What is the solution for y^{''}-5/t y^'+(18)/(t^2)y=0 ?

The solution for y^{''}-5/t y^'+(18)/(t^2)y=0 is y=t^3(c_{1}cos(3ln(t))+c_{2}sin(3ln(t)))

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y^{''}-5/t y^'+(18)/(t^2)y=0

y^{^{''}} - \frac{5}{t} y^{^{'}} + \frac{18}{t ^{2}} y = 0

y = t^{3} (c_{1} cos (3 ln (t)) + c_{2} sin (3 ln (t)))

Solution steps

y^{''} (t) - \frac{5}{t} y^{'} (t) + \frac{18}{t ^{2}} y = 0

Solve linear ODE:

y = t^{3} (c_{1} cos (3 ln (t)) + c_{2} sin (3 ln (t)))

y = t^{3} (c_{1} cos (3 ln (t)) + c_{2} sin (3 ln (t)))

\int (2 cot^{2} (x) - 3 tan^{2} (x)) d x

\frac{d}{d y} (x e^{y})

\int x^{2} e^{y} d y

\int \frac{1}{x ^{2} 144 - x ^{2}} d x

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

What is the solution for y^{''}-5/t y^'+(18)/(t^2)y=0 ?
The solution for y^{''}-5/t y^'+(18)/(t^2)y=0 is y=t^3(c_{1}cos(3ln(t))+c_{2}sin(3ln(t)))