kamilarasuleva kamilarasuleva
  • 23-11-2019
  • Mathematics
contestada

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

Respuesta :

freckledspots freckledspots
  • 23-11-2019

Answer:

[tex]z=-2t^{\frac{-1}{2}}+-4.75[/tex]

Step-by-step explanation:

Integrate both sides:

[tex]z=\frac{t^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+C[/tex]

Simplify:

[tex]z=\frac{t^{\frac{-1}{2}}}{\frac{-1}{2}}+C[/tex]

[tex]z=-2t^{\frac{-1}{2}}+C[/tex]

Now to find [tex]C[/tex]. We are going to use [tex]z(64)=-5[/tex].

[tex]-5=-2(64)^{\frac{-1}{2}}+C[/tex]

[tex]-5=-2(\frac{1}{8})+C[/tex]

[tex]-5=\frac{-1}{4}+C[/tex]

Add 1/4 on both sides:

[tex]-4.75=C[/tex]

So the equation is:

[tex]z=-2t^{\frac{-1}{2}}+C[/tex]

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