solve the differential equation y"-3y'-4y=3x+1 using method of undetermined coefficient
To solve the given differential equation y"-3y'-4y=3x+1 using the method of undetermined coefficients, we first find the complementary function and then find the particular solution.
Step 1: Find the complementary function (the solution to the homogeneous equation y"-3y'-4y=0)
The characteristic equation for this homogeneous equation is r^2 - 3r - 4 = 0. Solving for r, we get r = 4 and r = -1. Thus, the complementary function is given by:
yc(x) = c1 * e^(4x) + c2 * e^(-x)
Step 2: Find the particular solution (the solution to the non-homogeneous equation y"-3y'-4y=3x+1)
Assume a particular solution of the form:
yp(x) = Ax + B (since the right side of the differential equation is a linear function)
Now, find the first and second derivatives of yp(x):
yp'(x) = A
yp"(x) = 0
Plug these derivatives into the given differential equation:
0 -3A - 4(Ax+B) = 3x+1
This simplifies to:
-4Ax - 4B - 3A = 3x + 1
Now, we match coefficients of x terms and constant terms on both sides:
-4A = 3 => A = -3/4
-4B - 3A = 1 => -4B + 9/4 = 1 => B = 1/4
So, the particular solution is:
yp(x) = -3/4x + 1/4
Step 3: Add the complementary function and the particular solution to get the general solution:
y(x) = yc(x) + yp(x) = c1 * e^(4x) + c2 * e^(-x) - (3/4)x + 1/4