What is the solution for x^2y^{''}-4xy^'+6y=0 ?

The solution for x^2y^{''}-4xy^'+6y=0 is y=c_{1}x^3+c_{2}x^2

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x^2y^{''}-4xy^'+6y=0

x^{2} y^{^{''}} - 4 x y^{^{'}} + 6 y = 0

y = c_{1} x^{3} + c_{2} x^{2}

Solution steps

x^{2} y^{''} (x) - 4 x y^{'} (x) + 6 y = 0

Solve linear ODE:

y = c_{1} x^{3} + c_{2} x^{2}

y = c_{1} x^{3} + c_{2} x^{2}

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 3

\frac{\partial}{\partial x} (2 (x - y) e^{2 y + x^{2}})

\int 8 x cos^{2} (x) d x

\frac{\partial}{\partial x} ((x - 7)^{2} + (y - 1)^{2} + 29)

x \to 4 lim (\frac{3}{x - 4})

What is the solution for x^2y^{''}-4xy^'+6y=0 ?
The solution for x^2y^{''}-4xy^'+6y=0 is y=c_{1}x^3+c_{2}x^2