Question

Find the solution of the differential equation that satisfies the given initial condition.

Answer 1

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)