Solving Bernoulli Equation with y'+3y=e^(-3x)*y^4

[Laura1321412]

Homework Statement

y'+3y=e^(-3x)*y^4 , IC: y(1) = (12/4e^-3)^(-1/3)

Homework Equations

Bernoulli Method

The Attempt at a Solution

So n=4, i can substitue u=y^-3

u'+(-3)(3)u=(-3)e^(-3x)
determine an integrating factor of e^-9x, then integrate both sides

ue^(-9x)=e^(-3x) +C return to y

y^(-3)*e^(-9x)=e^(-3x) Now, to find constant

(4e^-3)/12 *e^(-9) = e^-3 +C

(e^-12)/3=e^-3 +C
(e^-12/3)-(e^-3)= C

SO,

y^(-3)*e^(-9x)=e^(-3x)+(e^-12)/3) - (e^-3)

y^-3 = e^6x + e^(9x-12)/3 - e^9x-3

y=e^(-2x)+e^(-3x+4)*(3^1/3)-e^(-3x+1)

But, apparently this is wrong...

I have no idea where i went wrong and I've repeated this question numerous times... I don't know if i have an issue with the method itself or what... Any suggestions??

Thanks :)

 

[dextercioby]

Don't despair.

$$y^{-4}y' + 3y^{-3} = e^{-3x}$$

$$u = y^{-3}$$

$$u'= -3y^{-4} y' \Rightarrow y^{-4} y' = -\frac{1}{3}u'$$

$$u'-9u=-3e^{-3x}$$

$$\left(ue^{-9x}\right)' = -3 e^{-12x}$$

Can you continue from here ?

 

[Laura1321412]

OhhhHH, i forgot to multipy the g(x) term on the right by the integrating factor as well... Oh man.

THANK you!