What is the solution for y^{''}-3/t y^'+(20)/(t^2)y=0 ?

The solution for y^{''}-3/t y^'+(20)/(t^2)y=0 is y=t^2(c_{1}cos(4ln(t))+c_{2}sin(4ln(t)))

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y^{''}-3/t y^'+(20)/(t^2)y=0

y^{^{''}} - \frac{3}{t} y^{^{'}} + \frac{20}{t ^{2}} y = 0

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

Solution steps

y^{''} (t) - \frac{3}{t} y^{'} (t) + \frac{20}{t ^{2}} y = 0

Solve linear ODE:

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

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

arcsin (2 x - 3)

\int \frac{x}{sec ( x )} d x

\frac{d y}{d x}, y^{x} = x^{y}

\frac{d y}{d t} - y = 1, y (0) = 0

t y^{^{'}} + y = 1 + y + y^{2}

What is the solution for y^{''}-3/t y^'+(20)/(t^2)y=0 ?
The solution for y^{''}-3/t y^'+(20)/(t^2)y=0 is y=t^2(c_{1}cos(4ln(t))+c_{2}sin(4ln(t)))