\[y'+3x^2y=x^2\]
\[y'+\color{red}{3x^2}y=x^2\] we start by finding \(\mu\) in two steps
take the function of x in from of the y and integrate it wrt x
\[\int 3x^2dx\] is step 1
So x^3
right
next step, raise \(e\) to that power, that is your \(\mu\)
e^(x^3))
that is \(\mu\) now multiply both sides by your \(\mu\) and you will have, on the left, the derivative of \(\mu u\)
i mean the derivative of \(\mu y\)
e^(x^3))*y' + e^(2x^3)) = e^(x^3))*(x^2)
yes, and integrating on the left gives \[e^{3x}y\] so your only real job is to integrate on the right
So u-sub?
yes mental u sub
(1/3)e^(x^3)
I don't think thats right
i do
plus c, which is kind of important
Shouldn't the x come out front?
god help me
you were right, leave it be
Ok
\[e^{3x}y=\frac{1}{3}e^{3x}+c\] the c is real important
Okie
gives \[e=\frac{1}{3}\] but with it you get \[y=\frac{1}{3}+ce^{-3x}\]
that didnt come out right at all
Nope
i meant if you forget the c and solve for y you get \[y=\frac{1}{3}\]
but the correct answer is \[y=\frac{1}{3}+ce^{-3x}\]
Right
why this miracle works is explained on page 55 i have time for one more if you like, just so you get the hang of it
I appreciate it but I'm so tired. Tomorrow?
ok you try some though method is probably not clear yet but we did all the steps to find mu
I will do 2.3 tomorrow.
I should come annoy you Saturday or something
ok that is this section, do a ton i am not free saturday birthday party for a 3 year old
is your job monday through friday?
I will do my entire assigned homework problems. And well maybe a weekend soon. My exam is on the 27th of June so I have to annoy you sometime about it. (if you're free of course)
And yes it is. I come back to philly on the weekends
ok we will see we can do some tomorrow night if you like, send me a text
g'night
Absolutely
Thank you! Goodnight!
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