Determine the general solution of y′′ - 3y′ = 2e2x sin x

Determine the general solution of the second-order differential equation

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

The homogeneous equation related to this is an equation of the form $y'' + ay' + by = 0$ with $a = -3$ and $b = 0$ . This gives us $d = a^2 - 4b = 9$ and $k = \frac{1}{2} \sqrt{d} = \frac{3}{2}$ . So, the general solution of the homogeneous equation $y'' - 3y' = 0$ is given by

$\begin{align*} y &= e^{-\frac{ax}{2}} (c_1 e^{kx} + c_2 e^{-kx} ) \\ &= e^{\frac{3x}{2}} (c_1 e^{\frac{3}{2} x} + c_2 e^{-\frac{3}{2}x} ) \\ &= c_1 e^{3x} + c_2. \end{align*}$

Then we know that $v_1(x) = 1$ and $v_2(x) = e^{3x}$ are particular solutions of the homogeneous equation (taking $c_1 = 0, \ c_2 = 1$ and $c_1 = 0, \ c_2 = 1$ , respectively). The Wronskian of $v_1$ and $v_2$ is given by

$\begin{align*} W(x) &= v_1 v_2' - v_1' v_2 \\ &= 3e^{3x}. \end{align*}$

Thus, a particular solution of the non-homogeneous equation is given by (from Theorem 8.9)

$\begin{align*} y_1 &= t_1 v_1 + t_2 v_2 \\ t_1 &= -\int v_2 \frac{R}{W} \, dx \\ &= - \int e^{3x} \frac{ 2e^{2x} \sin x}{3e^{3x}} \, dx \\ &= - \frac{2}{3} \int e^{2x} \sin x \, dx \\ &= \frac{2}{15} e^{2x} (\cos x - 2 \sin x) \\ t_2 &= \int v_1 \frac{R}{W} \, dx \\ &= \int \frac{2e^{2x} \sin x}{3e^{3x}} \, dx \\ &= \frac{2}{3} \int e^{-x} \sin x \, dx \\ &= -\frac{1}{3} e^{-x} (\sin x + \cos x) \end{align*}$

This implies

$\begin{align*} y_1 &= \frac{2}{15} e^{2x} (\cos x - 2 \sin x) - \frac{1}{3} e^{2x} (\cos x + \sin x) \\ &= e^{2x} \left( \frac{2}{15} \cos x - \frac{4}{15} \sin x - \frac{1}{3} \cos x - \frac{1}{3} \sin x\right) \\ &= e^{2x} \left( -\frac{3}{15} \cos x - \frac{9}{15} \sin x \right) \\ &= -\frac{1}{5} e^{2x} (3 \sin x + \cos x). \end{align*}$

Therefore, the general solution is

$y =c_1 + c_2 e^{3x} - \frac{1}{5} e^{2x} (\cos x + 3 \sin x).$

Stumbling Robot

Determine the general solution of y′′ – 3y′ = 2e^2x sin x

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