Please Solve Bernoulli Equation (6x+1)y^2y^'+3x^2+2y^3=0

Question

All I need to know is what my V variable is. After simplifying the equation in the Bernoulli format I keep getting V = y^1-n = y^1-1. n cannot be 1 and I don't know what I did wrong.

Answer 1

Matthew W., you are correct that it cannot be n = 1; rather, n = -2 in your problem.

( 6x+1 ) * y² * y' + 3x² + 2y³ = 0

( 6x+1 ) * y² * y' + 2y³ = -3x²

y² * y' + [ 2 / ( 6x + 1 ) ] * y³ = - 3x² / ( 6x + 1 )

Standard Bernoulli Equation Format:

y^-n * y' + P(x) * y^1-n = Q(x)

Setting P(x) = 2 / ( 6x + 1 ) & Q(x) = - 3x² / ( 6x + 1 )

y² * y' + P(x) * y³ = Q(x)

Everything will match, so long as:

-n = 2 → n = -2

&

1 - n = 3 → 1 = n + 3 → n = -2

Thus, V = y^1-n = y^1-(-2) ∴ V = y³

I hope this helps! Message me in the comments with any questions, comments, or concerns you may have on what I did above!

Also, feel free to message me in the comments if you'd like to double-check your final answer after you've done the subsequent differential equation solving! I did it on a separate piece of paper that I could attempt to upload here if you so ask for it!