What is the solution for 9y^{''}-6y^'+y=cos(t) ?

The solution for 9y^{''}-6y^'+y=cos(t) is y=c_{1}e^{t/3}+c_{2}e^{t/3}t-3/50 sin(t)-2/25 cos(t)

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9y^{''}-6y^'+y=cos(t)

9 y^{^{''}} - 6 y^{^{'}} + y = cos (t)

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{\frac{t}{3}} t - \frac{3}{50} sin (t) - \frac{2}{25} cos (t)

Solution steps

9 y^{''} (t) - 6 y^{'} (t) + y = cos (t)

Solve linear ODE:

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{\frac{t}{3}} t - \frac{3}{50} sin (t) - \frac{2}{25} cos (t)

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{\frac{t}{3}} t - \frac{3}{50} sin (t) - \frac{2}{25} cos (t)

y^{^{''}} - 9 y = x + 2

y^{^{''}} - y = c^{x} + 2 c^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = x e^{2 x} + x^{3} e^{2 x}

y^{^{''}} - 11 y^{^{'}} + 10 y = 2 x^{2} - 44 x + 4

y^{^{''}} + 9 y = 11 sec^{2} (3 t)

What is the solution for 9y^{''}-6y^'+y=cos(t) ?
The solution for 9y^{''}-6y^'+y=cos(t) is y=c_{1}e^{t/3}+c_{2}e^{t/3}t-3/50 sin(t)-2/25 cos(t)