1. (20 pts) Find a general solution to y" - 2y' + y = 2et/(1 + t2).

Solution

Finding a solution takes several steps that I'll combine into two.

  1. Find and solve the characteristic equation. r2 - 2r + 1 = 0 has two roots both equal to 1. So we have two independent solutions to the homogeneous equation y" - 2y' + y = 0:

    y1 = et and y2 = tet and yc = C1et + C2tet.

  2. Use Variation of Parameters to find a particular solution to the nonhomogeneous equation y" - 2y' + y = 2et/(1 + t2).

    Let g(t) = 2et/(1 + t2). Then

    yp = u1(t)y1 + u2(t)y2,

    where u1(t) = - Integral(y2g·dt/W(y1, y2)), while u2(t) = Integral(y1g·dt/W(y1, y2)). The Wronskian W(y1, y2) = e2t. So we get

    u1(t) = - Integral(tet2etdt/(1 + t2)/e2t) = -Integral(2t·dt/(1 + t2)).

    Therefore we can take u1(t) = -ln(1 + t2). On the other hand,

    u2(t) = Integral(et2etdt/(1 + t2)/e2t) = Integral(2dt/(1 + t2)).

    Therefore we can take u2(t) = 2tan-1(t).

    We now have a general solution:

    Answer

    y(t) = et(C1 + C2t) - etln(1 + t2) + 2tettan-1(t).