- #1

- 203

- 0

[tex]y''+y'+y=e^{2x}[/tex], One would choose [tex]y=Ae^{2x}[/tex]

[tex]y''+y'+y=x[/tex], One would choose [tex]y=Ax+B[/tex]

But what am I supposed to choose if I have: [tex]y''+y'+y=xe^{2x}[/tex]?

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- #1

- 203

- 0

[tex]y''+y'+y=e^{2x}[/tex], One would choose [tex]y=Ae^{2x}[/tex]

[tex]y''+y'+y=x[/tex], One would choose [tex]y=Ax+B[/tex]

But what am I supposed to choose if I have: [tex]y''+y'+y=xe^{2x}[/tex]?

- #2

- 1,862

- 8

[tex]y''+y'+y=e^{2x}[/tex], One would choose [tex]y=Ae^{2x}[/tex]

[tex]y''+y'+y=x[/tex], One would choose [tex]y=Ax+B[/tex]

But what am I supposed to choose if I have: [tex]y''+y'+y=xe^{2x}[/tex]?

Well, you know that in case of:

Dy = exp(p x)

where D is a linear differential operator,

you would choose y = A exp(px)

You also know that a solution of:

Dy = f(x) + g(x)

can be obtained by solving

Dy = f(x)

and

Dy =g(x)

separately and adding up the solutions. Linearitity of D implies that this will work.

Then, in case of:

Dy = x exp(px)

you could simply consider the factor x to be the derivative w.r.t. p. So, if you simply solve:

Dy = exp(px)

by putting y = A(p)exp(px), then the linearity of D implies that the solution of

Dy = [exp[(p+epsilon)x] - exp(px)]/epsilon

is given by

[A(p+epsilon)exp[(p+epsilon)x] - A(p)exp(px)]/epsilon

So, taking the derivative of A(p)exp(px) w.r.t. p will do. By Leibnitz's rule, you see that this amounts to putting

y = (A + B x)exp(px)

- #3

- 116

- 0

for [itex] x^{2} e^{2x}[/tex] use [itex] (A + Bx +Cx^{2})e^{2x} [/itex]

and so on

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