Thursday
May 23, 2013

# Homework Help: Calc

Posted by Scott on Sunday, July 11, 2010 at 8:04am.

Solve the following separable equation.
y'=(4x^3-1)/y for y(1)= -3

• Calc - bobpursley, Sunday, July 11, 2010 at 8:12am

y dy=4x^3 dx-dx

1/2 y^2=x^4-x + C

y= sqrt(2x^2-2x) + C1
y(1)=sqrt(0)===> C1=-3

y=sqrt(2x^3-2x) -3

check this

• Calc - drwls, Sunday, July 11, 2010 at 8:15am

dy/dx * y = 4x^3 -1

integral of y dy = integral of (4x^3-1)dx

y^2/2 = x^4 - x + C

9/2 = 1 -1 + C
C = 9/2

y^2 = 2x^4 -2x +9
y = -sqrt(2x^4 -x +9)

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