Question

find the general solution of the differential equation, y'=x+y.

Answer 1

linear ODE

Answer 2

put it in standard form :

\(y' - y = x\)

Answer 3

heard of `Integrating factor` before ?

Answer 4

yes

Answer 5

IF = \(e^{\int -1 dx} = e^{-x}\)

Answer 6

multiply thru the equation by IF

Answer 7

\(y'e^{-x} - ye^{-x} = xe^{-x}\)

Answer 8

compress the left hand side using reverse product rule :

\((ye^{-x})' = xe^{-x}\)

Answer 9

integrate

Answer 10

i still can't find the answer

Answer 11

where are u stuck ?

Answer 12

\((ye^{-x})' = xe^{-x}\)

integrate :

\(\int (ye^{-x})'dx = \int xe^{-x}dx\)

Answer 13

on left side, by FTC, integral and derivative cancel each other out

Answer 14

\((ye^{-x})' = xe^{-x}\)

integrate :

\(\int (ye^{-x})'dx = \int xe^{-x}dx\)

\(ye^{-x} = \int xe^{-x}dx\)

Answer 15

evaluate the integral for right hand side,
and you're done.

Answer 16

i just dunno how to do the right side...

Answer 17

you can do it by parts

Answer 18

but il just use advanced guessing :


\(ye^{-x} = \int xe^{-x}dx\)

\(ye^{-x} = -xe^{-x} - e^{-x} + C\)

Answer 19

isolating y gives u :

\(y = Ce^x -x-1\)

Answer 20

thx a lot

Answer 21

np :) u should review `by parts` when u get time

Answer 22

yes i definitely will