Question

Hi fellas! Differential Equation SOS!!!!

Answer 1

\[y''+6y'-7y=x^2e^x+7e^(-7x)+e^(-7x)sinx\]

Answer 2

I need to write down the general and the private solution of the DE @Loser66

Answer 3

@freckles

Answer 4

and THANKS A LOT guys I truly appreciate that!

Answer 5

genuinely

Answer 6

I know how to get to the general solution, but the private one is killing me

Answer 7

\[yh=C1e ^{-7x}+C2e ^{x}+yp\]

Answer 8

can you help me out with the solution of the private solution?

Answer 9

The homogenious part give you the solutions, right? what are they?

Answer 10

\(y_h= C_1 e^{-7x}+C_2e^x\), right?

Answer 11

yes

Answer 12

hence \(e^x\) is available.
Now private part, the first term is \(x^2e^x\) , that is we have
\(y_{p1}= (Ax^2+Bx+C)*x(e^x)\)

\(y'_{p1}= you~~ do\)
\(y"_{p1} =.....\)
then apply to the equation y"+6y'+7y to solve for \(y_{p1}\)

Answer 13

not sure I'm following, let me have a look though

Answer 14

wow it's so long

Answer 15

a couple of minutes I need to sort it out

Answer 16

The second term is \(7e^{-7x}\) , I do it for you as a sample
We need find \(y_{p2}= Dte^{-7x}\\y'_{p2}= -7De^{-7x}\\y"_{p2}= 49 De^{-7x}\)
Now combine
\(y"_{p2}+6y'_{p2}-7y_{p2}= 49De^{-7x} -42De^{-7x}+7Dte^{-7x}= 7e^{-7x}\)

Answer 17

from that you have \(7De^{-7x}+7Dxe^{-7x}= 7e^{-7x}\)
Make a comparison, you have D =1
hence the solution for that is \(y_{p2}= xe^{-7x}\)
Sorry for t at the beginning, it is x.

Answer 18

sorry but what does D refer to?

Answer 19

Do the same with the last term\(e^{-x} sinx\)