What is the solution for 400y^{''}-9y=0 ?

The solution for 400y^{''}-9y=0 is y=c_{1}e^{(3t)/(20)}+c_{2}e^{-(3t)/(20)}

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400y^{''}-9y=0

400 y^{^{''}} - 9 y = 0

y = c_{1} e^{\frac{3 t}{20}} + c_{2} e^{- \frac{3 t}{20}}

Solution steps

400 y^{''} (t) - 9 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{3 t}{20}} + c_{2} e^{- \frac{3 t}{20}}

y = c_{1} e^{\frac{3 t}{20}} + c_{2} e^{- \frac{3 t}{20}}

8 y^{^{''}} + 2 y = 0

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = 0

y^{^{''}} + 10 y^{^{'}} + 23 y = 0

4 \frac{d ^{2} y}{d x ^{2}} + 12 \frac{d y}{d x} + 9 y = 0

y^{^{''}} - 6 y^{^{'}} + 5 y = 0

What is the solution for 400y^{''}-9y=0 ?
The solution for 400y^{''}-9y=0 is y=c_{1}e^{(3t)/(20)}+c_{2}e^{-(3t)/(20)}