What is the solution for (dr)/(dθ)=(r^2+4rθ)/(2θ^2) ?

The solution for (dr)/(dθ)=(r^2+4rθ)/(2θ^2) is r=(2θ^2)/(-θ+c_{1)}

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(dr)/(dθ)=(r^2+4rθ)/(2θ^2)

\frac{d r}{d θ} = \frac{r ^{2} + 4 r θ}{2 θ ^{2}}

r = \frac{2 θ ^{2}}{- θ + c _{1}}

Solution steps

\frac{d r}{d θ} = \frac{r ^{2} + 4 r θ}{2 θ ^{2}}

Solve Bernoulli ODE:

r = \frac{2 θ ^{2}}{- θ + c _{1}}

r = \frac{2 θ ^{2}}{- θ + c _{1}}

d er i v a t i v e sin (3 x^{2} + 2)

\frac{d}{d z} (\frac{1}{z ^{2}})

d er i v a t i v e ln (1 + e^{- a x})

\int (x + 1) (6 x - 5) d x

\int \frac{( x ^{2} + x + 1 )}{x} d x

What is the solution for (dr)/(dθ)=(r^2+4rθ)/(2θ^2) ?
The solution for (dr)/(dθ)=(r^2+4rθ)/(2θ^2) is r=(2θ^2)/(-θ+c_{1)}