Diff. Eq. - finding particular solution using the Method of Undetermined Coefficients

 Given the equation:

y'' - 4y' + 3y = e^(3x) * (6+8x)

(a) Find a particular solution of the equation by using the Method of Undetermined Coefficients.

(b) Find the general solution of the equation.

(Start by finding the auxiliary equation of the given diff. equation)

y'' - 4y' + 3y = 0  →  r^2 - 4r + 3y = 0...

► r = 3, 1

► y = e^x & e^3x

Based off your general solution answers, find a particular solution that matches the f(x) to the original function: (e^(3x) * (6+8x))

* Remember that e^(3x) was one of the roots, and we should avoid creating two roots that are exactly the same.

Thus, multiply by 'x'.

yp = xe^3x * (A+Bx) or, e^3x * (Ax+Bx^2)

(Let's start differentiating to get y'p and y''p)

* differentiate by using product rule.

y'p = [e^3x * (Ax+Bx^2)'] + [e^3x' * (Ax+Bx^2)]

►  = e^3x * (A+3Ax+2Bx+3Bx^2)

y''p = [e^3x * (A+3Ax+2Bx+3Bx^2)'] + [e^3x' * (A+3Ax+2Bx+3Bx^2)]

►  = e^3x * (6A+2B+9Ax+12Bx+9Bx^2)

Now, plug in your values into the given differential equation you were first handed.

y'' - 4y' + 3y = e^(3x) * (6+8x)

► (e^3x * (6A+2B+9Ax+12Bx+9Bx^2)) - 4[e^3x * (A+3Ax+2Bx+3Bx^2)] + 3[e^3x * (A+3Ax+2Bx+3Bx^2)] = e^(3x) * (6+8x)

Cancel / divide term e^(3x)

► ((6A+2B+9Ax+12Bx+9Bx^2)) - 4[(A+3Ax+2Bx+3Bx^2)] + 3[(A+3Ax+2Bx+3Bx^2)] = (6+8x)

Now distribute and simplify (this is a very tedious process!)

2A + 2B + 4Bx = 6+8x

►  (2A + 2B) = 6

►  4B = 8

→ B = 2, A = 1

Blue in A & B values for the yp and make your general solution as well.

Answers are....

(a) yp = xe^(3x) * (1+2x)

(b) y = C1e^x + C2e^3x + e^3x * (x+2x^2)