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

The solution for x^{''}-5x^'+6x=e^{2t} is x=c_{1}e^{3t}+c_{2}e^{2t}-e^{2t}t

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

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

x = c_{1} e^{3 t} + c_{2} e^{2 t} - e^{2 t} t

Solution steps

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

Solve linear ODE:

x = c_{1} e^{3 t} + c_{2} e^{2 t} - e^{2 t} t

x = c_{1} e^{3 t} + c_{2} e^{2 t} - e^{2 t} t

y = (x^{2} - 6 x + 10) 25 - x^{2}

\int 13 tan^{4} (x se) c^{6} x d x

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.1}} d x

f (x) = \frac{8 x}{x ^{2} + 1}, at x = 1

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

What is the solution for x^{''}-5x^'+6x=e^{2t} ?
The solution for x^{''}-5x^'+6x=e^{2t} is x=c_{1}e^{3t}+c_{2}e^{2t}-e^{2t}t