What is the solution for y^{''}+8y^'+16y=0,y(1)=0,y^'(1)=1 ?

The solution for y^{''}+8y^'+16y=0,y(1)=0,y^'(1)=1 is y=-e^{4-4t}+e^{4-4t}t

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y^{''}+8y^'+16y=0,y(1)=0,y^'(1)=1

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (1) = 0, y^{^{'}} (1) = 1

y = - e^{4 - 4 t} + e^{4 - 4 t} t

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = - e^{4 - 4 t} + e^{4 - 4 t} t

y = - e^{4 - 4 t} + e^{4 - 4 t} t

(x^{4} + 3 x^{2} - 2)^{5}

(x + 1) y^{^{'}} + (x + 2) y = 8 x e^{- x}

\int 8 x^{3} e^{x^{4}} d x

8 x sin (x)

1 - x^{2}, 13 - 7 x

What is the solution for y^{''}+8y^'+16y=0,y(1)=0,y^'(1)=1 ?
The solution for y^{''}+8y^'+16y=0,y(1)=0,y^'(1)=1 is y=-e^{4-4t}+e^{4-4t}t