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

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

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

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

y = 4 e^{t} - 8 e^{t} t

Solution steps

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

Solve linear ODE:

y = 4 e^{t} - 8 e^{t} t

y = 4 e^{t} - 8 e^{t} t

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 2) = 5

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 15

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

y^{^{''}} - 4 y^{^{'}} - a^{2} y = 0

0.5 y^{^{''}} + 2 y^{^{'}} + 6 y = 0

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