the first-order differential equation y by = f(x), y = y(x), b = const, can be solved by multiplying both sides by ebx. This equation then becomes e bxy = ebxf(x) and can be solved by integrating both sides. Note that b is the root of the associated linear equation r b = 0. This approach has an analogue for second-order inhomogeneous linear equations y + by + cy = f(x), y = y(x), b, c = const. (1) (a) If r1, r2 are the two roots of the quadratic equation r 2 +br+c = 0 associated to (1), show that e (r1r2)x (er1x y) = er2x (y + by + cy). (2) By (2), equation (1) is equivalent to e (r1r2)x (er1x y) = er2x f(x), y = y(x), (3) which can be solved by integrating twice. (b) Find the general solution y = y(x) to the differential equation y + 5y + 4y = ex . Hint 1: comparing this equation with equation (1) above, what are b, c, f(x), r1, and r2 here? How does the sentence following part (a) apply in this case? Hint 2: choosing the order of the roots wisely could simplify the computation. (c) Find the general solution y = y(x) to the differential equation y + 4y = 4 cos 2x. Hint 1: see the two hints above. Hint 2: replacing cos 2x by e2ix and then taking the real part of the resulting general solution would simplify the computation. This real part is the general (real) solution to the above equation because cos 2x is the real part of e2ix and all coefficients in the equation are real. Note: If you ask someone at MLC/RTC to help you with this problem, do not just point them to part (b) or (c), but ask them to read the introduction at the beginn