What is the solution for y^{''}-7y^'+12y=0,y(0)=0,y^'(0)=3 ?

The solution for y^{''}-7y^'+12y=0,y(0)=0,y^'(0)=3 is y=3e^{4t}-3e^{3t}

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

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

y = 3 e^{4 t} - 3 e^{3 t}

Solution steps

y^{''} (t) - 7 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = 3 e^{4 t} - 3 e^{3 t}

y = 3 e^{4 t} - 3 e^{3 t}

y^{^{''}} + \frac{5}{2} y^{^{'}} + 1.578125 y = 0

u^{^{''}} + \frac{2}{r} u^{^{'}} = 0

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

y^{^{''}} - 2 y^{^{'}} - 195 y = 0, y (0) = 10, y^{^{'}} (0) = - 46

y^{^{'''}} + 8 y^{^{''}} + 7 y^{^{'}} = 0

What is the solution for y^{''}-7y^'+12y=0,y(0)=0,y^'(0)=3 ?
The solution for y^{''}-7y^'+12y=0,y(0)=0,y^'(0)=3 is y=3e^{4t}-3e^{3t}