What is the solution for y^{''}-7y^'+12y=0,y(0)=-1/2 ,y^'(0)=-9/4 ?

The solution for y^{''}-7y^'+12y=0,y(0)=-1/2 ,y^'(0)=-9/4 is y=(-1/2-1/4)e^{4t}+1/4 e^{3t}

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y^{''}-7y^'+12y=0,y(0)=-1/2 ,y^'(0)=-9/4

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = - \frac{1}{2}, y^{^{'}} (0) = - \frac{9}{4}

y = (- \frac{1}{2} - \frac{1}{4}) e^{4 t} + \frac{1}{4} e^{3 t}

Solution steps

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

Solve linear ODE:

y = (- \frac{1}{2} - \frac{1}{4}) e^{4 t} + \frac{1}{4} e^{3 t}

y = (- \frac{1}{2} - \frac{1}{4}) e^{4 t} + \frac{1}{4} e^{3 t}

\int \frac{15 e ^{5 y}}{e ^{5 y} + 2} d y

x^{3} + y^{3} = 8 x y, (4, 4)

x \to 3 lim (1 0^{x^{2} - 3 \cdot x})

\frac{\partial}{\partial y} (x e^{9 x y})

x + 2, x^{2} - 4

What is the solution for y^{''}-7y^'+12y=0,y(0)=-1/2 ,y^'(0)=-9/4 ?
The solution for y^{''}-7y^'+12y=0,y(0)=-1/2 ,y^'(0)=-9/4 is y=(-1/2-1/4)e^{4t}+1/4 e^{3t}