Differentialgleichung Anfangs- / Randwertprobleme 2(y-1)y'= x^3 + 4x^2 + 2 ; y(0)= -1

Question

Löse folgende Anfangs-bzw. Randwertprobleme:

2(y-1)y'= x^3 + 4x^2 + 2   ; y(0)= -1

Answer 1

2·(y - 1)·y' = x^3 + 4·x^2 + 2 ; y(0) = -1

(2·y - 2) dy/dx = x^3 + 4·x^2 + 2

(2·y - 2) dy = (x^3 + 4·x^2 + 2) dx

∫ (2·y - 2) dy = ∫ (x^3 + 4·x^2 + 2) dx

y^2 - 2·y = 1/4·x^4 + 4/3·x^3 + 2·x + C

y^2 - 2·y + 1 = 1/4·x^4 + 4/3·x^3 + 2·x + 1 + C

 

Randwert: (-1)^2 - 2·(-1) + 1 = 1/4·0^4 + 4/3·0^3 + 2·0 + 1 + C --> C = 3

 

(y - 1)^2 = 1/4·x^4 + 4/3·x^3 + 2·x + 4

y - 1 = ± √(1/4·x^4 + 4/3·x^3 + 2·x + 4)

y = 1 ± √(1/4·x^4 + 4/3·x^3 + 2·x + 4)