What is the solution for y^'+y=e^{2t},y(0)=-1 ?

The solution for y^'+y=e^{2t},y(0)=-1 is y=(e^{2t})/3-4/(3e^t)

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y^'+y=e^{2t},y(0)=-1

y^{^{'}} + y = e^{2 t}, y (0) = - 1

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

Solution steps

y^{'} (t) + y = e^{2 t}

Solve linear ODE:

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

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

\int (1 - ln (x)) d x

\frac{\partial}{\partial y} (\frac{e ^{x} + e ^{y}}{2})

y^{^{''}} + 2 y^{^{'}} + y = 1, y (1) = 1, y^{^{'}} (1) = 0

x \to - 5 + lim (\frac{( x - 5 )}{x + 5})

\frac{d}{d x} ((6 x + 12)^{3})

What is the solution for y^'+y=e^{2t},y(0)=-1 ?
The solution for y^'+y=e^{2t},y(0)=-1 is y=(e^{2t})/3-4/(3e^t)