What is the solution for 144y^{''}-312y^'+169y=0 ?

The solution for 144y^{''}-312y^'+169y=0 is y=c_{1}e^{(13t)/(12)}+c_{2}e^{(13t)/(12)}t

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144y^{''}-312y^'+169y=0

144 y^{^{''}} - 312 y^{^{'}} + 169 y = 0

y = c_{1} e^{\frac{13 t}{12}} + c_{2} e^{\frac{13 t}{12}} t

Solution steps

144 y^{''} (t) - 312 y^{'} (t) + 169 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{13 t}{12}} + c_{2} e^{\frac{13 t}{12}} t

y = c_{1} e^{\frac{13 t}{12}} + c_{2} e^{\frac{13 t}{12}} t

y^{^{''}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} - 18 y^{^{'}} + 97 y = 0

\frac{d ^{3} y}{d x ^{3}} + 3 \frac{d ^{2} y}{d x ^{2}} = 0

y^{^{''}} + \frac{16}{3} y = 0

4 y^{^{''}} + 22 y^{^{'}} + 18 y = 0, y (0) = 3, y^{^{'}} (0) = 0

What is the solution for 144y^{''}-312y^'+169y=0 ?
The solution for 144y^{''}-312y^'+169y=0 is y=c_{1}e^{(13t)/(12)}+c_{2}e^{(13t)/(12)}t