What is the solution for y^'=e^{(5x+4y)} ?

The solution for y^'=e^{(5x+4y)} is y=-(ln(-4/5 e^{5x}-c_{1}))/4

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y^'=e^{(5x+4y)}

y^{^{'}} = e^{(5 x + 4 y)}

y = - \frac{ln ( - \frac{4}{5} e ^{5 x} - c _{1} )}{4}

Solution steps

y^{'} (x) = e^{(5 x + 4 y)}

Solve separable ODE:

y = - \frac{ln ( - \frac{4}{5} e ^{5 x} - c _{1} )}{4}

y = - \frac{ln ( - \frac{4}{5} e ^{5 x} - c _{1} )}{4}

What is the solution for y^'=e^{(5x+4y)} ?
The solution for y^'=e^{(5x+4y)} is y=-(ln(-4/5 e^{5x}-c_{1}))/4