Solving a coupled system of first order homogeneous differential equations with variable coefficients

Question

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?

Answer 1

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 $\theta(t)$.

Substituting our expressions in $\theta$ back in $(\ast)$ gives (provided that $R \neq 0$) that $$\theta' = -a,$$ hence $$\theta = -\int a \,dt$$