What is the solution for t^2y^{''}-3ty^'+4y=0 ?

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

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t^2y^{''}-3ty^'+4y=0

t^{2} y^{^{''}} - 3 t y^{^{'}} + 4 y = 0

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

Solution steps

t^{2} y^{''} (t) - 3 t y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial y} ((x + y)^{5} + (x - y)^{5})

f (x) = x^{2} + 5, (- 5, 30)

x y^{^{''}} - 2 y^{^{'}} + 4 = 0

(- 3.7) (- 4.12)

f (x) = x^{3} (3 - x)^{4}, at x = 2

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