solve for xy'' + 2y' + x = 1, y(1)=2, y'(1)=1

Question

how to soleve this equation:

xy'' + 2y' + x = 1, y(1)=2, y'(1)=1

Answer 1

y"+2y'/x=1/x-1 after dividing through by x.

e^∫2dx/x=e^ln(x²)=x². This is an integrating factor:

x²y"+2xy'=x-x²,

d(x²y')/dx=x-x²,

x²y'=½x²-⅓x³+c₁, where c₁ is a constant, divide through by x²:

dy/dx=½-⅓x+c₁x^-2,

y=½x-⅙x²-c₁/x+c₂.

y'(1)=1 so ½-⅓+c₁=1, c₁=⅚; y(1)=2 so:

y=½-⅙-⅚+c₂=2, c₂=5/2.

y=½x-⅙x²-5/(6x)+5/2.

CHECK

y'=½-⅓x+⅚x^-2; y"=-⅓-5/(3x³).

xy"=-⅓x-5/(3x²); 2y'=1-⅔x+5/(3x)²;

xy"+2y'+x=-⅓x-5/(3x²)+1-⅔x+5/(3x)²+x=1✔️