Given the coupled system of first order homogeneous ordinary differential equations with variable coefficients.
$$
\frac{dx}{dt} = a(t)z(t)
$$
$$
\frac{dz}{dt} = -a(t)x(t)
$$
How is the general solution below derived?
$$
x(t) = c_1 \cos\left(\int_1^t a(\tau) d\tau \right)+c_2 \sin\left(\int_1^t a(\tau) d\tau\right)
$$
$$
z(t) = -c_1 \sin\left(\int_1^t a(\tau) d\tau \right)+c_2 \cos\left(\int_1^t a(\tau) d\tau\right)
$$
I thought it might be something like guessing the solution takes the form
$$
x(t) = c_1 \cos\left(f(t)\right)+c_2 \sin\left(f(t)\right)
$$
$$
z(t) = \frac{1}{\left(\frac{df}{dt}\right)} \frac{dx}{dt} = -c_1 \sin\left(f(t)\right)+c_2 \cos\left(f(t)\right)
$$
And $\frac{df}{dt} = a(t)$
But I imagine there is a general approach to this type of problem?
Best Answer
Hint Notice that $$\phantom{(\ast)} \qquad (x^2 + z^2)' = 2 x x' + 2 z z' = 2 x (az) + 2 (-ax) z = 0 , \qquad (\ast)$$ so $(x(t), z(t))$ traces out an arc of a circle, say of radius $R \geq 0$, centered at the origin in the $xz$-plane. Using the standard parameterization of that circle gives that $$x(t) = R \cos \theta(t), \qquad z(t) = R \sin \theta(t)$$ for some function $\theta(t)$.