What is the solution for y^{''}-y^'+y=2 ?

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

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

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

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

Solution steps

y^{''} (t) - y^{'} (t) + y = 2

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} - 2 = 0

y^{^{''}} + 49 y = - 5 cos (7 t + π^{3}), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x} \cdot ln (9 + x^{2})

y^{^{''}} - 4 y = (x - 2) e^{x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 15

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