What is the solution for y^{''}-2y^'+y=0,y(0)=8,y^'(0)=3 ?

The solution for y^{''}-2y^'+y=0,y(0)=8,y^'(0)=3 is y=8e^t-5e^tt

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y^{''}-2y^'+y=0,y(0)=8,y^'(0)=3

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 8, y^{^{'}} (0) = 3

y = 8 e^{t} - 5 e^{t} t

Solution steps

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

Solve linear ODE:

y = 8 e^{t} - 5 e^{t} t

y = 8 e^{t} - 5 e^{t} t

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 5^{2} y = 0

12 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = 0

\frac{d ^{2} y}{d x ^{2}} (1) - \frac{d y}{d x} (10) = 0

y^{^{'''}} - 15 y^{^{''}} + 72 y^{^{'}} - 108 y = 0

\frac{1}{2} \frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 26 y = 0

What is the solution for y^{''}-2y^'+y=0,y(0)=8,y^'(0)=3 ?
The solution for y^{''}-2y^'+y=0,y(0)=8,y^'(0)=3 is y=8e^t-5e^tt