What is the solution for y^'=3cos(t)-2y,y(0)=0 ?

The solution for y^'=3cos(t)-2y,y(0)=0 is y=(3(sin(t)+2cos(t)))/5-6/(5e^{2t)}

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y^'=3cos(t)-2y,y(0)=0

y^{^{'}} = 3 cos (t) - 2 y, y (0) = 0

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} - \frac{6}{5 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{6}{5 e ^{2 t}}

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} - \frac{6}{5 e ^{2 t}}

y^{^{'}} + 9 y = t + e - 8 t

\frac{d}{d x} (x,)

n = 8 \sum \infty e^{7 - 6 n}

x y^{^{'}} = 2 x^{2} y + 3 x^{3} e^{x^{2}}, y (0) = 5

- x + 2, x^{2} - 1, [0, 1]

What is the solution for y^'=3cos(t)-2y,y(0)=0 ?
The solution for y^'=3cos(t)-2y,y(0)=0 is y=(3(sin(t)+2cos(t)))/5-6/(5e^{2t)}