What is the solution for (dx)/(dt)=t^{3/2},x(3)=7 ?

The solution for (dx)/(dt)=t^{3/2},x(3)=7 is x=(2t^{5/2})/5+7-(18*3^{1/2})/5

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(dx)/(dt)=t^{3/2},x(3)=7

\frac{d x}{d t} = t^{\frac{3}{2}}, x (3) = 7

x = \frac{2 t ^{\frac{5}{2}}}{5} + 7 - \frac{18 \cdot 3 ^{\frac{1}{2}}}{5}

Solution steps

\frac{d x}{d t} = t^{\frac{3}{2}}

Substitute

\frac{d x}{d t}

with

x^{'} (t)

x^{^{'}} (t) = t^{\frac{3}{2}}

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

then

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

x = \int t^{\frac{3}{2}} d t

\int t^{\frac{3}{2}} d t = \frac{2}{5} t^{\frac{5}{2}} + c_{1}

x = \frac{2}{5} t^{\frac{5}{2}} + c_{1}

Apply initial conditions:

x = \frac{2 t ^{\frac{5}{2}}}{5} + 7 - \frac{18 \cdot 3 ^{\frac{1}{2}}}{5}

x = \frac{2 t ^{\frac{5}{2}}}{5} + 7 - \frac{18 \cdot 3 ^{\frac{1}{2}}}{5}

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

\frac{d}{d x} (6 x^{- \frac{1}{2}})

y = 2 x^{2} + \frac{3}{x}

\frac{d}{d t} (0.1 cos (1 0^{5}) t)

\frac{d y}{d x} + \frac{y}{x} = \frac{1}{x ^{2}}

What is the solution for (dx)/(dt)=t^{3/2},x(3)=7 ?
The solution for (dx)/(dt)=t^{3/2},x(3)=7 is x=(2t^{5/2})/5+7-(18*3^{1/2})/5