What is the solution for x^2y^{''}-4xy^'+6y=0,y(1)=0.4,y^'(1)=0 ?

The solution for x^2y^{''}-4xy^'+6y=0,y(1)=0.4,y^'(1)=0 is y=-0.8x^3+1.2x^2

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x^2y^{''}-4xy^'+6y=0,y(1)=0.4,y^'(1)=0

x^{2} y^{^{''}} - 4 x y^{^{'}} + 6 y = 0, y (1) = 0.4, y^{^{'}} (1) = 0

y = - 0.8 x^{3} + 1.2 x^{2}

Solution steps

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

Solve linear ODE:

y = - 0.8 x^{3} + 1.2 x^{2}

y = - 0.8 x^{3} + 1.2 x^{2}

y d x - (x + y) d y = 0

(5, - 5), (- 3, 5)

y^{^{'''}} - 7 y^{^{''}} + 8 y^{^{'}} + 16 y = 0

\frac{d}{d x} (I n (sin (x)))

\int x^{2} x sin (x^{2}) d x

What is the solution for x^2y^{''}-4xy^'+6y=0,y(1)=0.4,y^'(1)=0 ?
The solution for x^2y^{''}-4xy^'+6y=0,y(1)=0.4,y^'(1)=0 is y=-0.8x^3+1.2x^2