What is the solution for y^{''}-6y^'+5y=te^{2t} ?

The solution for y^{''}-6y^'+5y=te^{2t} is y=c_{1}e^{5t}+c_{2}e^t-(e^{2t}t)/3+2/9 e^{2t}

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y^{''}-6y^'+5y=te^{2t}

y^{^{''}} - 6 y^{^{'}} + 5 y = t e^{2 t}

y = c_{1} e^{5 t} + c_{2} e^{t} - \frac{e ^{2 t} t}{3} + \frac{2}{9} e^{2 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 5 y = t e^{2 t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{t} - \frac{e ^{2 t} t}{3} + \frac{2}{9} e^{2 t}

y = c_{1} e^{5 t} + c_{2} e^{t} - \frac{e ^{2 t} t}{3} + \frac{2}{9} e^{2 t}

y^{^{''}} + 4 y = 6 sin (x)

y^{^{'''}} - 5 y^{^{''}} + 3 y^{^{'}} + 9 = 0

y^{^{''}} - 2 y^{^{'}} + y = - \frac{1}{x ^{2}} e^{x}

y^{^{''}} + \frac{1}{8} y^{^{'}} + 4 y = 3 cos (6 t)

y^{^{''}} + 7 y^{^{'}} + 12 y = 21 e^{3 t}

What is the solution for y^{''}-6y^'+5y=te^{2t} ?
The solution for y^{''}-6y^'+5y=te^{2t} is y=c_{1}e^{5t}+c_{2}e^t-(e^{2t}t)/3+2/9 e^{2t}