Question #164226

Find the general solution of the first-order linear partial differential equation

xux yuy = u and investigate first-order, quasi-linear and linear partial differential equations


1
Expert's answer
2021-02-24T07:24:06-0500

We have the linear differential equation


xux+yuy=uxu_x+yu_y=u


The system of eqautions is,

dxx=dyy=duu\dfrac{dx}{x}=\dfrac{dy}{y}=\dfrac{du}{u}


Now, taking first two terms

dxx=dyy\dfrac{dx}{x}=\dfrac{dy}{y}


Integrating Both the sides-


lnx=lny+C1lnxlny=lneC1lnx=lny+C1\\ lnx-lny=\text{ln}e^{C1}


(xy)=eC1xy=C1ϕ=xy=C1(\dfrac{x}{y})=e^{C1}\\ \dfrac{x}{y}=C1'\\ \phi=\dfrac{x}{y}=C1'


Now, taking last two terms

dyy=duulny=lnu+C2lny=lnu+lneC2ln(yu)=lneC2\dfrac{dy}{y}=\dfrac{du}{u}\\ lny=lnu+C2\\ lny=lnu+lne^{C2}\\\\ ln(\dfrac{y}{u})=lne^{C2}


yu=eC2\dfrac{y}{u}=e^{C2}


yu=C2\dfrac{y}{u}=C2'


ϵ=yu=C2\epsilon=\dfrac{y}{u}=C2'


Hence, the general solution is f(ϕ,ϵ)=f(xy,yu)=0f(\phi,\epsilon)=f(\dfrac{x}{y},\dfrac{y}{u})=0


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