What is the solution for x^2y^{''}+2xy^'-12y=0,y(1)=0,y^'(1)=-2 ?

The solution for x^2y^{''}+2xy^'-12y=0,y(1)=0,y^'(1)=-2 is y=-(2x^3)/7+2/(7x^4)

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x^2y^{''}+2xy^'-12y=0,y(1)=0,y^'(1)=-2

x^{2} y^{^{''}} + 2 x y^{^{'}} - 12 y = 0, y (1) = 0, y^{^{'}} (1) = - 2

y = - \frac{2 x ^{3}}{7} + \frac{2}{7 x ^{4}}

Solution steps

x^{2} y^{''} (x) + 2 x y^{'} (x) - 12 y = 0

Solve linear ODE:

y = - \frac{2 x ^{3}}{7} + \frac{2}{7 x ^{4}}

y = - \frac{2 x ^{3}}{7} + \frac{2}{7 x ^{4}}

n = 1 \sum \infty \frac{1}{n ^{0}}

x \frac{d y}{d x} + 4 y = x^{3} - x

(1.3) (4.2)

y \frac{d y}{d t} = 2 t + 7

y = x^{\frac{2}{x}}

What is the solution for x^2y^{''}+2xy^'-12y=0,y(1)=0,y^'(1)=-2 ?
The solution for x^2y^{''}+2xy^'-12y=0,y(1)=0,y^'(1)=-2 is y=-(2x^3)/7+2/(7x^4)