Differential Equations Exercise 17
Let’s solve the following non-homogeneous second-order differential equation:
$$ y''+y = x^2 $$
This is a complete linear differential equation.
The corresponding homogeneous equation is:
$$ ay''+ by' + cy = 0 $$
with a=1, b=0, and c=1
$$ y''+ y = 0 $$
To solve it, we look at the characteristic equation using the auxiliary variable t:
$$ t^2 + 1 = 0 $$
Its discriminant is negative:
$$ \Delta = b^2 - 4ac = 0^2 - 4(1)(1) = -4 $$
So the characteristic equation has two complex roots:
$$ t = \frac{0 \pm \sqrt{-4}}{2} $$
$$ t = \frac{\pm \sqrt{4i^2}}{2} $$
$$ t = \frac{\pm i \sqrt{4}}{2} $$
$$ t = \frac{\pm i \cdot 2}{2} = \begin{cases} 0 + i \\ \\ 0-i \end{cases} $$
That is, α=0 and β=1.
Hence, the general solution of the homogeneous equation is a complex solution of the form:
$$ y_o = c_1 e^{ \alpha x} \cos(\beta x) +c_2e^{\alpha x} \sin{\beta x} $$
Substituting α=0 and β=1, we get:
$$ y_o = c_1 \cos(x) +c_2 \sin{x} $$
Next, we need a particular solution yp of the differential equation.
We use the method of undetermined coefficients:
$$ ay''+by'+cy = f(x) $$
Here f(x)=x^2, with a=1, b=1, c=0.
Since the forcing term is a quadratic polynomial P2(x)=x2, we assume:
$$ y_p = B x^2 + Ax + C $$
Taking derivatives gives:
$$ y_p' = A + 2Bx \qquad y_p'' = 2B $$
Substituting into the original equation:
$$ 2B + (Ax+Bx^2+C) = x^2 $$
Grouping like terms:
$$ x^2(B) + x(A) + (2B+C) = x^2 $$
Matching coefficients yields A=0, B=1, and 2B+C=0:
$$ \begin{cases} A = 0 \\ \\ B = 1 \\ \\ 2B+C = 0 \end{cases} $$
Explanation. Comparing coefficients: the x term gives A=0. The x2 term gives B=1. The constant term gives 2B+C=0.
Solving the system:
$$ A = 0, \; B = 1, \; C = -2 $$
So the particular solution is:
$$ y_p = x^2 -2 $$
The general solution is then:
$$ y = y_o + y_p = c_1 \cos(x) +c_2 \sin{x} + x^2 - 2 $$
Alternative Approach
The same problem can also be solved using the Wronskian method. Starting from the homogeneous solution:
$$ y_o = c_
