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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int 2 (2 + x^{2})^{\frac{3}{2}} x d x

(5 - 2) (58)

\int \frac{2 x}{y} d x

y = 7 e^{x} cos (x)

sin (3 x)

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