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

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

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

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

y = 2 e^{2 t} (t - 1) + 3 e^{t}

Solution steps

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

Solve linear ODE:

y = 2 e^{2 t} (t - 1) + 3 e^{t}

y = 2 e^{2 t} (t - 1) + 3 e^{t}

sin (x y)

\int (\frac{5}{x} - 2 x^{3}) d x

\frac{d y}{d x} - e^{5 x + y} = 0

\frac{d}{d x} (\frac{- 15}{x})

\frac{\partial}{\partial x} (sin (x yz))

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