Answer to Question #164341 in Differential Equations for murtaza irfan

Question #164341

Q2. (c) 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
2021-02-24T07:39:14-0500

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=lneC1\text{lnx}=\text{lny}+C_1\\ \text{lnx}-\text{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\Rightarrow\dfrac{dy}{y}=\dfrac{du}{u}\\\Rightarrow lny=lnu+C_2\\ \Rightarrow lny=lnu+lne^{C_2}\\\\ \Rightarrow ln(\dfrac{y}{u})=lne^{C_2}


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


yu=C2\Rightarrow \dfrac{y}{u}=C_2'


ϵ=yu=C2\Rightarrow\epsilon=\dfrac{y}{u}=C_2'


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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Comments

Assignment Expert
16.07.21, 00:01

Dear Opeyemi Yahaya, please use the panel for submitting a new question.


Opeyemi Yahaya
03.07.21, 10:13

Solve (d²u)/(dxdt) = sin(x+t) at t=0, ux=1 and at x=0, x=(1-t)²

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