What is the solution for y^'-y=5te^{2t},y(0)=1 ?

The solution for y^'-y=5te^{2t},y(0)=1 is y=5e^{2t}(t-1)+6e^t

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y^'-y=5te^{2t},y(0)=1

y^{^{'}} - y = 5 t e^{2 t}, y (0) = 1

y = 5 e^{2 t} (t - 1) + 6 e^{t}

Solution steps

y^{'} (t) - y = 5 t e^{2 t}

Solve linear ODE:

y = 5 e^{2 t} (t - 1) + 6 e^{t}

y = 5 e^{2 t} (t - 1) + 6 e^{t}

n = 0 \sum \infty (1.075)^{n}

y = x + 1, (0, 2)

7 x^{3} - 6 x + \frac{3}{x ^{7}} + 9

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

\int_{1}^{4} \frac{x - 3}{x} d x

What is the solution for y^'-y=5te^{2t},y(0)=1 ?
The solution for y^'-y=5te^{2t},y(0)=1 is y=5e^{2t}(t-1)+6e^t