Math 307, section c, spring 2006, midterm i solutions: find the general solution of the di erential equation y + 4 t y = 6t 5. (10 points) This is a rst order linear di erential equation. (t) = er 4 t dt = e4 ln(t) = t4. 1 t4 (t6 t5 + c). or: solve the initial value problem y(t) = t2 t + C t4 . (10 points) x 7x + 12x = 0 x(0) = 3, x (0) = 1. This is a second order linear homogeneous equation with constant coe cients. The characteristic equation is r2 7r + 12 = 0. The roots are real and distinct therefore the general solution is. The initial conditions are x(t) = c1e3t + c2e4t x (t) = 3c1e3t + 4c2e4t. 3 = x(0) = c1 + c2 and. 1 = x (0) = 3c1 + 4c2 so. The solution to the initial value problem is.
