What is the solution for (cos(θ))dx+(1+e-x)(sin(θ))dθ=0 ?

The solution for (cos(θ))dx+(1+e-x)(sin(θ))dθ=0 is x(θ)=-(c_{1})/(cos(θ))+1+e

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(cos(θ))dx+(1+e-x)(sin(θ))dθ=0

(cos (θ)) d x + (1 + e - x) (sin (θ)) d θ = 0

x (θ) = - \frac{c _{1}}{cos ( θ )} + 1 + e

Solution steps

(cos (θ)) d x + (1 + e - x) (sin (θ)) d θ = 0

Solve separable ODE:

x (θ) = - \frac{c _{1}}{cos ( θ )} + 1 + e

x (θ) = - \frac{c _{1}}{cos ( θ )} + 1 + e

\frac{d r}{d q} = (50 - 4 q) e^{- \frac{r}{5}}

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

x y d x + x^{2} d y = 0

- 9 y^{^{'}} + 3 y = 8

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

What is the solution for (cos(θ))dx+(1+e-x)(sin(θ))dθ=0 ?
The solution for (cos(θ))dx+(1+e-x)(sin(θ))dθ=0 is x(θ)=-(c_{1})/(cos(θ))+1+e