What is the solution for x^{''}-7x^'+10x=0 ?

The solution for x^{''}-7x^'+10x=0 is x=c_{1}e^{5t}+c_{2}e^{2t}

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x^{''}-7x^'+10x=0

x^{^{''}} - 7 x^{^{'}} + 10 x = 0

x = c_{1} e^{5 t} + c_{2} e^{2 t}

Solution steps

x^{''} (t) - 7 x^{'} (t) + 10 x = 0

Solve linear ODE:

x = c_{1} e^{5 t} + c_{2} e^{2 t}

x = c_{1} e^{5 t} + c_{2} e^{2 t}

\frac{3}{4} x^{^{''}} + 9 x = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 6, y^{^{'}} (0) = 1

y^{^{''}} + 6 y^{^{'}} + 35 y = 0

(4 D^{3} - 15 D^{2} + 5 D + 6) y = 0

y^{^{''''}} - k^{2} y = 0

What is the solution for x^{''}-7x^'+10x=0 ?
The solution for x^{''}-7x^'+10x=0 is x=c_{1}e^{5t}+c_{2}e^{2t}