What is the solution for y^{''}-2y^'+1y=0,y(0)=8,y^'(0)=2 ?

The solution for y^{''}-2y^'+1y=0,y(0)=8,y^'(0)=2 is y=8e^t-6e^tt

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y^{''}-2y^'+1y=0,y(0)=8,y^'(0)=2

y^{^{''}} - 2 y^{^{'}} + 1 y = 0, y (0) = 8, y^{^{'}} (0) = 2

y = 8 e^{t} - 6 e^{t} t

Solution steps

y^{''} (t) - 2 y^{'} (t) + 1 y = 0

Solve linear ODE:

y = 8 e^{t} - 6 e^{t} t

y = 8 e^{t} - 6 e^{t} t

\frac{\partial}{\partial y} (- 2 e^{x} cos (yz))

\int (\frac{1}{2 x}) d x

\frac{d}{d x} (F (x^{4}))

f (x) = 2 x^{2} - x + 1

\int - 8 x^{2} d x

What is the solution for y^{''}-2y^'+1y=0,y(0)=8,y^'(0)=2 ?
The solution for y^{''}-2y^'+1y=0,y(0)=8,y^'(0)=2 is y=8e^t-6e^tt