Chapter D2.3, Problem 52E

Variation of parameters Finding a particular solution when a homogeneous solution appears in the right-side function is handled using a method called variation of parameters. (This method is also used to find particular solutions of equations that cannot be handled by undetermined coefficients.) The following steps show how variation of parameters is used to find the particular solution of one specific equation.

a.    Consider the equation y" – y – e^t. Show that the homogeneous solutions are y₁ = e^t and y₂ = e^–t. Note that the right–side function is a homogeneous solution.

b.    Assume a particular solution has the form

$y_p\left(t\right)=u_1\left(t\right)y_1\left(t\right)+u_2\left(t\right)y_2\left(t\right)=u_1\left(t\right)e^t+u_2\left(t\right)e^{-t},$

where the functions u₁ and u₂ are to be determined. Compute y_p' and impose the condition u₁'e^t + u₂'e^–t = 0 to show that y_pʹ = u₁e^t – u₂′e^–t.

c.    Compute y_p" and substitute it into the differential equation. After canceling several terms, show that the equation reduces to u₁ʹe^t – u₂ʹe^–t = e^t.

d.    Parts (c) and (d) give two equations for u₁ʹ and u₂ʹ Eliminate u₂ʹ and show that the equation for u₁ is $u_1{}^{\prime }=\frac{1}{2}$.

e.    Solve the equation in part (d) for u₁.

f.    Use part (e) to show that the equation for u₂ is $u_2{}^{\prime }=-\frac{1}{2}{e}^{2t}$.

g.    Solve the equation in part (f) for u₂.

h.    Now assemble the particular solution y_p(t) =u₁(t)e^t + u₂(t)e^–t and show that $y_p=\frac{1}{2}t{e}^t-\frac{1}{4}{e}^t$.