Second degree linear derivate with exponential function, which general solution? Suppose

James Dale 2022-01-17 Answered
Second degree linear derivate with exponential function, which general solution?
Suppose y4y=xe2x
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Steve Hirano
Answered 2022-01-18 Author has 34 answers
For any polynomial P, constant k and function u, P(D)ekxu=exp(kx)P(D+k)u (where D stands for derivative). In your case P(t)=t24,k=2,and P(t+2)=(t+2)24=t2+4t so if y=e2xu we have y4y=e2x(u+4u). Now you want y4y=xe2x so u+4u=x. Writing v=u, we have the first-order linear equation in v:v+4v=x, which has a solution v=x4116. An antiderivative of this is u=x28x16, corresponding to the particular solution y=e2x(x28x16) of your original equation.

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Medicim6
Answered 2022-01-19 Author has 33 answers
One way to get this mechanically is to use the method of Laplace Transforms.
Of course, I havent

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alenahelenash
Answered 2022-01-24 Author has 368 answers

You want to solve an equation Ly=xneλx, where L is a constant linear differential operator - in your case Ly=y4y. Let us solve Ly=e(λ+ϵ)x instead: if we substitute y=c(ϵ)e(λ+ϵ)x, we get c(ϵ)p(λ+ϵ)=1 (where p is the characteristic polynomial of L, in your case p(t)=t24),i.e.c(ϵ)=1/p(λ+ϵ). We now expand
Le(λ+ϵ)xp(λ+ϵ)=e(λ+ϵ)x
to a power series in ϵ, look at the term at ϵn, and get a solution of Ly=xneλx/n!.
In your case λ=2,p(t)=t24,and c(ϵ)=1/(ϵ(4+ϵ)). We thus get
L((ϵ141+(41x42)+ϵ(41x2/242x+43)+)e2x)=(1+ϵx+ϵ2x2/2+)e2x
Looking at the term at ϵ, we get
L((41x2/242x+43)e2x)=xe2x,
i.e. y=(41x2/242x+43)e2x is a solution of your equation (we can drop 43e2x from the solution, as it solves the homogeneous equation).

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