What is the solution for x^2y^{''}-2xy^'-4y=0,y(1)=0,y^'(1)=-5 ?

The solution for x^2y^{''}-2xy^'-4y=0,y(1)=0,y^'(1)=-5 is y=-x^4+1/x

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x^2y^{''}-2xy^'-4y=0,y(1)=0,y^'(1)=-5

x^{2} y^{^{''}} - 2 x y^{^{'}} - 4 y = 0, y (1) = 0, y^{^{'}} (1) = - 5

y = - x^{4} + \frac{1}{x}

Solution steps

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

Solve linear ODE:

y = - x^{4} + \frac{1}{x}

y = - x^{4} + \frac{1}{x}

\frac{d}{d x} (x^{4} - \frac{x ^{8}}{6} - x^{2} sin (x^{2}))

\int (5 x + 4)^{3} d x

\frac{\partial}{\partial x} (sin^{2} (t))

n = 0 \sum \infty \frac{n}{( \frac{3}{2} ) ^{n}}

\frac{( s + 1 ) ^{3}}{s ^{4}}

What is the solution for x^2y^{''}-2xy^'-4y=0,y(1)=0,y^'(1)=-5 ?
The solution for x^2y^{''}-2xy^'-4y=0,y(1)=0,y^'(1)=-5 is y=-x^4+1/x