Use method of variation of parameters to find the particular solution of (d^2)y/
ID: 3078491 • Letter: U
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Use method of variation of parameters to find the particular solution of (d^2)y/(dx^2)+y=tan(t)Explanation / Answer
example Use the method of variation of parameters to find a particular solution to the differential equation: y'' + 9y' + 20y= 2e^x having the form yparticular= u1y1 + u2y2 One finds y1 and y2 from solving the homogeneous DE y'' + 9y' + 20y = 0. This has characteristic equation r^2 + 9r + 20 = 0, which has roots r = -4, -5. So, the homogeneous solution is y = Ae^(-4x) + Be^(-5x). ==> Set y1 = e^(-4x) and y2 = e^(-5x). Next, we set y_p = u * e^(-4x) + v * e^(-5x). [I'll use u and v instead of u1 and u2 for simplicity.] So, y'_p = e^(-4x) u' - 4e^(-4x) u + e^(-5x) v ' - 5e^(-5x) v. **Set e^(-4x) u' + e^(-5x) v ' = 0. ==> y'_p = - 4e^(-4x) u - 5e^(-5x) v. Differentiating again: y''_p = -4e^(-4x) u' + 16e^(-4x) u - 5e^(-5x) v ' + 25 e^(-5x) v. Substituting back into the original DE, we get [-4e^(-4x) u' + 16e^(-4x) u - 5e^(-5x) v ' + 25 e^(-5x) v] + 9 [-4e^(-4x) u - 5e^(-5x) v] + 20 [e^(-4x) u + e^(-5x) v] = 2e^x This simplifies to - 4e^(-4x) u' - 5e^(-5x) v ' = 2e^x ==> 4e^(-4x) u' + 5e^(-5x) v ' = -2e^x Now we use e^(-4x) u' + e^(-5x) v ' = 0 4e^(-4x) u' + 5e^(-5x) v ' = -2e^x to find u' and v '. We find that -e^(-4x) u' = -2e^x ==> u' = 2e^(5x) e^(-5x) v ' = -2e^x ==> v ' = -2e^(6x). Integrating yields (ignoring constants) u = (2/5) e^(5x) and v = (-1/3) e^(6x). Thus, we have yp = [(2/5) e^(5x) ] * e^(-4x) + [(-1/3) e^(6x)] * e^(-5x) = (1/15) e^x. (It is easy to verify that this indeed satsifies the DE.) So, the general solution is given by y = Ae^(-4x) + Be^(-5x) + e^x / 15. ------------------------------------ We now use the initial conditions to solve for A and B. y(0) = 0 ==> 0 = A + B + 1/15. y' = -4Ae^(-4x) - 5Be^(-5x) + e^x / 15. y'(0) = 0 ==> 0 = -4A - 5B + 1/15. So, we solve A + B = -1/15 and 4A + 5B = 1/15. ==> A = -6/15 and B = 5/15. So, the final answer is y = [-6e^(-4x) + 5e^(-5x) + e^x] / 15. I hope this helps!
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