What is the solution for 4y^{''}+4y^'+y=0,y(2)=-6,y^'(2)=4 ?

The solution for 4y^{''}+4y^'+y=0,y(2)=-6,y^'(2)=4 is y=-8e^{1-t/2}+e^{1-t/2}t

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4y^{''}+4y^'+y=0,y(2)=-6,y^'(2)=4

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (2) = - 6, y^{^{'}} (2) = 4

y = - 8 e^{1 - \frac{t}{2}} + e^{1 - \frac{t}{2}} t

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + y = 0

Solve linear ODE:

y = - 8 e^{1 - \frac{t}{2}} + e^{1 - \frac{t}{2}} t

y = - 8 e^{1 - \frac{t}{2}} + e^{1 - \frac{t}{2}} t

t^{6} e^{-, at}

\int \frac{x}{a + b x} d x

\frac{d}{d x} ((x^{2} - 6) (x^{3} + x^{2} + x + 2))

\int 2 (x - 6)^{3} d x

2 x^{2} cos (x) cot (x)

What is the solution for 4y^{''}+4y^'+y=0,y(2)=-6,y^'(2)=4 ?
The solution for 4y^{''}+4y^'+y=0,y(2)=-6,y^'(2)=4 is y=-8e^{1-t/2}+e^{1-t/2}t