- #1

- 760

- 11

## Main Question or Discussion Point

Verify that given function is a solution.

y'' - 2y' + 2y = 0 , y=e^x(Acos x + Bsin x)

First I take derivative of y which is y+e^x(-Asin x + B cos x) then I asign e^x(-Asin x + B cos x) to y'-y. Then I take derivative of y' which is y'+(y'-y)-y which equals 2y'-2y=y'' then I use y'' as 2y'-2y which satifies. Do you have a faster or more exciting way?

Have a nice day.

y'' - 2y' + 2y = 0 , y=e^x(Acos x + Bsin x)

First I take derivative of y which is y+e^x(-Asin x + B cos x) then I asign e^x(-Asin x + B cos x) to y'-y. Then I take derivative of y' which is y'+(y'-y)-y which equals 2y'-2y=y'' then I use y'' as 2y'-2y which satifies. Do you have a faster or more exciting way?

Have a nice day.