What is the solution for -ye^{-x}+xe^{-x}y^'=0 ?

The solution for -ye^{-x}+xe^{-x}y^'=0 is y=c_{1}x

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-ye^{-x}+xe^{-x}y^'=0

- y e^{- x} + x e^{- x} y^{^{'}} = 0

y = c_{1} x

Solution steps

- y e^{- x} + x e^{- x} y^{'} (x) = 0

Solve separable ODE:

y = c_{1} x

y = c_{1} x

y^{^{'}} = \frac{x}{sin ( y )}

y^{^{'}} = - \frac{x 1 + y ^{2}}{y 1 + x ^{2}}

\frac{d v}{d t} = 5 - 2 \cdot v

\frac{d y}{d x} x = y^{2} - y

y^{^{'}} = - \frac{2 t ( y ^{2} + 1 )}{y}, y (0) = 1

What is the solution for -ye^{-x}+xe^{-x}y^'=0 ?
The solution for -ye^{-x}+xe^{-x}y^'=0 is y=c_{1}x