What is the answer to solvefor θ,r=φe^{-θ}+ωθe^{-θ} ?

The answer to solvefor θ,r=φe^{-θ}+ωθe^{-θ} is θ=-(ω\W(-r/(ωe^{φ/ω)})+φ)/ω

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solvefor θ,r=φe^{-θ}+ωθe^{-θ}

solve for

θ, r = φ e^{- θ} + ω θ e^{- θ}

θ = - \frac{ω W ( - \frac{r}{ω e ^{\frac{φ}{ω}}} ) + φ}{ω}

Solution steps

r = φ e^{- θ} + ω θ e^{- θ}

Prepare

r = φ e^{- θ} + ω θ e^{- θ}

for Lambert form:

r = e^{- θ} (φ + ω θ)

Rewrite the equation with

- θ - \frac{φ}{ω} = u

and

θ = - \frac{ω u + φ}{ω}

r = e^{- (- \frac{ω u + φ}{ω})} (φ + ω (- \frac{ω u + φ}{ω}))

Simplify

r = - ω e^{\frac{ω u + φ}{ω}} u

Rewrite

r = - ω e^{\frac{ω u + φ}{ω}} u

in Lambert form:

e^{u} u = - \frac{r}{ω e ^{\frac{φ}{ω}}}

Solve

e^{u} u = - \frac{r}{ω e ^{\frac{φ}{ω}}} : u = W (- \frac{r}{ω e ^{\frac{φ}{ω}}})

u = W (- \frac{r}{ω e ^{\frac{φ}{ω}}})

Substitute back

u = - θ - \frac{φ}{ω},

solve for

θ

Solve

- θ - \frac{φ}{ω} = W (- \frac{r}{ω e ^{\frac{φ}{ω}}}) : θ = - \frac{ω W ( - \frac{r}{ω e ^{\frac{φ}{ω}}} ) + φ}{ω}

θ = - \frac{ω W ( - \frac{r}{ω e ^{\frac{φ}{ω}}} ) + φ}{ω}

3^{x} = 81

6^{x} = 38

6^{x} = 15

5^{2 x + 1} = 3125

5^{m} - 4 (3^{m}) = 0

What is the answer to solvefor θ,r=φe^{-θ}+ωθe^{-θ} ?
The answer to solvefor θ,r=φe^{-θ}+ωθe^{-θ} is θ=-(ω\W(-r/(ωe^{φ/ω)})+φ)/ω