(dy/dx)-y=4y^5?

1 Answer
Andrea S.
May 4, 2018

#y= 1/sqrt2 1/root(4)( ce^(-4x) -1) #

Explanation:

This is a separable differential equation:

#dy/dx -y = 4y^5#

#dy/dx = y + 4y^5#

#dy/(y + 4y^5) = dx#

#int dy/(y + 4y^5) = x + c#

Solving the integral:

#int dy/(y + 4y^5) = int dy/(4y^5(1/(4y^4) + 1) #

Substitute:

#u = (1/(4y^4) + 1)#

#du = -1/y^5#

so:

#int dy/(y + 4y^5) = -1/4 int (du)/u =ln abs u + c#

#int dy/(y + 4y^5) = -1/4 ln abs (1/(4y^4) + 1)+c#

Then:

#x +c = -1/4 ln abs (1/(4y^4) + 1)#

#-4x +c = ln abs (1/(4y^4) + 1)#

#ce^(-4x) = 1/(4y^4) + 1#

#ce^(-4x) -1 = 1/(4y^4) #

#y^4 = 1/(4(ce^(-4x) -1) #

#y= 1/sqrt2 1/root(4)( ce^(-4x) -1) #

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