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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int (9 t^{2} + \frac{1}{t ^{3}}) d t

y^{^{'}} = e^{3 x} - y

\int 3 tan^{3} (x) d x

n = 0 \sum \infty \frac{7 ^{n}}{4}

x^{^{''}} + 2 x^{^{'}} + 2 x = 0

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