What is the solution for (dh)/(dt)=4-16t^2,\quad h(1)=0 ?

The solution for (dh)/(dt)=4-16t^2,\quad h(1)=0 is h=4t-(16t^3)/3+4/3

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(dh)/(dt)=4-16t^2,\quad h(1)=0

\frac{d h}{d t} = 4 - 16 t^{2}, h (1) = 0

h = 4 t - \frac{16 t ^{3}}{3} + \frac{4}{3}

Solution steps

\frac{d h}{d t} = 4 - 16 t^{2}

Substitute

\frac{d h}{d t}

with

h^{'} (t)

h^{^{'}} (t) = 4 - 16 t^{2}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

h = \int 4 - 16 t^{2} d t

\int 4 - 16 t^{2} d t = 4 t - \frac{16 t ^{3}}{3} + c_{1}

h = 4 t - \frac{16 t ^{3}}{3} + c_{1}

Apply initial conditions:

h = 4 t - \frac{16 t ^{3}}{3} + \frac{4}{3}

h = 4 t - \frac{16 t ^{3}}{3} + \frac{4}{3}

x \to 1 + lim (\frac{8}{x ^{3} - 1})

f (x) = (x^{3} - 9 x)^{10}, (- 3, 0)

\frac{d}{d x} (x^{y z})

2 cos^{2} (y) d x + tan (y) d y = 0

\frac{d}{d x} (\frac{3}{2 x})

What is the solution for (dh)/(dt)=4-16t^2,\quad h(1)=0 ?
The solution for (dh)/(dt)=4-16t^2,\quad h(1)=0 is h=4t-(16t^3)/3+4/3