# Second order homogenous coupled differential equations

I am trying to solve the following coupled differential equations, \begin{align} f''(x)+c_1(f(x)+g(x))&=0\\ g''(x)+c_2(f(x)+g(x))&=0 \end{align}

where $$c_1$$ and $$c_2$$ are positive constants. The problem is I don't know any boundary conditions. The only condition that I can think of is $$f(\infty)=g(\infty)=0$$. How can I solve this?

Equation 1 $$\times c_2$$ - Equation 2 $$\times c_1$$: $$f''(x) = \frac{c_1}{c_2} g''(x)$$ Hence, $$f(x) = \frac{c_1}{c_2} g(x) + Ax + B$$

Apply some boundary conditions to solve for $$A,B$$.

This decouples both the differential equations. Now second equation becomes: $$g''(x) = -c_2(\frac{c_1}{c_2} g(x) + Ax + B + g(x))$$ $$g''(x) = -(c_1+c_2) g(x) - A'x - B'$$

One possible solution is $$g(x) = e^{i \sqrt{c_1+c_2} x} + C_1 x^3 + C_2 x^2.$$ Substituting above $$g(x)$$ in differential equation, we get: $$6C_1 = -(c_1+c_2) C_1 -A'$$ $$2C_2 = -(c_1+c_2) C_2 -B'$$

Use $$A',B'$$ from boundary conditions and solve above equation to get the solution.

• @Gonçalo edited my answer to give a solution. May 19 at 4:01

Summing the two equations, one obtains $$(f+g)''+(c_1+c_2)(f+g)=0, \tag{1}$$ the general solution of which is $$f(x)+g(x)=A\cos(kx)+B\sin(kx),\qquad k=\sqrt{c_1+c_2}. \tag{2}$$ Also, notice that $$c_2f''-c_1g''=0, \tag{3}$$ which, integrated twice, yields $$c_2f(x)-c_1g(x)=Cx+D. \tag{4}$$ Solving the system of linear equations $$(2)$$ and $$(4)$$ one finally obtains \begin{align} f(x)&=\frac{1}{c_1+c_2}[c_1(A\cos(kx)+B\sin(kx))+Cx+D], \\ g(x)&=\frac{1}{c_1+c_2}[c_2(A\cos(kx)+B\sin(kx))-Cx-D]. \tag{5} \end{align}