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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{3 x ^{2}}{( 169 + x ^{2} ) ^{2}} d x

n = 1 \sum \infty \frac{4}{n + 1}

n = 0 \sum \infty n (\frac{4}{5})^{n}

\frac{\partial}{\partial x} (cos (2 x + 2 y))

n = 1 \sum \infty - \frac{3}{2 ^{n}}

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