What is the solution for y^{''}=6y^'-8y,y^'(0)=2 ?

The solution for y^{''}=6y^'-8y,y^'(0)=2 is y=(e^{4t}(1-c_{2}))/2+c_{2}e^{2t}

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y^{''}=6y^'-8y,y^'(0)=2

y^{^{''}} = 6 y^{^{'}} - 8 y, y^{^{'}} (0) = 2

y = \frac{e ^{4 t} ( 1 - c _{2} )}{2} + c_{2} e^{2 t}

Solution steps

y^{''} (t) = 6 y^{'} (t) - 8 y

Solve linear ODE:

y = \frac{e ^{4 t} ( 1 - c _{2} )}{2} + c_{2} e^{2 t}

y = \frac{e ^{4 t} ( 1 - c _{2} )}{2} + c_{2} e^{2 t}

(d^{3} - d^{2} + d - 1) y = 0

\frac{d ^{2} y}{d x ^{2}} + 3 y = 0

y^{^{'''}} + 6 \cdot y^{^{''}} = 0

y^{^{''}} - 4 y^{^{'}} + 40 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = - 3

What is the solution for y^{''}=6y^'-8y,y^'(0)=2 ?
The solution for y^{''}=6y^'-8y,y^'(0)=2 is y=(e^{4t}(1-c_{2}))/2+c_{2}e^{2t}