What is the solution for y^'=e^{5y+8x},y(0)=3 ?

The solution for y^'=e^{5y+8x},y(0)=3 is y=-(ln(-5/8 e^{8x}+1/(e^{15)}+5/8))/5

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y^'=e^{5y+8x},y(0)=3

y^{^{'}} = e^{5 y + 8 x}, y (0) = 3

y = - \frac{ln ( - \frac{5}{8} e ^{8 x} + \frac{1}{e ^{15}} + \frac{5}{8} )}{5}

Solution steps

y^{'} (x) = e^{5 y + 8 x}

Solve separable ODE:

y = - \frac{ln ( - \frac{5}{8} e ^{8 x} + \frac{1}{e ^{15}} + \frac{5}{8} )}{5}

y = - \frac{ln ( - \frac{5}{8} e ^{8 x} + \frac{1}{e ^{15}} + \frac{5}{8} )}{5}

\frac{d}{d x} (\frac{2}{e ^{x} + e ^{- x}})

\frac{d}{d x} (\frac{2 x ^{4} + 3 x ^{2} - 1}{x ^{2}})

x \to \frac{π}{2} lim (\frac{π}{2} - x)

\frac{2}{7 x ^{2}}

\int \frac{( x ^{2} + 2 x + 3 )}{( x ^{3} + 3 x ^{2} + 9 x )} d x

What is the solution for y^'=e^{5y+8x},y(0)=3 ?
The solution for y^'=e^{5y+8x},y(0)=3 is y=-(ln(-5/8 e^{8x}+1/(e^{15)}+5/8))/5