What is the solution for y^'=xe^{5y-3x},y(0)=1 ?

The solution for y^'=xe^{5y-3x},y(0)=1 is y=-(ln((5e^{-3x}x)/3+5/9 e^{-3x}+1/(e^5)-5/9))/(5)

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y^'=xe^{5y-3x},y(0)=1

y^{^{'}} = x e^{5 y - 3 x}, y (0) = 1

y = - \frac{ln ( \frac{5 e ^{- 3 x} x}{3} + \frac{5}{9} e ^{- 3 x} + \frac{1}{e ^{5}} - \frac{5}{9} )}{5}

Solution steps

y^{'} (x) = x e^{5 y - 3 x}

Solve separable ODE:

y = - \frac{ln ( \frac{5 e ^{- 3 x} x}{3} + \frac{5}{9} e ^{- 3 x} + \frac{1}{e ^{5}} - \frac{5}{9} )}{5}

y = - \frac{ln ( \frac{5 e ^{- 3 x} x}{3} + \frac{5}{9} e ^{- 3 x} + \frac{1}{e ^{5}} - \frac{5}{9} )}{5}

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

\frac{d y}{d s} = \frac{s}{y ^{2}}

2 y^{^{'}} + 10 y = 3 u, y (0) = - 2

2 y^{^{'}} + y = 0, y (0) = e^{\frac{- x}{2}}

y^{^{'}} = \frac{x ^{2} - 6}{y ^{2}}, y (0) = 2

What is the solution for y^'=xe^{5y-3x},y(0)=1 ?
The solution for y^'=xe^{5y-3x},y(0)=1 is y=-(ln((5e^{-3x}x)/3+5/9 e^{-3x}+1/(e^5)-5/9))/(5)