Second degree linear derivate with exponential function, which general solution? Suppose y−4y=xe2x

Name: Second degree linear derivate with exponential function, which general solution? Suppose y−4y=xe2x
Uploaded: 2022-02-23T17:22:57+00:00
Description: Second degree linear derivate with exponential function, which general solution? Suppose y−4y=xe2x

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Second degree linear derivate with exponential function, which general solution?  Suppose y−4y=xe2x

Steve Hirano · Accepted Answer

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)=t2−4,k=2,and P(t+2)=(t+2)2−4=t2+4t so if y=e2xu we have y″−4y=e2x(u″+4u′). Now you want y″−4y=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=x4−116. An antiderivative of this is u=x28−x16, corresponding to the particular solution y=e2x(x28−x16) of your original equation.

Medicim6 · Accepted Answer

One way to get this mechanically is to use the method of Laplace Transforms.   Of course, I havent

alenahelenash · Accepted Answer

You want to solve an equation Ly=xneλx, where L is a constant linear differential operator - in your case Ly=y″−4y. 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)=t2−4),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)=t2−4,and c(ϵ)=1/(ϵ(4+ϵ)). We thus get L((ϵ−14−1+(4−1x−4−2)+ϵ(4−1x2/2−4−2x+4−3)+⋯)e2x)=(1+ϵx+ϵ2x2/2+⋯)e2x Looking at the term at ϵ, we get L((4−1x2/2−4−2x+4−3)e2x)=xe2x, i.e. y=(4−1x2/2−4−2x+4−3)e2x is a solution of your equation (we can drop 4−3e2x from the solution, as it solves the homogeneous equation).

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

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