Question

solve the given equation by the method of undetermined coefficients
$y - 2y - 3y = -15$

the solution is $y = \square$

Explanation:

Step1: Solve the homogeneous equation

The homogeneous equation is \(y'' - 2y' - 3y = 0\). The characteristic equation is \(r^{2}-2r - 3=0\).
Factor the quadratic equation: \((r - 3)(r+ 1)=0\).
So, \(r_{1}=3\) and \(r_{2}=-1\).
The complementary function \(y_{c}=C_{1}e^{3x}+C_{2}e^{-x}\).

Step2: Find a particular solution

Since the non - homogeneous term is a constant (\(-15\)), assume a particular solution of the form \(y_{p}=A\) (where \(A\) is a constant to be determined).
Differentiate \(y_{p}\): \(y_{p}' = 0\) and \(y_{p}''=0\).
Substitute into the original non - homogeneous equation \(y'' - 2y' - 3y=-15\):
\(0-2\times0 - 3A=-15\).
Solve for \(A\): \(-3A=-15\), so \(A = 5\).

Step3: Write the general solution

The general solution of the differential equation \(y=y_{c}+y_{p}\).

Answer:

\(y = C_{1}e^{3x}+C_{2}e^{-x}+5\)