Question

SOS Differential Equations! y'+y=xy^3

Answer 1

\[y'+y=xy^3\]

Answer 2

the answer is \[x=cy+\ln^2y\]

Answer 3

it's clearly a Bernoulli equation

Answer 4

Looks like you answered your own question!

Answer 5

@chaise It's from the book. but I am very far from the solution

Answer 6

@IrishBoy123 ? Will you be my savior?

Answer 7

@SithsAndGiggles ?

Answer 8

Substitute \(t=y^{-2}\).

Answer 9

t?

Answer 10

you mean adter dividing by y^3?

Answer 11

after

Answer 12

Any letter or symbol will do. Dividing through by \(y^3\) yields
\[y^{-3}y'+y^{-2}=x\]
and substituting \(t=y^{-2}\) gives \(t'=-2y^{-3}y'\) via the chain rule, and so the ODE is linear in \(t\):
\[-\frac{1}{2}t'+t=x\]

Answer 13

oh yeah I did that

Answer 14

and from there I'm quite confused tbh

Answer 15

Well it's an ordinary linear equation, so you can solve it by finding an integrating factor, or undetermined coefficients, or some more nonstandard creative way.

Answer 16

It would be easier to multiply everything by -2, and then use integrating factor to solve:

\[t'-2t=-2x\]

Now apply integrating factor.

Answer 17

Let p(x)=e^int(-2.dt)=e^(-2t)

Now apply reverse product rule. This video might help:

https://www.youtube.com/watch?v=eWLgWazTc-0

Answer 18

well I did went from bernoulli to a linear equation by the book, and it's very far away from the solution... can you solve it to see if you have the same problem?

Answer 19

@SithsAndGiggles I did the same way, but I'm not sure if the problem is with me or with the answer

Answer 20

@wio?

Answer 21

Any idea?

Answer 22

\[
\mu (x) = e^{\int p(x) dx} = e^{\int -2\;dx} = Ce^{-2x}
\]

Answer 23

Multiply by integrating factor: \[
\mu(x)t' + -2t\mu(x) = \mu(x)2x \implies (\mu(x)t)' = \mu(x) 2x \\
Ce^{-2x}t = \int Ce^{-2x}(-2x)\;dx
\]Then integrate it.

Answer 24

why did you go from \[Ce^-2x=\int\limits?\]

Answer 25

\[
e^{\int -2\;dx} = e^{-2x+k} = Ce^{-2x}
\]

Answer 26

oh ok

Answer 27

Basically going backwards: \[
(Ce^{-2x}t) ' = (Ce^{-2x})'t + Ce^{-2x}t' = (Ce^{-2x})(-2t) +(Ce^{-2x})t'
\]

Answer 28

Then I integrate both sides. The integrating factor is designed this way.

Answer 29

I'm a bit confused tbh, I don't see it leads me to the same answer as in the book..., I've just looked and I did just the same

Answer 30

I guess then there's a problem with the answer in the book

Answer 31

Multiply by integrating factor: \[
Ce^{-2x}t = \int Ce^{-2x}(-2x)\;dx
\]Then integrate it.
We get \[
Ce^{-2x}t = Ce^{-2x}\left(x+\frac{1}{2}\right) + K \implies t = x+\frac 12+ K'e^{2x}
\]I'm guessing at this point we would undo our substitution.

Answer 32

see? I did get that as well!!!! Horray!

Answer 33

then it probably means the answer in the book is wrong

Answer 34

thanks a lot! @wio !!!!

Answer 35

@Hipocampus well done, i go that too but i thought i'd stay out while you had all the big guns helping you out.

those homogeneous eqns you were doing a few weeks back must seem real simple now!

Answer 36

@IrishBoy123 thanks to you they are!!!!!!

Answer 37

without your help I don't know what I would have done!

Answer 38

wherever you are on the globe, thank YOU!