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

The solution for y^'-y=9te^{2t},y(0)=1 is y=9e^{2t}(t-1)+10e^t

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y^'-y=9te^{2t},y(0)=1

y^{^{'}} - y = 9 t e^{2 t}, y (0) = 1

y = 9 e^{2 t} (t - 1) + 10 e^{t}

Solution steps

y^{'} (t) - y = 9 t e^{2 t}

Solve linear ODE:

y = 9 e^{2 t} (t - 1) + 10 e^{t}

y = 9 e^{2 t} (t - 1) + 10 e^{t}

x \to 0 - lim (\frac{1}{x \cdot e ^{\frac{1}{x}}})

x \to 2 lim (3 x^{3} - 12 x - 4)

\int sin (6 x - 10) d x

t an g e n t f (x) = x^{2} + 2, (- 4, 18)

\frac{d}{d x} (sec^{2} (x) + csc^{2} (x))

What is the solution for y^'-y=9te^{2t},y(0)=1 ?
The solution for y^'-y=9te^{2t},y(0)=1 is y=9e^{2t}(t-1)+10e^t