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

The solution for (dr(θ))/(r(θ))=(r(θ)dθ)/(θ^2) is θ(r)=c_{1}r,θ(r)=(c_{1})/r

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(dr(θ))/(r(θ))=(r(θ)dθ)/(θ^2)

\frac{d r ( θ )}{r ( θ )} = \frac{r ( θ ) d θ}{θ ^{2}}

θ (r) = c_{1} r, θ (r) = \frac{c _{1}}{r}

Solution steps

\frac{d r ( θ )}{r ( θ )} = \frac{r ( θ ) d θ}{θ ^{2}}

Let

θ

be the dependent variable. Divide by

d r

\frac{\frac{d θ}{d r}}{r \frac{d θ}{d r}} = \frac{r \frac{d θ}{d r} \cdot \frac{d θ}{d r}}{θ ^{2}}

Substitute

\frac{d θ}{d r}

with

θ^{'} (r)

\frac{θ ^{^{'}} ( r )}{r θ ^{^{'}} ( r )} = \frac{r θ ^{^{'}} ( r ) θ ^{^{'}} ( r )}{θ ( r ) ^{2}}

Isolate

θ^{'} (r) : θ^{'} (r) = \frac{θ ( r )}{r}, θ^{'} (r) = - \frac{θ ( r )}{r}

θ^{'} (r) = \frac{θ ( r )}{r} : θ (r) = c_{1} r

θ^{'} (r) = - \frac{θ ( r )}{r} : θ (r) = \frac{c _{1}}{r}

θ (r) = c_{1} r, θ (r) = \frac{c _{1}}{r}

(1 - y^{2}) + x (\frac{d y}{d x}) = 0

y^{^{'}} = 0.2 x y

y^{^{'}} = \frac{y cos ( x )}{1 - 2 y}

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

x^{2} y^{^{'}} = \frac{1}{y} x^{5}, y (0) = 2

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