2. Find the solution of the initial value problem y² + 4y = g(t) = { 16 0 4. y"+by+cy=gi(t); y(0) = a, on interval 0 < t < to obtain the solution, say, y(t). (7) y"+by+cy=92(t); (-) y'(0) B Then you can solve the IVP ✓(7) = × (7) on interval t > to obtain the solution, say, y2(t). (You are able to evaluate y₁ (7) and y(), the initial values at t = 4, since you have obtained y₁ (t) already.) The solution of the original IVP then is <= 4/1 y(t) = { ½(t), + > 0

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