What is the solution for y^{''}-11y^'+24y=0,y^'(0)=12 ?

The solution for y^{''}-11y^'+24y=0,y^'(0)=12 is y=(e^{8t}(12-3c_{2}))/8+c_{2}e^{3t}

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y^{''}-11y^'+24y=0,y^'(0)=12

y^{^{''}} - 11 y^{^{'}} + 24 y = 0, y^{^{'}} (0) = 12

y = \frac{e ^{8 t} ( 12 - 3 c _{2} )}{8} + c_{2} e^{3 t}

Solution steps

y^{''} (t) - 11 y^{'} (t) + 24 y = 0

Solve linear ODE:

y = \frac{e ^{8 t} ( 12 - 3 c _{2} )}{8} + c_{2} e^{3 t}

y = \frac{e ^{8 t} ( 12 - 3 c _{2} )}{8} + c_{2} e^{3 t}

\frac{d ^{8} y}{d x ^{8}} - 8 \frac{d ^{4} y}{d x ^{4}} + 16 y = 0

49 y^{^{''}} - y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 4 y - 3 y^{^{'}} = 0

- 100 y + 65 y^{^{'}} - 14 y^{^{''}} + y^{^{'''}} = 0

x^{^{''}} + 4 x^{^{'}} = - 2 x^{^{'}} - 8 x + 3 x

What is the solution for y^{''}-11y^'+24y=0,y^'(0)=12 ?
The solution for y^{''}-11y^'+24y=0,y^'(0)=12 is y=(e^{8t}(12-3c_{2}))/8+c_{2}e^{3t}