Question

Find a solution to
f′′− 2f′− 3f = 2 cos(x)
which also satisfies f(0) = 2 and f′(0) = 3.

-4/11sinx-2/11cox
solves the equation without the conditions..

Answer 1

\[\begin{vmatrix}W_1&W&W_2\\y_1&0&y_2\\
y_1^{'}&2cos(x)&y_{2}^{'}\end{vmatrix}\]

Answer 2

if you take the cross product/ determinant of that you will get your Wronskian values; and y1 y2 are the solutions to your homogenous part

Answer 3

only difference is you dont need to negate the W part :)

Answer 4

ok thank you!! how would i go about finding the solutions to the homogeonous part?!

Answer 5

by applying the characteristic equation associated with a second order homogenous linear diffyQ

Answer 6

assume e^rx is an answer and apply it to the problem in place of y y' and y''

Answer 7

what you end up with is something looking like a quadratic equation to find the roots to:
\[e^{r}(r^2-2r-3)=0\]

solve for "r"

Answer 8

ohh thank you!!!

Answer 9

yep

Answer 10

sorry can i ask how to go abotu solving e^r = 0?

Answer 11

since e^r is never zero; its not really important. find the zeros of the quadratic part and you are good to go

Answer 12

so r = -3, 1? and those are the solutions to put into the matrix?!

Answer 13

close, those are solutions of e^rx that are put into the matrix

(r-3)(r+1) = 0 ; r = -1,3
\[y=y_h+y_p\]\[y_h=c_1e^{3x}+c_2e^{-x}\]\[y_p=e^{3x}\int \frac{W_1}{W}+e^{-x} \int \frac{W_2}{W}\]