The first equation is separable:
dy/dx = x/y ⇒ y dy = x dx
(provided that y ≠ 0)
Integrating both sides yields
1/2 y² = 1/2 x² + C
Given that y(0) = -8, we find
1/2 • (-8)² = 1/2 • 0² + C ⇒ C = 32
so that the particular solution is
1/2 y² = 1/2 x² + 32
Solving for y explicitly, we have
y² = x² + 64
y = ± √(x² + 64)
but since y(0) is negative, we take the negative solution:
y = - √(x² + 64)
The second equation is also separable:
du/dt = (2t + sec²(t)) / (2u) ⇒ 2u du = (2t + sec²(t)) dt
Integrate both sides:
u² = t² + tan(t) + C
Given u(0) = -6, we have
(-6)² = 0² + tan(0) + C ⇒ C = 36
and so
u² = t² + tan(t) + 36
u = ± √(t² + tan(t) + 36)
but again we take the negative root here to agree with the initial condition, so
u = - √(t² + tan(t) + 36)
