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Find all solutions of y′′ + 4y′ + 5y = 0 for given initial values

Find the particular solution of the differential equation

    \[ y'' + 4y' + 5y = 0 \]

satisfying the initial condition y = 2 and y' = y'' when x = 0.


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

    \[ y'' + ay' + by = 0 \qquad \text{with} \qquad a= 4, \ \ b = 5. \]

These values of a and b give us d = a^2 - 4b = -4. So, d < 0 and k = \frac{1}{2} \sqrt{-d} = 1. 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 = \cos (kx), \quad u_2 = \sin (kx) \\ \implies && y&= e^{-2x} (c_1 \cos x + c_2 \sin x) \end{align*}

Then computing the first two derivatives, we have

    \begin{align*}  y' &= -2e^{-2x} (c_1 \cos x + c_2 \sin x) + e^{-2x} (c_2 \cos x - c_1 \sin x) \\  &= e^{-2x} (-2c_1 \cos x - 2c_2 \sin x - c_1 \sin x + c_2 \cos x) \\  y'' &= e^{-2x} (3c_1 \cos x + 3c_2 \sin x + 4c_1 \sin x - 4c_2 \cos x). \end{align*}

We then use the initial condition y'(0) = y''(0),

    \begin{align*}  y'(0) = y''(0) && \implies && -2c_1 + c_2 &= 3c_1 - 4c_2 \\  && \implies && c_1 = c_2. \end{align*}

Then, we use the other initial condition, y(0) = 2,

    \[ y(0) = 2 \quad \implies \quad c_1 = 2 \quad \implies c_2 = 2. \]

Therefore,

    \[ y = e^{-2x}(2\cos x + 2 \sin x). \]

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