Question

Solve the initial value problem dz/dt = t^-3/2, z(64)=-5

Answer 1

`z=-2t^{(-1)/(2)}+-4.75`

Explanation:

Integrate both sides:

`z=\frac{t^{(-3)/(2)+1}}{(-3)/(2)+1}+C`

Simplify:

`z=\frac{t^{(-1)/(2)}}{(-1)/(2)}+C`

`z=-2t^{(-1)/(2)}+C`

Now to find
`C`. We are going to use
`z(64)=-5`.

`-5=-2(64)^{(-1)/(2)}+C`

`-5=-2((1)/(8))+C`

`-5=(-1)/(4)+C`

Add 1/4 on both sides:

`-4.75=C`

So the equation is:

`z=-2t^{(-1)/(2)}+C`