# Math Help - Solve y'=y(1 + xy^4)

1. ## Solve y'=y(1 + xy^4)

2. It's Bernoulli, with standard form

$y'-y=xy^{5}$

3. ok

4. ## bernulli solution problematic points

$y'-y=5xy^5\\$
$\frac{y'}{y^5}-\frac{1}{y^4}=5x \, \, \, \, \, \,z=\frac{1}{y^4} \, \, \, \, \, \,\, \,dz=\frac{1}{y^5}dy$
how does from $dz=\frac{1}{y^5}dy$ we get z' and y' we only have dz and dy here???
in order to get y' we need to have dy/dx we dont have it here.

then
$z'-z=5x \, \, \, \, \, \, a=-1 \, \, \, \, \, A=-z \, \, \, \, \, \ e^{-z}(z'-z)=e^{-z} 5z \, \, \, \, \,$ $(e^{-z}z)'=e^{-z} 5z \, \, \, \, \, e^{-z}z=\int ^z e^{-t} 5t +c$ after solving by parts i get
$e^{-z}z=-5ze^{-z}+5e^{-z} +c$

$c=e^{-z}z+5ze^{-z}-5e^{-z}$
what to do now?

5. First off, if you let $z = y^{-4}$ then $\displaystyle \frac{dz}{dy} = -4y^{-5}$.

6. ok i missed that -4 in the defferentiation

7. Well, you need to transform the DE which is in terms of $\displaystyle \frac{dy}{dx}$ into a DE in terms of $\displaystyle \frac{dz}{dx}$.

You have $z = y^{-4}$.

By the chain rule, $\displaystyle \frac{dz}{dx} = \frac{dz}{dy}\,\frac{dy}{dx}$.

So $\displaystyle \frac{dz}{dx} = -4y^{-5}\,\frac{dy}{dx}$ and thus $\displaystyle \frac{dy}{dx} = -\frac{1}{4}y^5\,\frac{dz}{dx}$.

Substituting everything into the DE

$\displaystyle \frac{dy}{dx} - y = 5x\,y^5$

$\displaystyle -\frac{1}{4}y^5\,\frac{dz}{dx} - y = 5x\,y^5$

$\displaystyle -\frac{1}{4}\,\frac{dz}{dx} - y^{-4} = 5x$

$\displaystyle -\frac{1}{4}\,\frac{dz}{dx} - z = 5x$

$\displaystyle \frac{dz}{dx} + 4z = -20x$.

This is now first order linear, which you can solve using an integrating factor.

8. i got a similiar lenear differential equation
and i solved it by multiplying in exponet etc
and it doesnt come out
??

9. Your integrating factor is $e^{\int{4\,dx}} = e^{4x}$.

So multiplying through by the integrating factor gives

$\displaystyle e^{4x}\,\frac{dz}{dx} + 4e^{4x}z = -20x\,e^{4x}$

$\displaystyle \frac{d}{dx}(e^{4x}z) = -20x\,e^{4x}$

$\displaystyle e^{4x}z = \int{-20x\,e^{4x}\,dx}$

$\displaystyle e^{4x}z = -20\int{x\,e^{4x}\,dx}$

$\displaystyle e^{4x}z = -20\left(\frac{1}{4}x\,e^{4x} - \int{\frac{1}{4}e^{4x}\,dx}\right)$

$\displaystyle e^{4x}z = -5x\,e^{4x} + 5\int{e^{4x}\,dx}$

$\displaystyle e^{4x}z = -5x\,e^{4x} + \frac{5}{4}e^{4x} + C$

$\displaystyle z = -5x + \frac{5}{4} + Ce^{-4x}$

$\displaystyle y^{-4} = -5x + \frac{5}{4} + Ce^{-4x}$

$\displaystyle y = \left(-5x + \frac{5}{4} + Ce^{-4x}\right)^{-\frac{1}{4}}$.