The problem is to solve the second-order non-homogeneous linear differential equation: $2y'' + y' - 3y = e^x$

1. Problem Description

The problem is to solve the second-order non-homogeneous linear differential equation:

$2y'' + y' - 3y = e^x$

2. Solution Steps

To solve the given differential equation $2y'' + y' - 3y = e^x$, we first find the homogeneous solution $y_h$ by solving the homogeneous equation:

$2y'' + y' - 3y = 0$

The characteristic equation is:

$2r^2 + r - 3 = 0$

Factoring the quadratic equation, we get:

$(2r + 3)(r - 1) = 0$

The roots are $r_1 = 1$ and $r_2 = -\frac{3}{2}$.

Therefore, the homogeneous solution is:

$y_h = c_1 e^x + c_2 e^{-\frac{3}{2}x}$

where $c_1$ and $c_2$ are arbitrary constants.

Next, we find a particular solution $y_p$ for the non-homogeneous equation $2y'' + y' - 3y = e^x$.

Since the right-hand side is $e^x$, and $e^x$ is part of the homogeneous solution, we assume a particular solution of the form:

$y_p = Axe^x$

where $A$ is a constant.

Now, we find the first and second derivatives of $y_p$:

$y_p' = Ae^x + Axe^x = A(1+x)e^x$

$y_p'' = Ae^x + Ae^x + Axe^x = A(2+x)e^x$

Substitute $y_p$, $y_p'$, and $y_p''$ into the original differential equation:

$2(A(2+x)e^x) + (A(1+x)e^x) - 3(Axe^x) = e^x$

$A(4+2x)e^x + A(1+x)e^x - 3Axe^x = e^x$

$A(4+2x+1+x-3x)e^x = e^x$

$A(5)e^x = e^x$

$5A = 1$

$A = \frac{1}{5}$

Thus, the particular solution is:

$y_p = \frac{1}{5}xe^x$

Finally, the general solution is the sum of the homogeneous and particular solutions:

$y = y_h + y_p = c_1 e^x + c_2 e^{-\frac{3}{2}x} + \frac{1}{5}xe^x$

3. Final Answer

The general solution of the differential equation is:

$y = c_1 e^x + c_2 e^{-\frac{3}{2}x} + \frac{1}{5}xe^x$