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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

y \frac{d y}{d x} = 5 x (y^{2} + 1)^{6}, y (8) = 2

x^{^{'}} = t \cdot (1 + x^{2})

\frac{d y}{d x} = \frac{4 y ^{\frac{3}{4}}}{x ^{2} + 1}

\frac{d y}{d x} = y^{2} - 1

x d x - 4 y d y = 0

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