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Bernoulli ODE

  • #1

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 :)
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
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Don't despair.

[tex] y^{-4}y' + 3y^{-3} = e^{-3x} [/tex]

[tex] u = y^{-3} [/tex]

[tex] u'= -3y^{-4} y' \Rightarrow y^{-4} y' = -\frac{1}{3}u' [/tex]

[tex] u'-9u=-3e^{-3x} [/tex]

[tex] \left(ue^{-9x}\right)' = -3 e^{-12x} [/tex]

Can you continue from here ?
 
  • #3
OhhhHH, i forgot to multipy the g(x) term on the right by the integrating factor as well.... Oh man.

THANK you!!!
 

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