Question

find y''=y'

Answer 1

The solution of this homogeneous DE can be obtained by solving its auxiliary equation,
\(m^2-m=0 \implies m(m-1)=0\), which gives the zeros m=1 and m=0.
Thus, the general solution is \(y=c_1+c_2e^{x}\).

Answer 2

do it by series solution

Answer 3

I'm not sure if it can be done that way.

Answer 4

i have answer a0-a1+a1e^x

Answer 5

It's known that the solution of a DE of the given form is something of the form \(e^{mx}\).

Answer 6

That's right! You can consider \(a_0+a_1\) as another constant, call it \(a_2\). It's the same thing.

Answer 7

ri8 but what about series procedure i have to solve it by series solution or laplace

Answer 8

It can be solved by Laplace only if you're given its initial value, i.e y(0). I don't such a DE is solved using series solutions though.

Answer 9

y(0) and y'(0) actually.

Answer 10

1 and o respectively

Answer 11

Then you can use Laplace! Or just plug these values in the solution I gave above to get c_1 and c_2.

Answer 12

yeah i got it thanx :)

Answer 13

You're welcome!