What is the solution for (dy)/(dt)=e^{2(t)}-3y,y(0)=1 ?

The solution for (dy)/(dt)=e^{2(t)}-3y,y(0)=1 is y=(e^{2t})/5+4/(5e^{3t)}

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(dy)/(dt)=e^{2(t)}-3y,y(0)=1

\frac{d y}{d t} = e^{2 (t)} - 3 y, y (0) = 1

y = \frac{e ^{2 t}}{5} + \frac{4}{5 e ^{3 t}}

Solution steps

\frac{d y}{d t} = e^{2 (t)} - 3 y

Solve linear ODE:

y = \frac{e ^{2 t}}{5} + \frac{4}{5 e ^{3 t}}

y = \frac{e ^{2 t}}{5} + \frac{4}{5 e ^{3 t}}

\int_{- 1}^{0} 3 u + 4 d u

\int x^{9} x^{5} + 5 d x

y = (3 + 5 x)^{6} (3 x^{7} + 2)^{3}

\frac{d}{d x} (7 x sin (π x))

\frac{d}{d x} ((7 x^{2} - 14 x + 14) e^{x})

What is the solution for (dy)/(dt)=e^{2(t)}-3y,y(0)=1 ?
The solution for (dy)/(dt)=e^{2(t)}-3y,y(0)=1 is y=(e^{2t})/5+4/(5e^{3t)}