What is the solution for y^'-y=8te^{2t},y(0)=1 ?

The solution for y^'-y=8te^{2t},y(0)=1 is y=8e^{2t}(t-1)+9e^t

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y^'-y=8te^{2t},y(0)=1

y^{^{'}} - y = 8 t e^{2 t}, y (0) = 1

y = 8 e^{2 t} (t - 1) + 9 e^{t}

Solution steps

y^{'} (t) - y = 8 t e^{2 t}

Solve linear ODE:

y = 8 e^{2 t} (t - 1) + 9 e^{t}

y = 8 e^{2 t} (t - 1) + 9 e^{t}

\int_{0}^{0.5} 8000 x d x

n \to \infty lim (\frac{n}{2 n})

y^{^{'}} - x y = x

\int \frac{3 x + 8}{x ( x - 2 ) ^{2}} d x

\frac{d}{d x} (\frac{2 x ^{2} - 1}{3 x + 5})

What is the solution for y^'-y=8te^{2t},y(0)=1 ?
The solution for y^'-y=8te^{2t},y(0)=1 is y=8e^{2t}(t-1)+9e^t