Solve.
x^3 y^'''+5x^2 y^''+7xy^'+8y = t+e^(-t)
To solve the given differential equation, we will use the method of undetermined coefficients. This method involves assuming the form of the particular solution and solving for its coefficients.
Step 1: Find the homogeneous solution.
To find the homogeneous solution of the given equation, we set t = 0 and solve the resulting equation:
x^3y'''+5x^2y''+7xy'+8y = 0
Assume the form of the homogeneous solution as y_h = x^r. Substituting this into the equation, we get:
r(r-1)(r-2)x^(r-3) + 5r(r-1)x^(r-2) + 7rx^(r-1) + 8x^r = 0
Divide through by x^r to simplify:
r(r-1)(r-2) + 5r(r-1) + 7r + 8 = 0
Simplifying further, we obtain a cubic equation:
r^3 + 2r^2 - 7r - 8 = 0
We can factorize the equation as follows:
(r - 1)(r + 4)(r + 2) = 0
Thus, the roots of the cubic equation are r = 1, r = -4, and r = -2.
Therefore, the homogeneous solution is:
y_h = C1*x + C2*x^(-4) + C3*x^(-2)
Step 2: Find the particular solution.
To find the particular solution, we assume it has the form y_p = At + Be^(-t), where A and B are constants to be determined.
Substitute y_p into the given differential equation and solve for the coefficients A and B. We get:
x^3(0) + 5x^2(0) + 7x(0) + 8(At + Be^(-t)) = t + e^(-t)
8At + 8Be^(-t) = t + e^(-t)
From this equation, we can equate the coefficients of the terms with t and e^(-t) to solve for A and B:
8A = 1 (coefficient of t)
8B = 1 (coefficient of e^(-t))
Solving the above equations, we get A = 1/8 and B = 1/8.
Therefore, the particular solution is:
y_p = (1/8)t + (1/8)e^(-t)
Step 3: Find the general solution.
The general solution of the given differential equation is the sum of the homogeneous and particular solutions:
y = y_h + y_p
Substituting the values obtained for y_h and y_p, we get:
y = C1*x + C2*x^(-4) + C3*x^(-2) + (1/8)t + (1/8)e^(-t)
where C1, C2, and C3 are arbitrary constants.
This is the general solution to the given differential equation.