Find the solution of 2y′′ + 3y′ = 0 for given initial values

Find the particular solution of the differential equation

$2y'' + 3y' = 0$

satisfying the initial condition $y = 1$ and $y' = 1$ when $x = 0$ .

First, we rewrite the equation as

$2y'' + 3y' = 0 \quad \implies \quad y'' + \frac{3}{2} y' = 0.$

Therefore, this is a second-order linear differential equation of the form

$y'' + ay' + by = 0 \qquad \text{with} \qquad a= \frac{3}{2}, \ \ b = 0.$

These values of $a$ and $b$ give us $d = a^2 - 4b = \frac{9}{4}$ . So, $d> 0$ and $k = \frac{1}{2} \sqrt{d} = \frac{3}{4}$ . By Theorem 8.7 (pages 326-327 of Apostol) we then have

$\begin{align*} &&y &= e^{-\frac{ax}{2}} (c_1 u_1 + c_2 u_2) & u_1 = e^{kx}, \quad u_2 = e^{-kx} \\ \implies && y&= e^{-\frac{3x}{4}} \left( c_1 e^{\frac{3x}{4}} + c_2 e^{-\frac{3x}{4}} \right) \\ \implies && y&= c_1 + c_2e^{-\frac{3x}{2}}. \end{align*}$

Therefore,

$y' = -\frac{3}{2} c_2 e^{-\frac{3x}{2}}.$

We then use the initial conditions $y(0) = 1$ and $y'(0) = 1$ to solve for $c_1$ and $c_2$ ,

$\begin{align*} 1 = y'(0) &= -\frac{3}{2} c_2 & \implies && c_2 = -\frac{2}{3} \\ 1 = y(0) &= c_1 + c_2 & \implies && c_1 = \frac{5}{3}. \end{align*}$

Therefore,

$y = \frac{5}{3} - \frac{2}{3} e^{-\frac{3x}{2}}.$

Stumbling Robot

Find the solution of 2y′′ + 3y′ = 0 for given initial values

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