Question

find the general solution of the differential equation:
y' + y(3/t) = t^2

Answer 1

When I compute it, it becomes a real mess. So maybe you can help me?

Answer 2

Are you having trouble with the homogeneous solution or the particular solution or...?

Answer 3

I have problems with: finding K' if you see what I mean

Answer 4

\(K'\) is likely a book specific thing. What else is it called?

Answer 5

Actually I have no problems with finding y and y'. But then when I place what I got for y and y' in y' + y(3/t) = t^2 and try to find the function K(x) (this comes from the constant C) i get into trouble...

Answer 6

What did you get for \(y(t)\)?

Answer 7

(Kt)^-3

Answer 8

K being e^C

Answer 9

So that is the solution to the homogeneous equation: \[
y' + y(3/t) =0
\]Does that make sense?

Answer 10

I'm not saying anything to help, just making sure we're on the same page here.

Answer 11

yes that's right

Answer 12

Now we want to find the particular solution.

We need to make a guess at what \(y(t)\) would be.
Notice that we know it is some sort of polynomial since \(t^2\) is a polynomial.

Answer 13

So my guess would be \(At^n\), and we want to solve for \(n\) and \(A\).
Does this guess make sense to you?

Answer 14

I'm not sure but ok

Answer 15

Yes, I know it sounds stupid, but we do have to actually guess at this stage.

Answer 16

I dont understand how you come to that guess...?

Answer 17

\[
y(t) = At^n \\
y'(t) = nAt^{n-1}
\]

Answer 18

Okay basically \(t^2\) is a polynomial, right?

Answer 19

yes

Answer 20

A polynomial plus a polynomial is a polynomial.

Answer 21

The derivative of a polynomial is a polynomial.

Answer 22

Oooookay I think your A is my K

Answer 23

\(A\) is just some constant we need to solve for, so is \(n\).

Answer 24

yes i understand

Answer 25

\[
y(t) = At^n \\
y'(t) = nAt^{n-1}
\]Plugging this in gets: \[
nAt^{n-1}+\frac{3At^n}{t}=t^2 \\
nAt^{n-1}+3At^{n-1}=t^2 \\
(nA+3A)t^{n-1}=t^2 \\
\]Okay so there are two neat equations here: \[

(nA+3A)t^{\color{red} {n-1}}=t^\color{red} {2} \implies n-1=2 \\
\color{red}{(nA+3A)}t^{n-1}=\color{red}{1}t^2 \implies nA+3A = 1 \\
\]

Answer 26

We can see right away \(n=3\).
Putting that into the second equaiton: \[
nA+3A = (3)A+3A = 1
\]

Answer 27

Can you solve for \(A\)?

Answer 28

1/6 sorry but I think I have to get a little bit sleep see you tomorrow

Answer 29

ok

Answer 30

my brain is functioning to slowly

Answer 31

It's correct.\[
\frac{1}{6}t^3
\]

Answer 32

Full solution is \[
kt^{-3}+\frac{1}{6}t^3
\]

Answer 33

But that isn't the answer, is it?

Answer 34

It is a family of solutions because we need initial value like \(t(0)=1\) to solve for \(k\).

Answer 35

that's right, i'll try to understand it tomorrow

Answer 36

Whoops, I mean we need something like \(y(1) = 0\) or something.

Answer 37

thanx a lot

Answer 38

yes sure i didn't even notice

Answer 39

Now that i've fully understood the concept I got the same result than you!!! Thanx for your help