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Find all solutions of y′′ – 4y′ – y = 0 for given initial values

Find the particular solution of the differential equation

    \[ y'' - 4y' - y = 0 \]

satisfying the initial condition y = 2 and y' = -1 when x = 1.


This is a second-order linear differential equation of the form

    \[ y'' + ay' + by = 0 \qquad \text{with} \qquad a= -4, \ \ b = -1. \]

These values of a and b give us d = a^2 - 4b = 20. So, d> 0 and k = \frac{1}{2} \sqrt{d} = \sqrt{5}. By Theorem 8.7 (pages 326-327 of Apostol) we then have

    \begin{align*}  &&y &= e^{-\frac{ax}{2}} (c_1 u_1 + c_2 u_2) & u_1 = e^{kx}, \quad u_2 = e^{-kx} \\ \implies && y&= e^{2x} \left( c_1 e^{\sqrt{5} x} + c_2 e^{-\sqrt{5} x} \right) \\ \implies && y&= c_1e^{(2+\sqrt{5})x} + c_2e^{(2-\sqrt{5})x} \\ \implies && y' &= (2+\sqrt{5}) c_1 e^{(2+\sqrt{5})x} + (2-\sqrt{5})c_2 e^{(2-\sqrt{5})x}. \end{align*}

We then use the initial conditions y(1) = 2 and y'(1) = -1 to solve for c_1 and c_2, letting a = 2 - \sqrt{5} and b = 2 + \sqrt{5},

    \begin{align*} 2 = y(1) &= c_1e^b + c_2 e^a& \implies && c_1 e^b = 2 - c_2 e^a \\ -1 = y'(1) &= bc_1 e^b + ac_2 e^a & \implies && c_2 = \frac{2b+1}{b-a} e^{-a} \\  && \implies && c_1 = \left( \frac{2a+1}{a-b} \right) e^{-b}.  \end{align*}

Then, we can simplify these expressions for c_1 and c_2,

    \begin{align*}  c_1 &= \frac{2a+1}{a-b} e^{-b} \\  &= \frac{2(2-\sqrt{5}) + 1}{2 - \sqrt{5} - 2 - \sqrt{5}} e^{-b}\\  &= \frac{5 - 2\sqrt{5}}{-2\sqrt{5}} e^{-b}\\  &= \frac{1}{2} \cdot \frac{2\sqrt{5} - 5}{\sqrt{5}} e^{-b}\\  &= \frac{1}{2} \cdot \frac{10 - 5 \sqrt{5}}{5} e^{-b}\\  & =\frac{a}{2}e^{-b}. \end{align*}

And

    \begin{align*}  c_2 &= \frac{2b+1}{b-a} e^{-a} \\  &= \frac{2(2+\sqrt{5}) + 1}{2 + sqrt{5} - 2 + \sqrt{5}} e^{-a}\\  &= \frac{1}{2} \cdot \frac{5 + 2 \sqrt{5}}{\sqrt{5}} e^{-a}\\  &= \frac{1}{2} \cdot \frac{5\sqrt{5} + 10}{5} e^{-a}\\  &= \frac{b}{2} e^{-a}. \end{align*}

Therefore, we have

    \[ y = \frac{a}{2} e^{(b-1)x} + \frac{b}{2} e^{(a-1)x}. \]

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