Solve the following differential equation.

{eq}y'' + 4y = \sin 2x {/eq}

Question:

Ordinary Differential Equation:

The general solution of a second-degree ordinary differential equation will have two arbitrary constants of integration. The complete solution will have two parts added together, a complementary function and a particular integral. There are rules and formulas that can be applied to find these two solutions. The following formulas will be applied to evaluate the general solution of an ODE.

{eq}\begin{align} &\hspace{1cm} D=\frac{d}{dx} \\[0.3cm] &\hspace{1cm} D^2=\frac{d^2}{dx^2} \\[0.3cm] &\hspace{1cm} \frac{1 }{\phi(D) } e^{ax} = \frac{e^{ax} }{\phi(a) }, \phi(a) \neq 0\\[0.3cm] &\hspace{1cm} \frac{1 }{(D-a)^r }e^{ax} = \frac{ x^r }{r! } e^{ax} \\[0.3cm] &\hspace{1cm} \frac{1 }{f(D^2) } \sin (ax+b) = \frac{1 }{f(-a^2) } \sin (ax+b) \; , f(-a^2) \neq 0\\[0.3cm] &\hspace{1cm} \frac{1 }{f(D^2) } \cos (ax+b) = \frac{1 }{f(-a^2) } \cos (ax+b) \; , f(-a^2) \neq 0\\[0.3cm] &\hspace{1cm} \frac{1 }{f(D) } x^m = \left[f(D) \right]^{-1} x^m \\[0.3cm] \end{align} {/eq}

Answer and Explanation: 1