Question

Second Order Nonhomogeneous ODE IVP:
8y''-6y'+y=6cosh(x), y(0)=0.2, y'(0)
I have no idea how to find the particular solution y_p; I have not seen a table with cosh in it.

Answer 1

Try something like: \[
y = A\sinh(x)+B\cosh(x)
\]

Answer 2

Find \(y'\) and \(y''\).

Answer 3

The book's answer suggests that y_p includes \[e ^{x}\] and \[e ^{-x}\], not any form on sinh(x) or cosh(x)

the solution is \[y=e^{x/4}-2e^{x/2}+\frac{ 1 }{ 5 }*e^{-x}+e^{x}\]

I am also wondering where the -2 came from in the homogeneous solution because I got the right coefficients (1/4 and 1/2), but I just wrote it as \[y_h=e^{x/4}+e^{x/2}\]

Answer 4

the homogeneous equation is \[y=c_1e^{x/4}+c_2e^{x/2}\]
the particular solution is \[y_p=Ae^x + Be^{-x}\] that satisfies the DE
\[\large 8y_p''-6y_p '+y_p=6\cosh x=6(1/2)(e^x+e^{-x})=3ex^x+3e^{-x}\]
Solve A and B

use the IVP on \[\large y=c_1e^{x/4}+c_2e^{x/2}+Ae^x+Be^{-x}\]to solve \(c_1\) and \(c_2\)

Answer 5

Awesome, this helped a lot. Didn't know \[\cosh{x}\] was the same as \[\frac{ 1 }{ 2 }(e^{x}+e^{-x})\]