What is the solution for y^{''}-3y^'=18t+26sin(2t) ?

The solution for y^{''}-3y^'=18t+26sin(2t) is y=c_{1}e^{3t}+c_{2}-3t^2-2t-2sin(2t)+3cos(2t)

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y^{''}-3y^'=18t+26sin(2t)

y^{^{''}} - 3 y^{^{'}} = 18 t + 26 sin (2 t)

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

Solution steps

y^{''} (t) - 3 y^{'} (t) = 18 t + 26 sin (2 t)

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 32 sin (4 t)

\frac{d}{d x} (\frac{2}{3} x + \frac{4}{3 x ^{2}})

\frac{d}{d x} (f (x) \circ g (x))

x \to 1 - lim (\frac{x ^{3} - 1}{x - 1})

(x - 7)^{4} y^{^{'}} + 5 (x - 7)^{3} y = x + 7

What is the solution for y^{''}-3y^'=18t+26sin(2t) ?
The solution for y^{''}-3y^'=18t+26sin(2t) is y=c_{1}e^{3t}+c_{2}-3t^2-2t-2sin(2t)+3cos(2t)