Differential Equation, Bernoulli

[H2instinct]

Homework Statement

3y^2*y'+y^3= e^-x

The Attempt at a Solution

I am using Bernoulli's equation to substitute V in and I keep coming out with
Y^3= e^x(-e^-x+c)

my V = y^3
Y = v^(1/3)
and dy/dv=(1/3)v^(-2/3)

I peeked to see if I was correct, the right answer is supposed to be Y^3 = e^-x (x+c)

Where did I go wrong?

 

[Dick]

v=y^3 is certainly a good substitution to use, and v=e^(-x)*(x+c) is certainly the correct answer. What I don't understand is how you used Bernoulli's equation to get a wrong answer. Can you explain?

 

[H2instinct]

v=y^3 is certainly a good substitution to use, and e^(-x)*(x+c) is certainly the correct answer. What I don't understand is how you used Bernoulli's equation to get a wrong answer. Can you explain?

Ok I will try and explain what I did:

After getting those substitutions I applied them all to the original equation and got out:

(1/3)v^(-2/3)* Dv/Dx + (1/3)v^(1/3) = (1/(3e^x))*v^(-2/3)

Multiplied by the inverse of dy/dv to get:

dv/dx + v = 1/(e^x)

This is where I got confused, the function I am supposed to put into Bernoulli's formula usually lies in front of V, so I figured I would use 1, so I did:

e^{integral(1, dx)} = e^x

Multiplied both sides by this and integrated to get the answer that I got. I hope I better explained what I did. Do you see where I went wrong, because I still do not.

EDIT: When I multiplied both sides by e^x I seemed to have spaced and not multiplied the RHS... Thus giving me 1 on the right hand side which integrates to be x+C, thus giving me the correct answer... Having me go back through it really helped me figure out what I did wrong. Thanks for helping.

 

[H2instinct]

P.S. I really love this forum, everyone is so helpful!

 

[Dick]

Ok, so now you've got e^x*v'+e^x*v=1, right? That's (e^x*v)'=1. Integrate both sides then solve for v.