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

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