What is the solution for y^{''}-8y^'+16y=0,y(0)=4 ?

The solution for y^{''}-8y^'+16y=0,y(0)=4 is y=4e^{4t}+c_{2}te^{4t}

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y^{''}-8y^'+16y=0,y(0)=4

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4

y = 4 e^{4 t} + c_{2} t e^{4 t}

Solution steps

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

Solve linear ODE:

y = 4 e^{4 t} + c_{2} t e^{4 t}

y = 4 e^{4 t} + c_{2} t e^{4 t}

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 3 y = 0

y^{^{''}} - 5 y^{^{'}} - 50 y = 0

8 y^{^{''}} - 14 y^{^{'}} + 3 y = 0

y^{^{''}} - 12 y^{^{'}} + 32 y = 0, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''''}} - 34 y^{^{'''}} + 289 y^{^{''}} = 0

What is the solution for y^{''}-8y^'+16y=0,y(0)=4 ?
The solution for y^{''}-8y^'+16y=0,y(0)=4 is y=4e^{4t}+c_{2}te^{4t}