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

The solution for y^'+2y=2x+e^{4x} is y=x-1/2+(e^{4x})/6+(c_{1})/(e^{2x)}

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

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

y = x - \frac{1}{2} + \frac{e ^{4 x}}{6} + \frac{c _{1}}{e ^{2 x}}

Solution steps

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

Solve linear ODE:

y = x - \frac{1}{2} + \frac{e ^{4 x}}{6} + \frac{c _{1}}{e ^{2 x}}

y = x - \frac{1}{2} + \frac{e ^{4 x}}{6} + \frac{c _{1}}{e ^{2 x}}

\int 2 x cos (9 x) d x

f (x) = \frac{3}{x - 2}, at x = 3

\int_{0}^{4} \frac{3 t}{( t - 5 ) ^{2}} d t

f (x) = \frac{x}{x ^{2} + 6}

x \to 2 lim (5 x^{3} - 9 x^{2} + 1)

What is the solution for y^'+2y=2x+e^{4x} ?
The solution for y^'+2y=2x+e^{4x} is y=x-1/2+(e^{4x})/6+(c_{1})/(e^{2x)}