What is the solution for y^'=3cos(t)-2y ?

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

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y^'=3cos(t)-2y

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

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} + \frac{c _{1}}{e ^{2 t}}

Solution steps

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

Solve linear ODE:

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} + \frac{c _{1}}{e ^{2 t}}

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} + \frac{c _{1}}{e ^{2 t}}

\frac{3}{x + 2}

\int_{1}^{3} (4 x^{3} - 3 x^{2}) d x

\frac{1}{s ^{2} + s}

\frac{d ^{2} y}{d t ^{2}} - e^{2} y = 0

\int \frac{u}{u ^{2} + 2} d u

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