What is the solution for y^{''}-y^'+y=4,y^'(0)=-1,y(0)=4 ?

The solution for y^{''}-y^'+y=4,y^'(0)=-1,y(0)=4 is y=-2/(sqrt(3))e^{t/2}sin((sqrt(3)t)/2)+4

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y^{''}-y^'+y=4,y^'(0)=-1,y(0)=4

y^{^{''}} - y^{^{'}} + y = 4, y^{^{'}} (0) = - 1, y (0) = 4

y = - \frac{2}{3} e^{\frac{t}{2}} sin (\frac{3 t}{2}) + 4

Solution steps

y^{''} (t) - y^{'} (t) + y = 4

Solve linear ODE:

y = - \frac{2}{3} e^{\frac{t}{2}} sin (\frac{3 t}{2}) + 4

y = - \frac{2}{3} e^{\frac{t}{2}} sin (\frac{3 t}{2}) + 4

y^{^{''}} + y - 2 y = 2 + 5 cos (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{- t}

y^{^{''}} - 2 y - 4 x^{2} e^{2 x} = 0

4 y^{^{''}} + 20 y^{^{'}} + 169 y = 3, y^{^{'}} (0) = 1

y^{^{''}} - 4 y = 12 e^{2 x}

What is the solution for y^{''}-y^'+y=4,y^'(0)=-1,y(0)=4 ?
The solution for y^{''}-y^'+y=4,y^'(0)=-1,y(0)=4 is y=-2/(sqrt(3))e^{t/2}sin((sqrt(3)t)/2)+4