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

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

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

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

y = \frac{e ^{3 t} ( 1 - 2 c _{2} )}{3} + c_{2} e^{2 t}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{3 t} ( 1 - 2 c _{2} )}{3} + c_{2} e^{2 t}

y = \frac{e ^{3 t} ( 1 - 2 c _{2} )}{3} + c_{2} e^{2 t}

10 y^{^{''}} - 50 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 1

x^{^{''}} - 3 x^{^{'}} + 2 x = 0, x^{0} = 1, x^{^{'}} 0 = - 1

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''''}} - y^{^{'''}} - 3 y^{^{''}} + 17 y^{^{'}} - 30 y = 0

(d^{3} - d^{2} - 4 d + 4) y = 0

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