How do I solve the differential equation: y′′+y′=x2 ?

Answer 1

y" + y' =x^2

let y=e^r x y'=r e^r x y"=r^2 e^r x

solve the homogeneous equation first y" +y' =0 or e^r x (r^2 +r)= 0 and r(r+1)= 0 r=0,-1

then yh = c1e^0 x + c2 e^-1 x =c1 + c2 e^-x

now for the particular solution yp let yp= D x + E x^2 + F x^3 then yp'=D+ 2 E x + 3 F x^2 yp"= 2 E +6 F x

so y" +y' =x^2 becomes 2 E +6 F x + D + 2 E x + 3 F x^2 =x^2 equating coefficients F= 1/3 6 F + 2 E =0 and

2 E + D = 0 hence 6(1/3) + 2 E =2 +2 E =0 and E= -1 2(-1) + D= -2 + D=0 and D= 2

so yp = 2 x - x^2 + x^3/3 and the total solution is y = c1 +c2 e^-x + 2 x -x^2 +x^3/3

Answer 2

one way is to multiply by e^x

exy” + exy’ = x2ex

d/dx(exy’) = x2ex

then integrat using integration by parts, divide by ex, then interate again to go from y’ to y

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