Answer to Question #297053 in Differential Equations for Tokyo

Question #297053

<e> Solve using the method of separation of variables the pde partial du/dt+du/dx+2e^tu=0


1
Expert's answer
2022-02-18T13:07:49-0500

By the method of separation of variables we suppose that uu is of the form u(t,x)=ϕ(t)ψ(x)u(t,x)=\phi(t)\psi(x) which gives us :

ϕ(t)ψ(x)+ϕ(t)ψ(x)+2etϕ(t)ψ(x)=0\phi'(t) \psi(x)+\phi(t)\psi'(x)+2e^t \phi(t) \psi(x)=0

dividing both parts by u(t,x)=ϕ(t)ψ(x)u(t,x)=\phi(t)\psi(x) gives us

ϕ(t)/ϕ(t)+2et+ψ(x)/ψ(x)=0\phi'(t)/\phi(t) + 2e^t+\psi'(x)/\psi(x)=0

ϕ(t)/ϕ(t)+2et=ψ(x)/ψ(x)\phi'(t)/\phi(t)+2e^t=-\psi'(x)/\psi(x)

As the left part is a function of tt and the right is the function of xx the equality gives us

{ϕ/ϕ+2et=λψ/ψ=λ\begin{cases} \phi'/\phi +2e^t=\lambda \\ -\psi'/\psi = \lambda \end{cases} with λR\lambda\in \mathbb{R} a constant.

The second equation easily gives us ψλ(x)=Aeλx\psi_\lambda (x)= Ae^{-\lambda x}, AA a constant

The first one gives us

lnϕ(t)lnϕ(0)=λt2et+2\ln \phi(t) - \ln\phi(0)=\lambda t -2e^t+2

ϕλ(t)=Beλt2et\phi_\lambda(t)=Be^{\lambda t - 2e^t} , BB a constant

Combining two solutions gives us

uλ(t,x)=Ceλ(tx)e2etu_\lambda (t,x)= Ce^{\lambda(t-x)} e^{-2e^t}


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