One integration gives v0 x22A where A is a constant another gives v x36 Ax B. Y c c 1e r1x c 2e r2x c 1e x. 1 0 For this problem we apply the techniques from an elementary ODE class.
The other part of the solution wxt satis es wtt wxx.
For v0 0 we need B 0 and then for v1 0 we need 16A 0 or A 16. Namely we know that the general solution is the sum of the general solution of the homogenous problem h and any particular solution 00 p. The general solution of this nonhomogeneous differential equation is In this solution c1y1 x c2y2 x is the general solution of the corresponding homogeneous differential equation. So the solution is x 1 y 4 z 4.