Finding Particular Solutions using Variation of Parameters (continued)

3.

Finally substitute yp(x), dyp(x)/dx , d^2yp(x)/dx^2 into d^2y/dx^2 + b dy/dx + cy = f(x) The result is a second equation to determine du1/dx and du2/dx . This second equation has the form: du1(x)/dx dy1(x)/dx + du2(x)/dx dy2(x)/dx = f(x) (equation 2) where we used that both y1(x) and y2(x) satisfy the homogeneous d.e. a d^2y/dx^2 + b dy/dx + cy = 0

4.

The next step is to solve equations (1) and (2), using algebra, for du1(x)/dx and du2(x)/dx .Note, you have two equations in two unknowns. The unknowns are the derivatives of u1(x) and u2(x) . They equations are as follows: (repeated for you here) du1(x)/dx y1(x) + du2(x)/dx y2(x) = 0 (equation 1) du1(x)/dx dy1(x)/dx + du2(x)/dx dy2(x)/dx = f(x) (equation 2) Integration then gives u1(x) and u2(x). It turns out that the constants of integration end up being absorbed in the complementary solution. Note: This integration may not be simple. Finally, the particular solution is yp (x) = u1(x) y1(x) + u2(x) y2(x) Click here for an example.

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