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

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

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y^'-y=3te^{2t},y(0)=1

y^{^{'}} - y = 3 t e^{2 t}, y (0) = 1

y = 3 e^{2 t} (t - 1) + 4 e^{t}

Solution steps

y^{'} (t) - y = 3 t e^{2 t}

Solve linear ODE:

y = 3 e^{2 t} (t - 1) + 4 e^{t}

y = 3 e^{2 t} (t - 1) + 4 e^{t}

2 sin (t)

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

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 4 x}{x + 1})

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

f (x) = 2 - 10 x, at x = - 1

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