Bernoulli equation solution methods

[nhrock3]

$$y'-y=5xy^5\\$$
$$\frac{y'}{y^5}-\frac{1}{y^4}=5x \, \, \, \, \, \,z=\frac{1}{y^4} \, \, \, \, \, \,\, \,dz=\frac{1}{y^5}dy$$
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 don't 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?

 

[tiny-tim]

hi nhrock3!

in order to get y' we need to have dy/dx we don't have it here.

yes you do … y' and z' are dy/dx and dz/dx

$$z'-z=5x$$

unfortunately, after this you somehow changed the 5x into a 5z

start again …

you need to find all the solutions to z' - z = 0 first (the "complementary solutions"),

then you need to find one "particular solution" (looks like it'll be a polynomial, doesn't it? ) to the whole equation, to add to all the complementary solutions