What is the solution for y^{''}-1/t y^'+(26)/(t^2)y=0 ?

The solution for y^{''}-1/t y^'+(26)/(t^2)y=0 is y=t(c_{1}cos(5ln(t))+c_{2}sin(5ln(t)))

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y^{''}-1/t y^'+(26)/(t^2)y=0

y^{^{''}} - \frac{1}{t} y^{^{'}} + \frac{26}{t ^{2}} y = 0

y = t (c_{1} cos (5 ln (t)) + c_{2} sin (5 ln (t)))

Solution steps

y^{''} (t) - \frac{1}{t} y^{'} (t) + \frac{26}{t ^{2}} y = 0

Solve linear ODE:

y = t (c_{1} cos (5 ln (t)) + c_{2} sin (5 ln (t)))

y = t (c_{1} cos (5 ln (t)) + c_{2} sin (5 ln (t)))

\int 7 cos (x) d x

(\frac{1}{x} + x^{2} - 2)^{^{'}}

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What is the solution for y^{''}-1/t y^'+(26)/(t^2)y=0 ?
The solution for y^{''}-1/t y^'+(26)/(t^2)y=0 is y=t(c_{1}cos(5ln(t))+c_{2}sin(5ln(t)))