Answer:
Explanation:
To solve the differential equation x^2y'' − 3xy' + 3y = x^4e^x using variation of parameters, we can follow these steps:
Step 1: Find the complementary solution y_c by solving the homogeneous equation:
x^2y'' − 3xy' + 3y = 0
The characteristic equation is r(r-1) - 3r + 3 = 0, which simplifies to r^2 - 4r + 3 = 0. Factoring, we get (r-1)(r-3) = 0, so the roots are r=1 and r=3. Therefore, the complementary solution is:
y_c = c1x + c2x^3
where c1 and c2 are arbitrary constants.
Step 2: Find the particular solution y_p using variation of parameters. We assume that the particular solution has the form:
y_p = u1(x)*y1(x) + u2(x)*y2(x)
where y1 and y2 are solutions to the homogeneous equation (in this case, y1=x and y2=x^3) and u1 and u2 are functions to be determined.
We need to find the derivatives of y1 and y2:
y1' = 1, y1'' = 0
y2' = 3x^2, y2'' = 6x
Substituting into the differential equation and simplifying, we get:
x^2(u1''y1 + u2''y2) + x(u1'y1 + u2'y2) = x^4e^x
Plugging in y1 and y2 and simplifying, we get:
x^2(u1''*x + u2''*x^3) + x(u1'*1 + u2'*3x^2) = x^4e^x
We can write this equation in matrix form as:
[x^2x, x^2x^3] * [u1'', u2''] + [x1, x3x^2] * [u1', u2'] = [0, x^4e^x]
Taking the determinant of the coefficient matrix [x^2x, x^2x^3; x1, x3x^2], we get:
(x^2x)(x3x^2) - (x^2x^3)*x = 2x^4
Therefore, the inverse of the coefficient matrix is:
[3x, -x^3] / (2x^4)
Multiplying by the column vector [0, x^4e^x] gives:
[u1', u2'] = [3x, -x^3] * (1/(2x^4)) * [0, x^4e^x]
Simplifying, we get:
u1' = (3/2)*e^(-x), u2' = -(1/2)xe^(-x)
Integrating u1' and u2' with respect to x, we get:
u1 = -(3/2)*e^(-x), u2 = -(1/2)*xe^(-x) + C
where C is an arbitrary constant.
Therefore, the particular solution is:
y_p = -(3/2)xe^(-x) + Cx^3e^(-x)
where C is an arbitrary constant.
Step 3: The general solution of the differential equation is the sum of the complementary solution and the particular solution:
y = y_c + y_p