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

The solution for y^{''}-7y^'+12y=0,y(0)=3,y^'(0)=0 is y=-9e^{4t}+12e^{3t}

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

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

y = - 9 e^{4 t} + 12 e^{3 t}

Solution steps

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

Solve linear ODE:

y = - 9 e^{4 t} + 12 e^{3 t}

y = - 9 e^{4 t} + 12 e^{3 t}

\frac{d ^{2}}{d x ^{2}} ((u (x)) e^{2 x}) = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y^{^{'}} = 13, y (0) = 3

2 y^{^{''}} - 4 y^{^{'}} + 4 y = 0

x^{^{''}} + 5 x^{^{'}} - 6 x = 0

y^{^{''}} + 4 y^{^{'}} + 10 y = 0

What is the solution for y^{''}-7y^'+12y=0,y(0)=3,y^'(0)=0 ?
The solution for y^{''}-7y^'+12y=0,y(0)=3,y^'(0)=0 is y=-9e^{4t}+12e^{3t}