What is the solution for y^'-y=10te^2t,y(0)=1 ?

The solution for y^'-y=10te^2t,y(0)=1 is y=-10e^2(t^2+2(t+1))+(1+20e^2)e^t

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y^'-y=10te^2t,y(0)=1

y^{^{'}} - y = 10 t e^{2} t, y (0) = 1

y = - 10 e^{2} (t^{2} + 2 (t + 1)) + (1 + 20 e^{2}) e^{t}

Solution steps

y^{'} (t) - y = 10 t e^{2} t

Solve linear ODE:

y = - 10 e^{2} (t^{2} + 2 (t + 1)) + (1 + 20 e^{2}) e^{t}

y = - 10 e^{2} (t^{2} + 2 (t + 1)) + (1 + 20 e^{2}) e^{t}

\frac{5}{t}

\frac{1}{x ( 9 + ln ( x ) )}

ln (x), - 1

y = (2 x - 7), (0, 7)

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

What is the solution for y^'-y=10te^2t,y(0)=1 ?
The solution for y^'-y=10te^2t,y(0)=1 is y=-10e^2(t^2+2(t+1))+(1+20e^2)e^t