Separable Equations

Question

Find the function y = y(x) (for x > 0) which satisfies the separable differential equation dy/dx = (7 +18x) / (xy^2); x>0 with the initial condition y(1) = 2. Y=

Answer 1

dy/dx = (7+18x) / (xy²)

y²dy = (7x^-1 + 18)dx

∫ y²dy = ∫(7x^-1 + 18)dx

(1/3)y³ = 7ln lxl + 18x + C

So, y³ = 21lnlxl + 54x+ C

Since y(1) = 2, 8 = 21(0) + 54 + C. So, C = -46.

y³ = 21(lnx) + 54x - 46 (Note: since x > 0 by assumption, ln l x l = ln x)

y = (21(lnx) + 54x - 46)^1/3

Answer 2

The trick is to get all your y's on one side and all your x's on the other.

dy/dx = (7 + 18x)/(xy^2)

Multiply by both denominators. In other words, cross-multiply.

(dy/dx)*dx*(xy^2) = (7 + 18x)/(xy^2) * dx * (xy^2)

On the left, the two copies of dx cancel. On the right, the two copies of (xy^2) cancel.

(xy^2) dy = (7 + 18x) dx

The left side still has a factor of x, so divide both sides by x.

(y^2) dy = (7 + 18x)/x dx

Before we take the antiderivative, it helps to split the fraction on the right side into two simpler fractions.

(y^2) dy = (7/x + 18) dx

Now we can integrate.

∫(y^2) dy = ∫(7/x +18) dx

(1/3)*y^3 = 7*ln |x| + 18x + C

y^3 = 3(7*ln |x| + 18x + C) or 21*ln |x| + 54x + C

y = cubic root of [3(7*ln |x| + 18x + C)] or (21*ln |x| + 54x + C)^(1/3)