Question

Solve the equation using partial fractions, find an explicit solution if possible. Equation: dy/dx = (y^2-2y-3)dx

Answer 1

\[y-3/y+1 = C \exp (4x)\]

Answer 2

...so how did you get that?

Answer 3

dy = (y-3)(y+1)dx
dy/(y-3)(y+1) = dx
1/4(1/(y-3) -1/(y+1))dy = dx (partial fractions)
integration will give you
1/4(ln(y-3) - ln(y+1)) = x + integration constant
ln(y-3/y+1) = 4x + 4(integration constant)
y-3/y+1 = e^(4x + K)= e^k (e^4x)
y-3/y+1 = C e^4x

Answer 4

okay awesome thanks :)