What is the solution for (dy)/(dx)=3+e^{-3x+y+4} ?

The solution for (dy)/(dx)=3+e^{-3x+y+4} is y=-ln(-x-c_{1})+3x-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=3+e^{-3x+y+4}

\frac{d y}{d x} = 3 + e^{- 3 x + y + 4}

y = - ln (- x - c_{1}) + 3 x - 4

Solution steps

\frac{d y}{d x} = 3 + e^{- 3 x + y + 4}

Solve using substitution:

y = - ln (- x - c_{1}) + 3 x - 4

y = - ln (- x - c_{1}) + 3 x - 4

\int (x^{2} + 1) x - 2 d x

\int_{0}^{1} \frac{1}{t ^{2} + 1} d t

\frac{2 x + 2 y - 3}{- x - y - 1} = \frac{d y}{d x}

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = x^{3} e^{x}

\int \frac{5 x}{9 x ^{2} - 4} d x

What is the solution for (dy)/(dx)=3+e^{-3x+y+4} ?
The solution for (dy)/(dx)=3+e^{-3x+y+4} is y=-ln(-x-c_{1})+3x-4