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

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

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(2r-1)/t dr+(r-2r^2)/(t^2-1)dt=0,r(2)=4

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

r (t) = c_{1} t^{2} - 1

Solution steps

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

Solve separable ODE:

r (t) = c_{1} t^{2} - 1

r (t) = c_{1} t^{2} - 1

\frac{\partial}{\partial r} (\frac{1}{r ^{2}})

\frac{d}{d x} (\frac{128}{( 1 + 4 x ) ^{3}})

\frac{d ^{2}}{d x ^{2}} (∣ x ∣)

\frac{d}{d t} (t e^{- 3 t})

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