Answer:
The method of variation of parameters is used to solve second-order linear differential equations with non-homogeneous terms. The given differential equation is:
y'' + 2y' + y = e^(-x)ln(x)
First, let's find the complementary solution (homogeneous solution) for the associated homogeneous equation, which is:
y'' + 2y' + y = 0
The characteristic equation is:
r^2 + 2r + 1 = 0
This equation has a repeated root of -1:
(r + 1)(r + 1) = 0
So, the complementary solution is:
y_c(x) = (A + Bx)e^(-x)
Now, we need to find the particular solution (y_p) for the non-homogeneous equation. You can use the method of variation of parameters for this. Assume the particular solution is of the form:
y_p(x) = u(x)(A + Bx)e^(-x)
Now, find y_p'' and y_p':
y_p'' = u''(A + Bx)e^(-x) + 2u'(A + Bx)e^(-x) + u(A + Bx)e^(-x)
y_p' = u'(A + Bx)e^(-x) + u(A + Bx)e^(-x)
Now, substitute y_p, y_p', and y_p'' into the original equation:
[u''(A + Bx)e^(-x) + 2u'(A + Bx)e^(-x) + u(A + Bx)e^(-x)] + 2[u'(A + Bx)e^(-x) + u(A + Bx)e^(-x)] + [u(A + Bx)e^(-x)] = e^(-x)ln(x)
Now, simplify and collect terms with the same coefficients of u, u', and u'':
u''(A + Bx)e^(-x) + 4u'(A + Bx)e^(-x) + 3u(A + Bx)e^(-x) = e^(-x)ln(x)
Now, you need to choose u(x) so that the term with u'' simplifies to e^(-x)ln(x). Let's choose:
u''(A + Bx)e^(-x) = e^(-x)ln(x)
So, we have:
u'' = ln(x)
Integrate u'' to find u':
u' = ∫ln(x) dx
Integrating ln(x), we get:
u' = xln(x) - ∫x(1/x) dx
u' = xln(x) - ∫dx
u' = xln(x) - x + C
Now, integrate u' to find u:
u = ∫(xln(x) - x + C) dx
u = ∫xln(x) dx - ∫x dx + ∫C dx
Integrating xln(x) and x, we get:
u = (1/2)x^2ln(x) - (1/2)∫x^2 dx + Cx
Now, you have u(x), and you can find y_p:
y_p(x) = u(x)(A + Bx)e^(-x)
y_p(x) = [(1/2)x^2ln(x) - (1/2)∫x^2 dx + Cx](A + Bx)e^(-x)
Now, you can simplify and find the constants A, B, and C to match initial conditions or boundary conditions if provided. This will give you the complete solution to the differential equation.
PLEASE MARK ME BRAINLIEST