What is the solution for y^'+2y=1+e^{3x} ?

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

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y^'+2y=1+e^{3x}

y^{^{'}} + 2 y = 1 + e^{3 x}

y = \frac{1}{2} + \frac{e ^{3 x}}{5} + \frac{c _{1}}{e ^{2 x}}

Solution steps

y^{'} (x) + 2 y = 1 + e^{3 x}

Solve linear ODE:

y = \frac{1}{2} + \frac{e ^{3 x}}{5} + \frac{c _{1}}{e ^{2 x}}

y = \frac{1}{2} + \frac{e ^{3 x}}{5} + \frac{c _{1}}{e ^{2 x}}

4 x, x, x^{3}

\frac{\partial}{\partial x} ((x^{2} + 3 y^{2})^{7})

\frac{\partial}{\partial x} (x^{4} y^{2} - x^{3} y)

\int 16 x (8 x^{2} + 2)^{2} d x

y^{^{'}} + 8 y = 4

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