Question

I have to use Wronskian to find out the particular solution of ODE, but I cannot get the answer, please help
y"-5y'+6y=2e^t

Answer 1

I got y_1 =e^2t; y_2 =e^3t, and Wronskian is e^5s then ???

Answer 2

plug all of them into the formula, never get the right answer which is e^t. Please, help

Answer 3

one more question: when calculate Wronskian, if I let e^3t first then e^2t, the W(s) = -e^5s. which is different when I let e^2t the e^3t , the W(s) = e^5s. Why??

Answer 4

Swapping columns of a matrix changes the sign of the determinant; this makes intuitive sense if you look at it as an "oriented" volume of the parallelipid spanned by its columns in regular linear algebra

Answer 5

Oh, yah, so we should be "flexible" in choosing y_1 y_2 to get the W(s) positive, right?

Answer 6

right; sign is arbitrary and depends on which you call the "first" solution

Answer 7

can someone show me, please

Answer 8

\[
y_p= - y_1\int \frac{y_2 e^{2t}}{e^{5t}}dt+ y_2\int \frac{y_1 e^{2t}}{e^{5t}}dt
\]

Answer 9

yes

Answer 10

there is 2 in numerator since g(s) = 2e^t

Answer 11

the solution in book and in Wolframalpha is e^t. :( !! I didn't get it.

Answer 12

Oh Sorry, I read \[ g(t) = e^{2t}\]

Answer 13

Let me fix that

Answer 14

\[
y_p= - e^{2t}\int \frac{e^{3t} 2 e^{t}}{e^{5t}}dt+ e^{3t}\int \frac{e^{2t} 2 e^{t}}{e^{5t}}dt=\\

-e^{2t}\int2 e^{-t} dt+e^{3t}\int2 e^{-2t}dt=2 e^{t}- e^{t}=e^{t}\\
\]

Answer 15

so, we don't have to plug any arbitrary limits to the integral to find it out? Anyway, thanks a ton... I am stuck with it for 8 hours.

Answer 16

No, that is the correct way of doing it. Enjoy.

Answer 17

my prof asked us do something like