What is the solution for y^'=e^{(2+0)x+(1+5)y} ?

The solution for y^'=e^{(2+0)x+(1+5)y} is y=-(ln(-3e^{2x}-c_{1}))/6

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y^'=e^{(2+0)x+(1+5)y}

y^{^{'}} = e^{(2 + 0) x + (1 + 5) y}

y = - \frac{ln ( - 3 e ^{2 x} - c _{1} )}{6}

Solution steps

y^{'} (x) = e^{(2 + 0) x + (1 + 5) y}

Solve separable ODE:

y = - \frac{ln ( - 3 e ^{2 x} - c _{1} )}{6}

y = - \frac{ln ( - 3 e ^{2 x} - c _{1} )}{6}

d y = e^{2 y} + 1 d x

\frac{d y}{d x} y = e^{- x^{2}} cos (x)

\frac{d y}{d x} = e^{x} sin (2 y)

\frac{d y}{d x} = \frac{x}{y}, y (3) = 5

cos (x) (1 + y^{2}) d x + d y = 0

What is the solution for y^'=e^{(2+0)x+(1+5)y} ?
The solution for y^'=e^{(2+0)x+(1+5)y} is y=-(ln(-3e^{2x}-c_{1}))/6