Find the general solution of the differential equation y '' + 2Y '- 48y = e ^ X

Find the general solution of the differential equation y '' + 2Y '- 48y = e ^ X

Generally speaking, the simplest form of a differential equation is the derivative form of only one variable on both sides of the equation. Then divide the equation into two parts
(1) . y '' + 2Y '- 48y = 0 (right = 0)
(2) . y = a * e ^ x (the right side is the right side of the original equation, if it is in the form of x ^ n, then it is y = a * x ^ (n + J), where j is the highest order of the derivative of Y)
Just add the solutions of the two equations
Because the derivative of e ^ x is still e ^ x, and the derivative of e ^ NX is n * e ^ NX, the solution of equation (1) can be written in various combinations of e ^ NX. The solution of the characteristic equation is n in the e ^ NX coefficient of the corresponding differential equation
So it's like the solution upstairs
The characteristic equation is: x ^ 2 + 2x-48 = 0, two roots: 6, - 8
Therefore, the general solution can be set as: C1 * e ^ (6x) + C2 * e ^ (- 8x) + A * e ^ X
Substituting, a + 2a-48a = 1, a = - 1 / 45
Finally, the general solution is: C1 * e ^ (6x) + C2 * e ^ (- 8x) - 1 / 45 * e ^ X
This is more clear