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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 8 y = 0

y^{^{''}} - k \cdot y = 0

(6 d^{3} + 5 d^{2}) y = 0

y^{^{'''}} + 7 y^{^{''}} + 14 y^{^{'}} + 8 y = 0

\frac{d ^{4} y}{d t ^{4}} + 2 y^{^{''}} + y = 0

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