Differential Equations Exercise 23

We want to solve the following differential equation:

$$ y'' - y' = x $$

This is a non-homogeneous second-order linear differential equation.

The standard approach is to split the problem into two parts: first find the homogeneous solution, then a particular solution, and finally combine them.

Homogeneous Solution

The associated homogeneous equation is:

$$ y'' - y' = 0 $$

The corresponding characteristic equation is:

$$ z^2 - z = 0 $$

Factoring gives:

$$ z(z-1) = 0 $$

so the two distinct roots are:

$$ z = \begin{cases} 0 \\ \\ 1 \end{cases} $$

Therefore, the general solution of the homogeneous equation is:

$$ y = c_1 e^{0 \cdot x} + c_2 e^{1 \cdot x} $$

$$ y = c_1 + c_2 e^{x} $$

Particular Solution

The non-homogeneous term on the right-hand side suggests using the method of undetermined coefficients:

$$ y'' - y' = x $$

Since the forcing term is a first-degree polynomial, and with c=0 and b≠0 in the general form ay''+by'+cy = f(x), an appropriate trial solution is:

$$ y_p = Ax^2 + Bx $$

Computing derivatives:

$$ y'_p = 2Ax + B \qquad y''_p = 2A $$

Substitute into the differential equation:

$$ y''_p - y'_p = x $$

which becomes:

$$ 2A - (2Ax + B) = x $$

$$ 2A - 2Ax - B = x $$

Comparing coefficients on both sides gives:

$$ \begin{cases} -2A = 1 \\ \\ 2A - B = 0 \end{cases} $$

Explanation. - The coefficient of x is - 2A on the left and 1 on the right, so - 2A=1. - The constant term is 2A - B on the left and 0 on the right, so 2A - B=0.

Solving yields:

$$ A = -\tfrac{1}{2}, \quad B = -1 $$

Substituting back into the trial function gives:

$$ y_p = - \tfrac{x^2}{2} - x $$

General Solution

The general solution of the original equation is the sum of the homogeneous and particular solutions:

$$ y = y_o + y_p $$

That is:

$$ y = c_1 + c_2 e^{x} - \tfrac{x^2}{2} - x $$

This completes the solution.