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}

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)