What is the solution for 7y^'+1y=8e^3x,y(0)=9 ?

The solution for 7y^'+1y=8e^3x,y(0)=9 is y=8e^3(x-7)+(9+56e^3)/(e^{x/7)}

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7y^'+1y=8e^3x,y(0)=9

7 y^{^{'}} + 1 y = 8 e^{3} x, y (0) = 9

y = 8 e^{3} (x - 7) + \frac{9 + 56 e ^{3}}{e ^{\frac{x}{7}}}

Solution steps

7 y^{'} (x) + 1 y = 8 e^{3} x

Solve linear ODE:

y = 8 e^{3} (x - 7) + \frac{9 + 56 e ^{3}}{e ^{\frac{x}{7}}}

y = 8 e^{3} (x - 7) + \frac{9 + 56 e ^{3}}{e ^{\frac{x}{7}}}

f (x) = sin (3 x) cos (5 x)

\frac{1}{1 + 4 x}

\frac{d}{d t} (\frac{t + π}{t} cos (t))

x \to 0 lim (\frac{1}{x ^{2}} - \frac{1}{x})

\int 2 t + 1 d t

What is the solution for 7y^'+1y=8e^3x,y(0)=9 ?
The solution for 7y^'+1y=8e^3x,y(0)=9 is y=8e^3(x-7)+(9+56e^3)/(e^{x/7)}