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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 15 y^{^{'}} + 44 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 5

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 8 y = 0

\frac{d ^{2} y}{d x ^{2}} - 2 \cdot \frac{d y}{d x} = 0

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y (0) = - \frac{9}{4}

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