What is the solution for y^'-4y=-3e^{3t},y(0)=9 ?

The solution for y^'-4y=-3e^{3t},y(0)=9 is y=3e^{3t}+6e^{4t}

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y^'-4y=-3e^{3t},y(0)=9

y^{^{'}} - 4 y = - 3 e^{3 t}, y (0) = 9

y = 3 e^{3 t} + 6 e^{4 t}

Solution steps

y^{'} (t) - 4 y = - 3 e^{3 t}

Solve linear ODE:

y = 3 e^{3 t} + 6 e^{4 t}

y = 3 e^{3 t} + 6 e^{4 t}

\int - \frac{1}{2} e^{- 3 x} d x

x \to 0 lim (\frac{- cos ( x )}{x})

\int - x + 6 d x

\int (\frac{e ^{3 x}}{3} + 3 x \cdot 2^{x^{2}} - x^{4}) d x

\frac{d}{d x} (cos (6 x + 11))

What is the solution for y^'-4y=-3e^{3t},y(0)=9 ?
The solution for y^'-4y=-3e^{3t},y(0)=9 is y=3e^{3t}+6e^{4t}