What is the solution for y^{''}-5y^'+6y=0,y(0)=1,y^'(0)=-5 ?

The solution for y^{''}-5y^'+6y=0,y(0)=1,y^'(0)=-5 is y=-7e^{3t}+8e^{2t}

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y^{''}-5y^'+6y=0,y(0)=1,y^'(0)=-5

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

y = - 7 e^{3 t} + 8 e^{2 t}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = - 7 e^{3 t} + 8 e^{2 t}

y = - 7 e^{3 t} + 8 e^{2 t}

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

0.05 y^{^{''}} + 2 y^{^{'}} + \frac{1}{0.01} y = 0

3 y^{^{''}} - 2 y^{^{'}} - y = 0

4 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + y = 0

\frac{1}{2} x^{^{''}} + \frac{1}{2} x^{^{'}} + 6 x = 0

What is the solution for y^{''}-5y^'+6y=0,y(0)=1,y^'(0)=-5 ?
The solution for y^{''}-5y^'+6y=0,y(0)=1,y^'(0)=-5 is y=-7e^{3t}+8e^{2t}