Question

Solve the given differential equations:

х 1. y' = y(0) = -2 y+x2y 2. 3x²y dx – (x3 + y3)dy = 0 , y(1) = -2

Answer 1

1) Solution y' = y+xży; y(0) = -2 Standard form of a first order separable ODE : N(y) y' = M(x) х y+x2y Multiply both sides by y y'y 1+x yy' х 1+x2 х yy lette N(y) = y; M(x) = 1 + x2 →yy' = 1 If N(y). y' = M(x), y then / N(y) dy = / M(x) dx, up to a constant ydy = ' Azdx + idx Apply u substitution : 1 = 1 + x2 = 1 du

= } .l du = { In (u) Substitute back u = 1 + x2 = { In (1 + x²)+c1 = / ydy = + 02 1 ſ ydy == 1+1 == + C2 = { in (1 + x²) + C1 Multiply both sides by 2 2y = 2 . 3 In (1 + x²) +201 y2 = ln (1 + x²) + 2C1 For x2 = f (a) the solutions are x = f(a), -f(a) y = In (1 + x²) + 21, y = - In (1 + x²) + 2C1 y = In (1 + x2)+ci, y=-In (1 + x2)+ci