Question

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

Answer 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)

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

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

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

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

Answer 2

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

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 :-

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