What is the solution for y^'+7y=e^{6x} ?

The solution for y^'+7y=e^{6x} is y=(e^{6x})/(13)+(c_{1})/(e^{7x)}

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y^'+7y=e^{6x}

y^{^{'}} + 7 y = e^{6 x}

y = \frac{e ^{6 x}}{13} + \frac{c _{1}}{e ^{7 x}}

Solution steps

y^{'} (x) + 7 y = e^{6 x}

Solve linear ODE:

y = \frac{e ^{6 x}}{13} + \frac{c _{1}}{e ^{7 x}}

y = \frac{e ^{6 x}}{13} + \frac{c _{1}}{e ^{7 x}}

\int_{0}^{2 π} sin^{2} (x) d x

n = 1 \sum \infty n^{- \frac{1}{5}}

x \to 2 - lim (\frac{5 x}{x - 6})

f (x) = (x^{3} + 2) e^{x}

x \to - 2 lim (x^{2} - 4 x + 6)

What is the solution for y^'+7y=e^{6x} ?
The solution for y^'+7y=e^{6x} is y=(e^{6x})/(13)+(c_{1})/(e^{7x)}