What is the solution for y^'=(x-4)e^{-2y},y(4)=ln(4) ?

The solution for y^'=(x-4)e^{-2y},y(4)=ln(4) is y=(ln(x^2-8x+32))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(x-4)e^{-2y},y(4)=ln(4)

y^{^{'}} = (x - 4) e^{- 2 y}, y (4) = ln (4)

y = \frac{ln ( x ^{2} - 8 x + 32 )}{2}

Solution steps

y^{'} (x) = (x - 4) e^{- 2 y}

Solve separable ODE:

y = \frac{ln ( x ^{2} - 8 x + 32 )}{2}

y = \frac{ln ( x ^{2} - 8 x + 32 )}{2}

\frac{d}{d x} (2 x^{- \frac{14}{5}})

x \to 3 lim (\frac{6 x}{x ^{2} - 9})

\int - t^{2} d t

\int_{1}^{5} \frac{1}{3 x + 1} d x

\frac{d}{d x} (\frac{4 x ^{2} + 2 x + 2}{x})

What is the solution for y^'=(x-4)e^{-2y},y(4)=ln(4) ?
The solution for y^'=(x-4)e^{-2y},y(4)=ln(4) is y=(ln(x^2-8x+32))/2