Find the solution of the following Differential Equation

Question

y'' - 4y' + y = t^2 - 2t + 3

a. y = c1 e^(2+√3)t + c2 e^(2-√3)t + t^2 - 25t + 6

b. y = c1 e^(2+√3)t + c2 e^(2-√3)t + t^2 + 6t - 25

c. y = c1 cos √3 t + c2 sin √3 t + t^2 - 25t + 6

d. y = c1 cos √3 t + c2 sin √3 t + t^2 + 6t - 25

Answer 1

This equation has a solution that can be found using the method of "Undetermined Coefficients".

1) find the Complementary solution by first finding the roots of the characteristic equation, r1 and r2 to be 2+/-sqrt(3) and using those roots to form yc(t)=C1e^^r1t+C2e^^r2t.

2) find the form of the Particular solution noting g(t) is a simple polynominal thus yp=At^2+Bt+C.

3) Substitute yp into the original problem to find y''=2A, y'=2At+B, y=at^2+Bt+C and set the result equal to t^2-2t+3. Setting the coefficients of like terms equal to each other to find A=1, B=6, C=25.

4) The total solution is yc+yp=y, the answer is b)