Two coupled second order differential equations

Question

I have two coupled equations in the form:

$$f''(x) + g'(x) + f(x) = 0$$ $$g''(x) + f'(x) + g(x) = 0$$

Looking at the form, i can guess a relation of the form $g(x) = \lambda f(x)$. where $\lambda$ is some constant. I can find the constant by replacing $g(x)$ in the above equations and comparing the coefficients of every derivative. Finally i'm left with a single equation which is easily solvable.

The question is: is this procedure legal? and is the solution that i get is the most general one?

Thanks

Answer 1

Hint

Make a substitute $$ f' = p \\ g' = q $$ then you get a system \begin{align} f' &= p \\ g' &= q \\ p' &= -f - q \\ q' &= -g - p \end{align} or in matrix form $$ \left[ \begin{array}{c} f \\ g \\ p \\ q \end{array}\right ]' = \left [ \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & -1 \\ 0 & -1 & -1 & 0 \end{array}\right ] \cdot \left[ \begin{array}{c} f \\ g \\ p \\ q \end{array}\right ] $$ which is in form of $$ \mathbf y' = \mathbf A \cdot \mathbf y $$ Can you take it from here?

Answer 2

Setting $$ u=f+g,\quad v=f-g, $$ we have $$\tag{1} \left\{ \begin{array}{lcl} u''+u'+u&=&0\\ v''-v'+v&=&0 \end{array}\right.. $$ The solutions of the first and second equations of (1) is given respectively by \begin{eqnarray} u(x)&=&e^{-\frac{x}{2}}\left[a_1\cos\left(\frac{\sqrt{3}}{2}x\right)+a_2\sin\left(\frac{\sqrt{3}}{2}x\right)\right],\\ v(x)&=&e^{\frac{x}{2}}\left[b_1\cos\left(\frac{\sqrt{3}}{2}x\right)+b_2\sin\left(\frac{\sqrt{3}}{2}x\right)\right], \end{eqnarray} where $a_1,a_2,b_1,b_2$ are real constants.

Since $$ f=\frac{u+v}{2},\quad g=\frac{u-v}{2}, $$ we deduce that \begin{eqnarray}\ f(x)&=&\frac12\left(a_1e^{-\frac{x}{2}}+b_1e^{\frac{x}{2}}\right)\cos\left(\frac{\sqrt{3}}{2}x\right)+\frac12\left(a_2e^{-\frac{x}{2}}+b_2e^{\frac{x}{2}}\right)\sin\left(\frac{\sqrt{3}}{2}x\right),\\ g(x)&=&\frac12\left(a_1e^{-\frac{x}{2}}-b_1e^{\frac{x}{2}}\right)\cos\left(\frac{\sqrt{3}}{2}x\right)+\frac12\left(a_2e^{-\frac{x}{2}}-b_2e^{\frac{x}{2}}\right)\sin\left(\frac{\sqrt{3}}{2}x\right). \end{eqnarray}