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I have the equation

$dy/dt + ty = t/y^3$

My steps are as follows

$ y^3dy/dt + ty^4 = t$

Let $v = y^4$

$v' = 4y^3dy/dx$

$ 1/4v' + tv = t $

$v' + 4tv = 4t$

Integrating factor = $e^{\int4t dt} = e^{2t^2}$

$(e^{2t^2}v)' = 4te^{2t^2}$

$\int(e^{2t^2}v)'dt = \int4te^{2t^2}dt$

$e^{2t^2}v = \int4te^{2t^2}dt$

Let $u = 4t$

$u' = 4 $

$v' = e^{2t^2}$

Now how do I integrate $e^{2t^2}$ to get $v$ ?

Are my steps correct or did I go wrong somewhere?

Finance
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  • I don't understand what you're doing on the step $v = y^4 v' = 4y^3dy/dx$. Some substitution? Which? Then somehow you end up with $v + 4tv = 4t$. Was it meant to be $v' + 4tv = 4t$? Then everything is fine up to $\displaystyle e^{2t^2}v = \int 4te^{2t^2}$ at which point it is enough to note that $4te^{2t^2}=(2t^2)'e^{2t^2}$, thus $\displaystyle \int 4te^{2t^2}\mathrm dt=e^{2t^2}+C$. Plus you should always add $\mathrm dt$ (or whatever variable is at hand) to the integrals. – Git Gud Apr 27 '14 at 11:20
  • Sorry, there were a few typos which I've corrected now. – Finance Apr 27 '14 at 11:42
  • Could you explain the last step a little further? Even if you changed the $4t$ to that, how does it then come out to what you have written? Also could it be done by integration by parts also, or is it not possible? – Finance Apr 27 '14 at 11:43
  • First problem with the current version of the question: "Integrating factor" should read $\mathrm e^{-2t^2}$, not $\mathrm e^{2t^2}$. Later on (and once the first problem is corrected), you shall want to use that $(\mathrm e^{2t^2})'=4t\mathrm e^{2t^2}$. And everything should roll... – Did Apr 27 '14 at 11:44
  • @Did I don't recall exactly what the definition of integrating factor is, but see this. What the OP did is correct. – Git Gud Apr 27 '14 at 11:51
  • @Amzoti I'm used to doing it this way because Seperation of Variables doesn't always work, so I didn't look for it. Thanks for pointing it out. – Finance Apr 27 '14 at 12:08

1 Answers1

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You have this differential equation: $\frac{1}{4}u'+tu=t \ \ \ |-tu$

$\frac{1}{4}u'=t-tu\ \ \ \text{factor out t on the rhs}$

$\frac{1}{4}u'=t(1-u) \ \ \ | \ :(1-u)$

$\frac{1}{4}\frac{1}{1-u} \, du=t\, dt$

Now integrate both sides.

callculus42
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