What is the solution for (d^2y)/(dt^2)-5(dy)/(dt)+6y=0,y(0)=2 ?

The solution for (d^2y)/(dt^2)-5(dy)/(dt)+6y=0,y(0)=2 is y=(2-c_{2})e^{3t}+c_{2}e^{2t}

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(d^2y)/(dt^2)-5(dy)/(dt)+6y=0,y(0)=2

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 y = 0, y (0) = 2

y = (2 - c_{2}) e^{3 t} + c_{2} e^{2 t}

Solution steps

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 y = 0

Solve linear ODE:

y = (2 - c_{2}) e^{3 t} + c_{2} e^{2 t}

y = (2 - c_{2}) e^{3 t} + c_{2} e^{2 t}

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 10

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = - 4, y^{^{'}} (0) = - \frac{28}{3}

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 3.65

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

y^{^{'''}} - 4 y^{^{'}} - 5 y^{^{'}} = 0

What is the solution for (d^2y)/(dt^2)-5(dy)/(dt)+6y=0,y(0)=2 ?
The solution for (d^2y)/(dt^2)-5(dy)/(dt)+6y=0,y(0)=2 is y=(2-c_{2})e^{3t}+c_{2}e^{2t}