What is the function #y# if #\frac { d y } { d x } = \frac { 3x ^ { 2} + 4x + 2} { 2( y - 1) }# and #y ( 0) = - 1#?

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Answer 1 · 2017-02-02T09:36:00

#y = 1 - sqrt(x^3 + 2x^2 + 2x + 4)#

#\frac { d y } { d x } = \frac { 3x ^ { 2} + 4x + 2} { 2( y - 1) }#

This is separable!

#int 2( y - 1) d y = int 3x ^ { 2} + 4x + 2 d x #

#y^2 - 2y = x^3 + 2x^2 + 2x + C #

using the IV:

#(-1)^2 - 2(-1) = C implies C = 3 #

#implies y^2- 2y - ( x^3 + 2x^2 + 2x + 3) = 0 #

From the Quadratic Formula:

#y = (2 pm sqrt(4 + 4 ( x^3 + 2x^2 + 2x + 3)))/(2)#

#y = 1 pm sqrt(x^3 + 2x^2 + 2x + 4)#

Checking the IV again:

#-1 = 1 pm sqrt(0+ 4)#

And so the solution is the negative radical ie

#y = 1 - sqrt(x^3 + 2x^2 + 2x + 4)#