What is the solution for y^'=e^{5y+8x},y(0)=6 ?

The solution for y^'=e^{5y+8x},y(0)=6 is y=-(ln(-5/8 e^{8x}+1/(e^{30)}+5/8))/5

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y^'=e^{5y+8x},y(0)=6

y^{^{'}} = e^{5 y + 8 x}, y (0) = 6

y = - \frac{ln ( - \frac{5}{8} e ^{8 x} + \frac{1}{e ^{30}} + \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 ^{30}} + \frac{5}{8} )}{5}

y = - \frac{ln ( - \frac{5}{8} e ^{8 x} + \frac{1}{e ^{30}} + \frac{5}{8} )}{5}

y^{^{'}} = 0.3 y, y (0) = 2.1

y^{^{'}} sec (x) = 6 y^{2}, y (\frac{17 π}{2}) = \frac{1}{7}

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

\frac{d y}{d x} + (sin (x)) y = sin (x)

\frac{d v}{d t} = 70 - \frac{8 v}{100}

What is the solution for y^'=e^{5y+8x},y(0)=6 ?
The solution for y^'=e^{5y+8x},y(0)=6 is y=-(ln(-5/8 e^{8x}+1/(e^{30)}+5/8))/5