Exercise on Second-Order Differential Equations 18
We are asked to solve the following differential equation:
$$ y'' - 2y' - 3y = e^{4x} $$
This is a second-order linear nonhomogeneous differential equation.
The associated homogeneous equation is:
$$ ay'' + by' + cy = 0 $$
where \( a = 1 \), \( b = -2 \), and \( c = -3 \):
$$ y'' - 2y' - 3y = 0 $$
To solve it, we begin by analyzing the characteristic equation using an auxiliary variable \( t \):
$$ t^2 - 2t - 3 = 0 $$
The discriminant is positive:
$$ \Delta = b^2 - 4ac = (-2)^2 - 4(1)(-3) = 16 $$
Therefore, the equation has two distinct real roots:
$$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-2) \pm \sqrt{16}}{2} = \begin{cases} t_1 = \frac{2 - 4}{2} = -1 \\ \\ t_2 = \frac{2 + 4}{2} = 3 \end{cases} $$
The general solution to the homogeneous equation is:
$$ y_o = c_1 e^{-x} + c_2 e^{3x} $$
We now look for a particular solution \( y_p \) to the original equation using the method of undetermined coefficients.
Here, the nonhomogeneous term is exponential: \( f(x) = e^{4x} \). Since \( \lambda = 4 \) is not a root of the characteristic equation \( \lambda^2 - 2\lambda - 3 = 0 \), we assume a particular solution of the form:
$$ y_p = A \cdot e^{\lambda x} = A \cdot e^{4x} $$
Compute the first and second derivatives:
$$ y_p' = D_x[A e^{4x}] = 4A e^{4x} $$
$$ y_p'' = D_x[4A e^{4x}] = 16A e^{4x} $$
Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the original differential equation:
$$ y'' - 2y' - 3y = e^{4x} $$
$$ 16A e^{4x} - 2(4A e^{4x}) - 3(A e^{4x}) = e^{4x} $$
$$ (16A - 8A - 3A) e^{4x} = e^{4x} $$
$$ 5A e^{4x} = e^{4x} $$
Divide both sides by \( e^{4x} \) to isolate \( A \):
$$ 5A = 1 \quad \Rightarrow \quad A = \frac{1}{5} $$
Substitute the value of \( A \) into the particular solution:
$$ y_p = \frac{1}{5} e^{4x} $$
Finally, combine the homogeneous and particular solutions to obtain the general solution to the original equation:
$$ y = y_o + y_p $$
$$ y = c_1 e^{-x} + c_2 e^{3x} + \frac{e^{4x}}{5} $$
And this completes the solution.
