Verified answer

dy/dx + 2y /x = (2x - 3)

u(x)dy/dx + u(x) 2y /x = u(x)(2x - 3)

we want u(x) such that 2u(x)/x = u'(x)

u(x)dy/dx + u'(x)y = u(x)(2x - 3)

(u(x)y)' = u(x)(2x - 3)

∫(u(x)y)' = ∫u(x)(2x - 3)

u(x)y - C = ∫u(x)(2x - 3)

y = (∫u(x)(2x - 3) dx + C) / u(x)

2u(x)/x = u'(x)

2/x = u'(x) / u(x)

2/x = ln(u(x))'

∫2/x = ∫ln(u(x))'

2ln(x) + k = ln(u(x))

u(x) = e^(2ln(x) + k)

= x²e^(k)

y = (∫x²e^(k))(2x - 3) dx+ C) / x²e^(k)

= (∫2x³e^(k) - 3x²e^(k)dx + C) / x²e^(k)

= (x^4e^(k)/2 - x³e^(k) + C) / x²e^(k)

= x²/2 - x + K/x²

But I've never done one of these before, so i don't know that I am right

d^2y/dx^2 + 3(dy/dx) - 10y = 10x^2 r^2 + 3r - 10 = 0 (r + 5)(r - 2) = 0 r + 5 = 0, r - 2 = 0 r = - 5, r = 2 y = c?e^(-5x) + c?e^(2x) fixing the coefficient for yp = Ax^2 + Bx + C yp = Ax^2 + Bx + C y'p = 2Ax + B y''p = 2A 2A + 3(2Ax + B) - 10(Ax^2 + Bx + C) = 10x^2 2A + 6Ax + 3B - 10Ax^2 - 10Bx - 10C = 10x^2 - 10Ax^2 + (6A - 10B)x + 2A + 3B - 10C = 10x^2 fixing the coefficient of x^2 - 10A = 10 A = - a million for x 6A - 10B = 0 6(-a million) - 10B = 0 - 6 - 10B = 0 - 10B = 6 B = - 6/10 or -3/5 for consistent 2A + 3B - 10C = 0 2(- a million) + 3( - 3/5) - 10C = 0 - 2 - 9/5 - 10C = 0 - 19/5 - 10c = 0 - 10C = 19/5 C = - 19/50 for this reason, the final answer of the lenear, 2nd- order ODE is y = c?e^(-5x) + c?e^(2x) - x^2 - 3/5x - 19/50 answer//

dy/dx + 2y/x = 2x - 3

Find the integrating factor:

M = e^(∫ 2/x dx)

M = e^(2ln|x|)

M = e^(ln|x²|)

M = x²

Multiply both sides by the integrating factor:

x² dy/dx + 2xy = 2x³ - 3x²

Reverse product rule on left-hand side:

d/dx [x² y] = 2x³ - 3x²

Integrate both sides:

∫ d/dx [x² y] = ∫ (2x³ - 3x²) dx

x² y = (x^4)/2 - x³ + C

Divide both sides by x²:

y = (x²)/2 - x + C/x²