What is the solution for y^{''}-7y^'+12y=0,y(0)=1,y^'(0)=-2 ?

The solution for y^{''}-7y^'+12y=0,y(0)=1,y^'(0)=-2 is y=-5e^{4t}+6e^{3t}

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y^{''}-7y^'+12y=0,y(0)=1,y^'(0)=-2

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y = - 5 e^{4 t} + 6 e^{3 t}

Solution steps

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

Solve linear ODE:

y = - 5 e^{4 t} + 6 e^{3 t}

y = - 5 e^{4 t} + 6 e^{3 t}

x \to 4 lim (\frac{5}{x - 3})

y = \frac{x - 2}{x - 2}

\frac{\partial}{\partial x} (2 e^{5 x y})

\int tan (8 x) d x

\frac{\partial}{\partial x} (\frac{y ^{2}}{1 + x ^{2}})

What is the solution for y^{''}-7y^'+12y=0,y(0)=1,y^'(0)=-2 ?
The solution for y^{''}-7y^'+12y=0,y(0)=1,y^'(0)=-2 is y=-5e^{4t}+6e^{3t}