What is the solution for (e^{2x}+4)y^'=y ?

The solution for (e^{2x}+4)y^'=y is y=(e^{(x+c_{1})/4})/(\sqrt[8]{e^{2x)+4}}

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(e^{2x}+4)y^'=y

(e^{2 x} + 4) y^{^{'}} = y

Solution steps

(e^{2 x} + 4) y^{'} (x) = y

Solve separable ODE:

\int (2 x + 1)^{5} d x

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 x^{2} - 5

\frac{d y}{d x} = \frac{y ^{2}}{e ^{x}}

y = 1 4^{- \frac{8}{x}}

\frac{d}{d x} (20 + x)

What is the solution for (e^{2x}+4)y^'=y ?
The solution for (e^{2x}+4)y^'=y is y=(e^{(x+c_{1})/4})/(\sqrt[8]{e^{2x)+4}}