What is the solution for e^{x+3y}dy=xdx ?

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

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e^{x+3y}dy=xdx

e^{x + 3 y} d y = x d x

y = \frac{ln ( - 3 x e ^{- x} - 3 e ^{- x} + c _{1} )}{3}

Solution steps

e^{x + 3 y} d y = x d x

Solve separable ODE:

y = \frac{ln ( - 3 x e ^{- x} - 3 e ^{- x} + c _{1} )}{3}

y = \frac{ln ( - 3 x e ^{- x} - 3 e ^{- x} + c _{1} )}{3}

\int \frac{8}{x ^{2} + 1} d x

\int \frac{x ^{2}}{x ^{2} + 2 x + 2} d x

\frac{\partial}{\partial x} (x + 2 y - 1)

- x^{2} - 3 x, [- 6, - 1]

\frac{\partial}{\partial y} (2 y)

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