrac{v}{x}=x}

\sf{\implies\,\dfrac{dv}{dx}-\dfrac{2}{x}\,v=2x}

Now,

\sf{I.F.=e^{-\displaystyle\int\dfrac{2}{x}\,dx}}

\sf{=e^{-2\displaystyle\int\dfrac{1}{x}\,dx}}

\sf{=e^{-2\ln(x)}}

\sf{=e^{\ln(x^{-2})}}

\sf{=\dfrac{1}{x^2}}

Now,

\sf{v\cdot\,\dfrac{1}{x^2}=\displaystyle\int\,2x\cdot\dfrac{1}{x^2}\,dx}

\sf{\implies\,\dfrac{v}{x^2}=\displaystyle\int\,\dfrac{2}{x}\,dx}

\sf{\implies\,\dfrac{v}{x^2}=2\displaystyle\int\,\dfrac{1}{x}\,dx}

\sf{\implies\,\dfrac{v}{x^2}=2\ln|x|+C}

Let C= ln(k)

\sf{\implies\,\dfrac{v}{x^2}=\ln(x^2)+\ln(k)}

\sf{\implies\,\dfrac{v}{x^2}=\ln(k\,x^2)}

\sf{\implies\,\dfrac{y^2}{x^2}=\ln(k\,x^2)}

\sf{\implies\,y^2=x^2\,\ln(k\,x^2)}

Answered by mathdude500
5

\large\underline{\sf{Solution-}}

Given Differential equation is

\rm :\longmapsto\:xy \: y' =  {x}^{2} +  {y}^{2}

\rm :\longmapsto\:y' = \dfrac{ {x}^{2} +  {y}^{2} }{xy}

can be further rewritten as

\rm :\longmapsto\:\dfrac{dy}{dx}  = \dfrac{ {x}^{2} +  {y}^{2} }{xy}

To evaluate this Homogeneous differential equation, we substitute

\rm :\longmapsto\:\boxed{ \tt{ \: y \:  =  \: vx}} -  -  -  - (1)

So, above differential equation can be rewritten as

\rm :\longmapsto\:\dfrac{d}{dx}(vx) =  \dfrac{ {x}^{2}  +  {(vx)}^{2} }{ v{x}^{2} }

\rm :\longmapsto\:v\dfrac{d}{dx}x + x\dfrac{d}{dx}v=  \dfrac{ {x}^{2}  +  {v}^{2}  {x}^{2} }{ v{x}^{2} }

\rm :\longmapsto\:v + x\dfrac{dv}{dx}=  \dfrac{ {x}^{2}(1  +  {v}^{2})}{ v{x}^{2} }

\rm :\longmapsto\:v + x\dfrac{dv}{dx}=  \dfrac{1  +  {v}^{2}}{v}

\rm :\longmapsto\:x\dfrac{dv}{dx}=  \dfrac{1  +  {v}^{2}}{v}  - v

\rm :\longmapsto\:x\dfrac{dv}{dx}=  \dfrac{1  +  {v}^{2} -  {v}^{2} }{v}

\rm :\longmapsto\:x\dfrac{dv}{dx}=  \dfrac{1}{v}

\rm :\longmapsto\:v \: dv \:  =  \: \dfrac{dx}{x}

On integrating both sides, we get

\rm :\longmapsto\:\displaystyle\int\rm v \: dv \:  =  \: \displaystyle\int\rm \dfrac{dx}{x}

\rm :\longmapsto\:\dfrac{ {v}^{2} }{2}  = log |x|  + c

\rm :\longmapsto\:\dfrac{ {y}^{2} }{2 {x}^{2} }  = log |x|  + c

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More to know :-

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}

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