Q: y'' - 2y' + y = e^t/(1+t^2)
for homogenous equation y'' - 2y +1 = 0
Characteristic equation:r^2 - 2r + 1 = 0
r1 = 1 , r2 = 1
Complementary solution:
yc(t) = c1y1(t) + c2te^t
For non-homogenous equation: y'' - 2y + y = e^t/(1+t^2)
p(t) = -2, q(t) = 1, g(t) = e^t/(1+t^2) are continous everywhere.
Now,
W[y1,y2](t) = e^(2t) != 0
Thus, y1(t) and y2(t) form a fundamental set of solutions.
Therefore,
u1(t) = -1/2ln(1+t^2)
u2(t) = arctant
yp(t) = -1/2e^tln(1+t^2) + te^t arctant
Therefore, the general solution is:
y(t) = yc(t) +yp(t)
= c1e^t +c2e^t -1/2e^tln(1+t^2) + te^t arctant
