What is the solution for 9y^{''}-12y^'+40y=0,y(0)=8,y^'(0)=3 ?

The solution for 9y^{''}-12y^'+40y=0,y(0)=8,y^'(0)=3 is y=e^{(2t)/3}(8cos(2t)-7/6 sin(2t))

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9y^{''}-12y^'+40y=0,y(0)=8,y^'(0)=3

9 y^{^{''}} - 12 y^{^{'}} + 40 y = 0, y (0) = 8, y^{^{'}} (0) = 3

y = e^{\frac{2 t}{3}} (8 cos (2 t) - \frac{7}{6} sin (2 t))

Solution steps

9 y^{''} (t) - 12 y^{'} (t) + 40 y = 0

Solve linear ODE:

y = e^{\frac{2 t}{3}} (8 cos (2 t) - \frac{7}{6} sin (2 t))

y = e^{\frac{2 t}{3}} (8 cos (2 t) - \frac{7}{6} sin (2 t))

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = 10

\frac{1}{y} y^{^{''}} = a, y (0) = 0

y^{^{''''}} - 3 y^{^{''}} + 2 y = 0

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

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

What is the solution for 9y^{''}-12y^'+40y=0,y(0)=8,y^'(0)=3 ?
The solution for 9y^{''}-12y^'+40y=0,y(0)=8,y^'(0)=3 is y=e^{(2t)/3}(8cos(2t)-7/6 sin(2t))