Solve each initial value problem.

Question

a.) (12x5 + 2x2 −6)dx −2yx3dy = 0 with y(1) = 6
b.) x^3 dy/dx −x^2y = −xy^3 with y(1) = 1

Answer 1

(12x^5 + 2x^2 −6)dx −2yx^3dy = 0
2y dy = (12x^2 + 2/x - 6/x^3 dx
y^2 = 4x^3 + 2lnx + 3/x^2 + C
since y(1) = 6,
4+3 = C
y^2 = 4x^3 + 2lnx + 3/x^2 + 7

Now you try the other. Post your work if you get stuck