Solve the particular solution of the following.

Question

a. xy'-3y = x⁴(e^x+cos(x))-2x² , y(π) = (e^π+(2/π))π³

b. 2xyy' = y²-2x³ , y(1) = 2

Answer 1

a) Dividing by x we have y' - 3/xy = x³(e^x + cosx).

Integrated factor µ = e^∫-3/xdx = e^-3lnx = x^-3.

Multiplaying by µ = x^-3 we get x^-3y' - 3x^-4y = e^x + cosx; (x^-3y)' = e^x + cosx; x^-3y = ∫(e^x + cosx)dx;

y = (e^x + sinx + C)x³_. To get C we have y(π) = (e^π + sinπ + C) · π³= (e^π+ 2/π)π³, so C = 2/π

y = (e^x + sinx + 2/π)x³;

_{b) Dividing by 2xy: y' - 1/(2x)y = - x}² _y^-1. This is Bernully equation: v = y², v' = 2yy'

yy'- 1/(2x)y² = - x², 1/2v' - 1/(2x)v = - x², v' - 1/xv = - 2x²; µ = e^∫-1/xdx = e^--lnx = x^--1

x^--1v^' - x^--2v = - 2x; (x^-1v)' = -2x, 1/xv = ∫-2xdx = - x² + C; v = - x³ + Cx, y² = Cx - x³; 4 = C - 1. C = 5

y² = 5x - x³