Is there an explicit ODE solution for this system?

Differential equations: find the general solution of a 2nd order linear ODE?

• find the general solution of a 2nd order linear ODE 1.) homogeneous 2nd order linear ODE (x^2)(y'') + xy' - 9y = 0 2.) non-homogeneous 2nd order linear ODE (x^2)(y'') - xy' + y = xlnx

• Answer:

  Previous question posted by you was deleted. Differential equations: Solve the initial value problem y'' - 4y = 9xe^x? y'' - 4y = 9xe^x, y(0) = 1, y'(0) = -7 Here is my answer (I spent some time answering, and it's quite frustrating having all that work thrown away, so I'll repost it here) First we find solution of associated homogeneous equation y'' − 4y = 0 Find roots of characteristic equation: r² − 4 = 0 (r − 2) (r + 2) = 0 r = 2, r = −2 Solution to associated homogeneous equation: y_h = c₁ e^(2x) + c₂ e^(−2x) ------------------------------ Next we find particular integrall of non-homogeneous equation using method of undetermined coefficients. y_p = (Ax + B) e^x y_p' = (Ax + B + A) e^x y_p'' = (Ax + B + 2A) e^x We substitute into differential equation: y_p'' − 4y = 9x e^x (Ax + B + 2A) e^x − 4(Ax + B) e^x = 9x e^x (−3A x + (2A − 3B)) e^x = (9x + 0) e^x Matching coefficients, we get −3A = 9 --------> A = −3 2A − 3B = 0 -----> −6 = 3B -----> B = −2 y_p = (−3x − 2) e^x ------------------------------ General solution to non-homogeneous equation: y = y_h + y_p y = c₁ e^(2x) + c₂ e^(−2x) − (3x + 2) e^x We use initial conditions to find c₁ and c₂ y(0) = 1 c₁ e^0 + c₂ e^0 − (0 + 2) e^0 = 1 c₁ + c₂ − 2 = 1 c₁ + c₂ = 3 y' = 2c₁ e^(2x) − 2c₂ e^(−2x) − (3x + 5) e^x y'(0) = −7 2c₁ e^0 − 2c₂ e^0 − (0 + 5) e^x = −7 2c₁ − 2c₂ − 5 = −7 2c₁ − 2c₂ = −2 Solving the two simultaneous equations, we get c₁ = 1, c₂ = 2 Solution to initial value problem: y = e^(2x) + 2 e^(−2x) − (3x + 2) e^x ============================== I'll wait a little bit and see if this question gets deleted (I'm not sure if that was you or someone else who reported it for some reason). I just don't want to put work into it right now, unless I'm sure it's going to stay around.