What is the solution for y^{''}-5/t y^'+(25)/(t^2)y=0 ?

The solution for y^{''}-5/t y^'+(25)/(t^2)y=0 is y=t^3(c_{1}cos(4ln(t))+c_{2}sin(4ln(t)))

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y^{''}-5/t y^'+(25)/(t^2)y=0

y^{^{''}} - \frac{5}{t} y^{^{'}} + \frac{25}{t ^{2}} y = 0

y = t^{3} (c_{1} cos (4 ln (t)) + c_{2} sin (4 ln (t)))

Solution steps

y^{''} (t) - \frac{5}{t} y^{'} (t) + \frac{25}{t ^{2}} y = 0

Solve linear ODE:

y = t^{3} (c_{1} cos (4 ln (t)) + c_{2} sin (4 ln (t)))

y = t^{3} (c_{1} cos (4 ln (t)) + c_{2} sin (4 ln (t)))

\int 12 x d x

\int (1 - 3 x)^{\frac{1}{3}} d x

x \to π lim (tan (\frac{2}{3} x))

f (x) = \frac{x ^{2}}{2 + x}

n \to \infty lim (\frac{n}{1})

What is the solution for y^{''}-5/t y^'+(25)/(t^2)y=0 ?
The solution for y^{''}-5/t y^'+(25)/(t^2)y=0 is y=t^3(c_{1}cos(4ln(t))+c_{2}sin(4ln(t)))