What is the solution for 121y^{''}-16y=0 ?

The solution for 121y^{''}-16y=0 is y=c_{1}e^{(4t)/(11)}+c_{2}e^{-(4t)/(11)}

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121y^{''}-16y=0

121 y^{^{''}} - 16 y = 0

y = c_{1} e^{\frac{4 t}{11}} + c_{2} e^{- \frac{4 t}{11}}

Solution steps

121 y^{''} (t) - 16 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{4 t}{11}} + c_{2} e^{- \frac{4 t}{11}}

y = c_{1} e^{\frac{4 t}{11}} + c_{2} e^{- \frac{4 t}{11}}

\frac{\partial}{\partial x} ((x + y)^{2})

n = 1 \sum \infty \frac{1}{n ^{2} + 3 n + 2}

\int \frac{( x + 1 )}{( x ^{2} + 2 x )} d x

\int \frac{6 x ^{3} - 12 x ^{2} + 8}{x ^{2} - 2 x} d x

\int_{0}^{1} ((x + 2) (x - 3)) d x

What is the solution for 121y^{''}-16y=0 ?
The solution for 121y^{''}-16y=0 is y=c_{1}e^{(4t)/(11)}+c_{2}e^{-(4t)/(11)}