What is the solution for (dr)/(dt)=6(t+1)^{1/2},r(0)=0 ?

The solution for (dr)/(dt)=6(t+1)^{1/2},r(0)=0 is r=4(t+1)^{3/2}-4

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(dr)/(dt)=6(t+1)^{1/2},r(0)=0

\frac{d r}{d t} = 6 (t + 1)^{\frac{1}{2}}, r (0) = 0

r = 4 (t + 1)^{\frac{3}{2}} - 4

Solution steps

\frac{d r}{d t} = 6 (t + 1)^{\frac{1}{2}}

Substitute

\frac{d r}{d t}

with

r^{'} (t)

r^{^{'}} (t) = 6 (t + 1)^{\frac{1}{2}}

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

then

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

r = \int 6 (t + 1)^{\frac{1}{2}} d t

\int 6 (t + 1)^{\frac{1}{2}} d t = 4 (t + 1)^{\frac{3}{2}} + c_{1}

r = 4 (t + 1)^{\frac{3}{2}} + c_{1}

Apply initial conditions:

r = 4 (t + 1)^{\frac{3}{2}} - 4

r = 4 (t + 1)^{\frac{3}{2}} - 4

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What is the solution for (dr)/(dt)=6(t+1)^{1/2},r(0)=0 ?
The solution for (dr)/(dt)=6(t+1)^{1/2},r(0)=0 is r=4(t+1)^{3/2}-4