### Second-order linear ordinary differential equation

Second-order linear ordinary differential equation of the form a(x)y'' + b(x)y' + c(x) = g(x) is *homogeneous *if g(x) = 0. We will focus on the homogeneous 2nd-order ODE where a,b,c are constants.

**Problem 1** Find the general solution of y''+2y'-3y=0.

**Solution** Let’s guess the solution is y=e^{rx} for some constant r. Then (r^2+2r-3)e^{rx} = 0 and so r^2+2r-3=0 (the characteristic equation) gives r_1 = 1, r_2 = -3. These two roots correspond to the solutions y_1 = e^x, y_2 = e^{-3x}. The general solution is thus y = c_1e^x + c_2y^{-3x}.

**Problem 2 **Find the homogeneous 2nd-order ODE whose general solution is y=c_1e^{2x} + c_2e^{-3x}.

**Solution **Since The roots of the characteristic equation are 2 and -3, the characteristic equation is (r-2)(r+3) = r^2+r-6, and so the ODE is y''+y'-6y=0.

**Problem 3 **Solve the IVP y''+8y'-9y=0, y(1) = 1, y'(1) = 0.

**Solution **The roots of the characteristic equation is r^2+8r-9 are r = 1, -9. The general solution is y=c_1e^x+c_2e^{-9x}. We also get y'=c_1e^x - 9c_2e^{-9x}. Using the initial conditions, we obtain c_1e + c_2e^{-9} = 1, c_1e-9c_2e^{-9} = 0. Solve the system of linear equations c_1 = \frac{9}{10e}, c_2 = \frac{e^9}{10}. The solution of the IVP is y = \frac{9}{10e} e^x + \frac{e^9}{10} e^{-9x}.

**Problem 4 **Solve the IVP y''-y'-2y=0, y(0) = a, y'(0) = 2 and find a so that the solution approaches 0 as x tends to infinity.

**Solution** The roots of the characteristic equation is r^2-r-2 are r = -1, 2. The general solution is y=c_1e^{-x} + c_2e^{2x}. We also get y' = -c_1e^{-x} + 2c_2e^{2x}. Using the initial conditions, we obtain c_1 + c_2 = a, -c_1 + 2c_2 = 2. Solve the system of linear equations c_1 = \frac{2a-2}{3}, c_2 = \frac{a+2}{3}. The solution approaches 0 as x tends to infinity only if c_2 = 0. Therefore a=-2.

**Problem 5 **Solve the IVP y'' + 5y' + 6y = 0, y(0) = 2, y'(0) = b, where b > 0 and determine the maximum point x_m of the solution.

**Solution** The roots of the characteristic equation is r^2 + 5r + 6 are r = -2, -3. The general solution is y = c_1e^{-2x} + c_2e^{-3x}. We also get y' = -2c_1e^{-2x} -3c_2e^{-3x}. Using the initial conditions, we obtain c_1 + c_2 = 2, -2c_1 - 3c_2 = b. Solve the system of linear equations c_1 = b+6, c_2 = -b-4. The maximum point x_m is given by y'(x)=0, that is -2c_1e^{-2x} -3c_2e^{-3x} = 0. Solve the last equation and get x_m = \ln \frac{-3c_2}{2c_1} = \ln\frac{3(b+4)}{2(b+6)}. We need to check, via the second derivative test, that y''(x_m) < 0. Notice that y''(x_m) + 5y'(x_m) + 6y(x_m) = 0 and y'(x_m) = 0. We have y''(x_m) = -6y(x_m). It suffices to show that y(x_m) > 0. Compute y(x_m) = c_1e^{-2x_m} + c_2e^{-3x_m} = e^{-3x_m}(c_1e^{x_m} + c_2) = e^{-3x_m}(c_1\frac{-3c_2}{2c_1}+c_2) = e^{-3x_m}\frac{-c_2}{2}. As c_2 = -b-4 < 0, y(x_m) > 0.

**Problem 6** Find the general solution of y''' - 6y'' + 11y' - 6y = 0.

**Solution **Let’s guess the solution y = e^{rx} for some constant r. Then (r^3 - 6r^2 + 11r - 6)e^{rx} = 0 and so r^3 - 6r^2 + 11r - 6 = 0 gives r = 1, 2, 3. These 3 roots correspond to the solutions y = e^x, e^{2x}, e^{3x}. The general solution is thus y = c_1e^x + c_2e^{2x} + c_3e^{3x}.