Question

solve the initial value problem y'+3y=2x+1,
y(1)=5

Answer 1

i know that the integrating factor is e^3x but i'm stuck after that.

Answer 2

multiply everything through by the integrating factor
what do you have after?

Answer 3

(e^3x)y=(e^3x)(2x-1)

Answer 4

you skipped a step, which is making you confused as to how to proceed
don't 'undo' the product rule on the left without leaving the proper derivative notation
what you have written above is not true, until you integrate the right...

Answer 5

so then its (e^3x)y'+(e^3x)y=2x-1?

Answer 6

\[y'+3y=2x+1\]\[y'e^{3x}+3ye^{3x}=e^{3x}(2x-1)\]here is the step where we notice the left is a product rule:\[(ye^{3x})'=e^{3x}(2x-1)\]you need to keep y in the parentheses

think about it, how could you apply the product rule to the expression as you have it with y' outside the parentheses?

now you can integrate both sides...

Answer 7

I did get to that step its the integrating that i am having trouble with because i know its integration by parts and i got the RHS = {(1/3)e^3x}(4/3 + 2x)+c
it just doesnt seem right to me

Answer 8

if that's not right it's very close
I'm trying to see if our answers simplify the same
I will do the integration...

Answer 9

\[\int(2x-1)e^{3x}dx=\frac13(2x-1)e^{3x}-\frac23\int e^{3x}dx\]\[=\frac13(2x-1)e^{3x}-\frac29e^{3x}+C=(\frac23x-\frac59)e^{3x}+C\]I think you're off by a 1 somewhere...

Answer 10

the equation starts with +1 not -1

Answer 11

ah so it does...
minor change, one sec.

Answer 12

\[\int(2x+1)e^{3x}dx=\frac13(2x+1)e^{3x}-\frac23\int e^{3x}dx\]\[=\frac13(2x+1)e^{3x}-\frac29e^{3x}+C=\frac19(6x+1)e^{3x}+C\]there we go :)

Answer 13

so now do i just divide by the integrating factor?

Answer 14

yes, then apply the initial conditions to find C

Answer 15

THANK YOU!!!!

Answer 16

very welcome :D

Answer 17

i got c=45/7 ?

Answer 18

no, there should be an e in it...
what do you have after dividing by the IF ?

Answer 19

i did 5=[(1/9)(6(1)+1)3^(3*1)]/e^3(1) + c/e^3

Answer 20

you need to be more careful, one step at a time
also, when dividing by e^(something) it's much nicer to just multiply both sides by e^(-something)
makes the expression much simpler-looking....

Answer 21

we had\[(ye^{3x})'=e^{3x}(2x+1)\]\[ye^{3x}=\frac19(6x+1)e^{3x}+C\]now multiply both sides by e^(-3):\[y=\frac19(6x+1)+Ce^{-3x}\]you forgot to cancel out that other e^(3x) earlier...
now initial conditions...

Answer 22

\[y(1)=\frac19(6+1)+Ce^{-3}=5\]\[Ce^{-3}=5-\frac79=\frac{38}9\to C=\frac{38}9e^3\]so we get\[y=\frac19(6x+1)+\frac{38}9e^3e^{-3x}\]and just to make it prettier we can simplify\[y=\frac19(6x+38e^{3(1-x)}+1)\]

Answer 23

*slight typo in the post before last:
after\[ye^{3x}=\frac19(6x+1)e^{3x}+C\]we multiply both sides by e^(-3x) to get...
(the rest is correct)

Answer 24

THANK YOU

Answer 25

you're welcome
here are some good notes on the subject:
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

Answer 26

I'm not really a stickler about it, but it's customary to give medals for good explanations so others know that the problem has been solved.
So feel free to click 'good answer' if you are satisfied with my explanation.

welcome to Open Study!