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Find the general solution of the equation y′′ – k2y = R(x)

by RoRi

For nonzero constant k, prove that

    \[ y_1 = \frac{1}{k} \int_0^x R(t) \sinh k(x-t) \, dt \]

is a particular solution of the equation

    \[ y'' - k^2 y = R(x). \]

Find the general solution of the equation

    \[ y'' - 9y = e^{3x}. \]

Proof. The general solution of the homogeneous equation y'' - k^2 y = 0 is y = c_1 e^{kx} + c_2 e^{-kx}. Then, particular solutions of the homogeneous equation are v_1(x) = e^{kx} and v_2 (x) = e^{-kx} (taking c_1 = 1, \ c_2 =0 and c_1 = 0, \ c_2 = 1, respectively). The Wronskian of v_1 and v_2 is then

    \[ W(x) = v_1 v_2' - v_1' v_2 = e^{kx}(-k e^{-kx}) - ke^{kx}e^{-kx} = -2k. \]

So, we have the functions (of Theorem 8.9) t_1 and t_2 given by

    \begin{align*} t_1 (x) &= -\int v_2 \frac{R(x)}{W(x)} \, dx \\ &= \frac{1}{2k} \int R(x) e^{-kx} \, dx \\ t_2 (x) &= \int v_1 \frac{R(x)}{W(x)} \, dx \\ &= -\frac{1}{2k} \int Re^{kx} \, dx \\ \end{align*}

Therefore we have a particular solution y_1 of the non-homogeneous equation given by

    \begin{align*} y_1 &= t_1 (x) v_1(x) + t_2(x) v_2(x) \\ &= \frac{e^{2k}}{2k} \int_0^x Re^{-kt} \, dt + \frac{-1}{2ke^{kx}} \int_0^x Re^{kt} \, dt \\ &= \frac{1}{k} \int_0^x R(x) \cdot \left( \frac{e^{k(x-t)} - e^{-k(x-t)}}{2} \right) \, dt \\ &= \frac{1}{k} \int_0^x R(t) \sinh (k(x-t)) \, dt. \qquad \blacksquare \end{align*}

Using this theorem, the general solution of

    \[ y'' - 9y = e^{3x} \]

is

    \begin{align*} y &= c_1 e^{3x} + c_2 e^{-3x} + \frac{1}{3} \int_0^x e^{3t} \sinh(3(x-t)) \, dt \\ &= c_1 e^{3x} + c_2 e^{-3x} + \frac{1}{3} \int_0^x e^{3t} \left( \frac{e^{3x-3t} - e^{3t-3x}}{2} \right) \, dt \\ &= c_1 e^{3x} + c_2 e^{-3x} + \frac{1}{6} \left( \int_0^x e^{3x} \, dt - \int_0^x e^{6t - 3x} \, dt \right) \\ &= c_1 e^{3x} + c_2 e^{-3x} + \frac{xe^{3x}}{6} - \frac{e^{3x}}{36} + \frac{e^{-3x}}{36} \\ &= c_1 e^{3x} + c_2 e^{-3x} + \frac{1}{6} xe^{3x} \end{align*}

(where we changed the values of the constants c_1 and c_2 in the last step to absorb the extra e^{3x} and e^{-3x} terms).

One comment

  1. Mohammad Azad says:

    At first I thought “this is nonsense! the integral clearly has a zero derivative and so it cannot be a solution” but then after some pondering I noticed that the integral is not an indefinite integral since the integrand is not fixed (it contains an x).

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