What is the solution for y^{''}-8y^'+16y=0,y(0)=0,y^'(0)=3 ?

The solution for y^{''}-8y^'+16y=0,y(0)=0,y^'(0)=3 is y=3e^{4t}t

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y^{''}-8y^'+16y=0,y(0)=0,y^'(0)=3

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y = 3 e^{4 t} t

Solution steps

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

Solve linear ODE:

y = 3 e^{4 t} t

y = 3 e^{4 t} t

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

\frac{d ^{2} y}{d x ^{2}} + 9 \frac{d y}{d x} + 9 y = 0

\frac{d ^{2} x}{d t ^{2}} + 16 \cdot x = 0

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = a, y^{^{'}} (0) = 5

x^{^{''}} + 4 x^{^{'}} + 29 x = 0

What is the solution for y^{''}-8y^'+16y=0,y(0)=0,y^'(0)=3 ?
The solution for y^{''}-8y^'+16y=0,y(0)=0,y^'(0)=3 is y=3e^{4t}t