Question

Find the general solution of dy/dx+3y=e^3x.

Answer 1

question: is it not first order linear equation and we should find out the integrating factor?

Answer 2

Oh this is first order :) derrrrp lol.
Bah I'm a little tired I think.

Answer 3

@zepdrix I didn't study it yet, I don't know how to solve.

Answer 4

integral p(x) = 3x

Answer 5

for diff equ
\[y' + p(x)y = q(x)\]
integrating factor is:
\[\large e^{\int\limits p(x) dx}\]

Answer 6

thank you

Answer 7

continue please, I save time,

Answer 8

ok, let me practice for next semester.
so integrating factor is e^3x

Answer 9

yes

Answer 10

time both sides by e^3x to gety'e^3x + 3e^3xy= (e^3x)^2

Answer 11

yes or no?

Answer 12

correct

Answer 13

factor to get d/dt (e^3x*3y)=e^3x)^2

Answer 14

looks like you have it pretty figured out...why dont you finish it

Answer 15

oh you dont need that extra "3"

--> d/dx(e^3x *y)

Answer 16

why? the original problem has 3y , so time integrating factor in , why 3 disappears?

Answer 17

because of chain rule
derivative of e^3x = 3*e^3x

Answer 18

ok so we have
\[\large e^{3x} y' +3e^{3x} y = e^{6x}\]
left side is in form of product rule
\[\large ->e^{3x}*y' + (e^{3x})' *y\]
so you can say
\[\large (e^{3x} y)' = e^{6x}\]
now integrate both sides
\[\large e^{3x} y = \frac{1}{6} e^{6x} +C\]
solve for "y"
\[\large y = \frac{1}{6}e^{3x} +C e^{-3x}\]

Answer 19

since we don't have initial condition, no need to solve for C. If we have y_0 we have to solve for C and plug back to this y , right?

Answer 20

where is @idealist

Answer 21

yes correct...its asking for general solution so constant is ok
also they would have given initial conditions otherwise

Answer 22

haha ok and yw

Answer 23

right, its a wierd system they have now...a lot of 99's out there

Answer 24

:)