Question

Find a particular solution of y"-4y'-5y=-6xe^(-x).

Answer 1

@freckles @CGGURUMANJUNATH

Answer 2

I just need to know the substitution yp. Any idea?

Answer 3

well probably (At+B)*exp(-x)

Answer 4

that t is suppose to be x

Answer 5

\[y''-4y'-5y=x \text{ if this was the differential equation my guess for } \\ \text {particular solution would be } y_p=Ax+B \\ y''-4y'-5y=e^{-x} \text{ if this was the differential equation my guess for the } \\ \text{ particular solution would be } y_p=Ce^{-x} \\ \text{ so my guess for } y''-4y'-5y=-6xe^{-x} \text{ would be } y_p=(Ax+B)e^{-x}\]

Answer 6

But why (Ax+B) for x?

Answer 7

because it works...
\[y''-4y'-5y=x \\ y_p=Ax+B \\ y_p'=A \\ y_p''=0 \\ 0-4A-5(Ax+B)=x \\ -4A-5Ax-5B=x \\ -4A-5B-5Ax=x \\ \text{ so } A=\frac{-1}{5} \text{ and } -4 \cdot \frac{-1}{5}-5B=0 \implies B=\frac{-1}{5}(-\frac{4}{5})=\frac{4}{25}\]

Answer 8

if you
had y''-4y'-5y=x^7
then I would do the particular solution:
\[y_p=Ax^7+Bx^6+Cx^5+ Dx^4+Ex^3+Fx^2+Gx+H\]

Answer 9

Okay, thanks for the hint.

Answer 10

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
this is a good source
took at the table and look at the few sentences below that table

Answer 11

"The more complicated functions arise by taking products and sums of the basic kinds of functions"

Answer 12

that is the sentence part I mainly want you to look at

Answer 13

in the table notice we have a polynomial*exponential
the polynomial we should a polynomial of the same degree with unknown coefficients*the exponential part

and actually they give an example after table exactly like the one you have here (well kind of exactly)