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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dr)/(dθ)=(r^2+4rθ)/(2θ^2)

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

r (θ) = \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 (θ) = \frac{2 θ ^{2}}{- θ + c _{1}}

r (θ) = \frac{2 θ ^{2}}{- θ + c _{1}}

f (x) = x^{4} + 3 x - 2, (1, 2)

x \to 0 - lim (2^{\frac{1}{x}})

sin (3 x^{2} + 2)

\int \frac{2 x - 1}{x ^{2} + 1} d x

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

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