What is the solution for y^{''}+2y^'+y=0,y(2)=1,y^'(2)=-2 ?

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

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y^{''}+2y^'+y=0,y(2)=1,y^'(2)=-2

y^{^{''}} + 2 y^{^{'}} + y = 0, y (2) = 1, y^{^{'}} (2) = - 2

y = 3 e^{2 - t} - e^{2 - t} t

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 0

Solve linear ODE:

y = 3 e^{2 - t} - e^{2 - t} t

y = 3 e^{2 - t} - e^{2 - t} t

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