Question

solve the equation, x^2 y''+4xy'+2y=e^x by Euler equations.​

Answer 1

Let v = lnx, dv/dx = 1/x, x = e(v), dx/dv = e(v)

dy/dx = dy/dv(dv/dx) and by the product rule,

d²y/dx² = d²y/dv²(dv/dx) + d²v/dx²

Hence, in general ax²y" + bxy' + cy = f(x)

= ax²[ d²y/dv²( 1/x² ) - 1/x² ] + bx[ dy/dv( 1/x ) ] + cy

= a[ d²y/dv² ] + ( b-a ) dy/dv + cy = g(v)

which is linear and has constant coefficients

In your example, a = 1, b = 4, c = 2

d²y/dv² + 3 dy/dv + 2y = e(x), λ = -1, -2

y_h = Ae(-v) + Be(-2v) = A/x + B/x²