What is the solution for t^2y^{''}-3ty^'+3y=0,y(1)=5,y^'(1)=3 ?

The solution for t^2y^{''}-3ty^'+3y=0,y(1)=5,y^'(1)=3 is y=-t^3+6t

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t^2y^{''}-3ty^'+3y=0,y(1)=5,y^'(1)=3

t^{2} y^{^{''}} - 3 t y^{^{'}} + 3 y = 0, y (1) = 5, y^{^{'}} (1) = 3

y = - t^{3} + 6 t

Solution steps

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

Solve linear ODE:

y = - t^{3} + 6 t

y = - t^{3} + 6 t

5 cos (4 x), 5 - 5 cos (4 x), [0, \frac{π}{4}]

\frac{\partial}{\partial x} (2 x - 4 y^{2} + 3 z^{2})

x \to 0 - lim (\frac{- \frac{1}{2}}{4 x})

e^{(- x^{2})}

\int 5^{- x - 2} d x

What is the solution for t^2y^{''}-3ty^'+3y=0,y(1)=5,y^'(1)=3 ?
The solution for t^2y^{''}-3ty^'+3y=0,y(1)=5,y^'(1)=3 is y=-t^3+6t