What is the solution for y^{''}-7/t y^'+(32)/(t^2)y=0 ?

The solution for y^{''}-7/t y^'+(32)/(t^2)y=0 is y=t^4(c_{1}cos(4ln(t))+c_{2}sin(4ln(t)))

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y^{''}-7/t y^'+(32)/(t^2)y=0

y^{^{''}} - \frac{7}{t} y^{^{'}} + \frac{32}{t ^{2}} y = 0

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

Solution steps

y^{''} (t) - \frac{7}{t} y^{'} (t) + \frac{32}{t ^{2}} y = 0

Solve linear ODE:

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

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

x^{2} \cdot y^{^{'}} - 4 x y = sin (3 x) \cdot x^{6}

x \to 0 lim (\frac{sin ( x ^{5} )}{x})

\int \frac{1}{1 + 9 x ^{2}} d x

\frac{- 3 e ^{(- s 2)}}{( s ^{2} - 5 s + 6 )}

x \to 0 + lim (x^{3} \cdot ln (x))

What is the solution for y^{''}-7/t y^'+(32)/(t^2)y=0 ?
The solution for y^{''}-7/t y^'+(32)/(t^2)y=0 is y=t^4(c_{1}cos(4ln(t))+c_{2}sin(4ln(t)))