What is the solution for (R+1)dR-R(in(c))dc=0 ?

The solution for (R+1)dR-R(in(c))dc=0 is R(c)=\W(1/(e^{(-inc^2-c_{1))/2}})

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(R+1)dR-R(in(c))dc=0

(R + 1) d R - R (in (c)) d c = 0

R (c) = W (\frac{1}{e ^{\frac{- in c ^{2} - c _{1}}{2}}})

Solution steps

(R (c) + 1) d R (c) - R (c) (in (c)) d c = 0

Solve separable ODE:

R (c) = W (\frac{1}{e ^{\frac{- in c ^{2} - c _{1}}{2}}})

R (c) = W (\frac{1}{e ^{\frac{- in c ^{2} - c _{1}}{2}}})

y^{^{'}} = (\frac{x + x y ^{2}}{4 y})

\frac{1}{x} \frac{d y}{d x} = y + 4

\frac{d y}{d x} = \frac{19 x y}{( ln ( y ) ) ^{10}}

(1 + y + y^{2}) d x + (x^{2} - 4) d y = 0

u^{^{'}} - \frac{2}{x} H (x) = \frac{- 2}{x}

What is the solution for (R+1)dR-R(in(c))dc=0 ?
The solution for (R+1)dR-R(in(c))dc=0 is R(c)=\W(1/(e^{(-inc^2-c_{1))/2}})