What is the solution for y^{''}-7y^'+12y=0,y(0)=-2,y^'(0)=4 ?

The solution for y^{''}-7y^'+12y=0,y(0)=-2,y^'(0)=4 is y=10e^{4t}-12e^{3t}

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y^{''}-7y^'+12y=0,y(0)=-2,y^'(0)=4

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = - 2, y^{^{'}} (0) = 4

y = 10 e^{4 t} - 12 e^{3 t}

Solution steps

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

Solve linear ODE:

y = 10 e^{4 t} - 12 e^{3 t}

y = 10 e^{4 t} - 12 e^{3 t}

\frac{d}{d x} (\frac{1}{x + sin ( x )})

\frac{d y}{d x} = \frac{4 e ^{4 x}}{e ^{y}}

\int 2 e^{- s t} d t

\frac{( s - 1 )}{s ^{2} + 2 s + 2}

x \to \infty lim (3 - f (x))

What is the solution for y^{''}-7y^'+12y=0,y(0)=-2,y^'(0)=4 ?
The solution for y^{''}-7y^'+12y=0,y(0)=-2,y^'(0)=4 is y=10e^{4t}-12e^{3t}