What is the solution for x^2y^{''}+7xy^'+5y=0,y(1)=0,y^'(1)=2 ?

The solution for x^2y^{''}+7xy^'+5y=0,y(1)=0,y^'(1)=2 is y= 1/(2x)-1/(2x^5)

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x^2y^{''}+7xy^'+5y=0,y(1)=0,y^'(1)=2

x^{2} y^{^{''}} + 7 x y^{^{'}} + 5 y = 0, y (1) = 0, y^{^{'}} (1) = 2

y = \frac{1}{2 x} - \frac{1}{2 x ^{5}}

Solution steps

x^{2} y^{''} (x) + 7 x y^{'} (x) + 5 y = 0

Solve linear ODE:

y = \frac{1}{2 x} - \frac{1}{2 x ^{5}}

y = \frac{1}{2 x} - \frac{1}{2 x ^{5}}

\frac{\partial}{\partial x} (x \cdot y^{2})

x \to 0 lim ((1 + \frac{x}{2})^{\frac{1}{x}})

- x^{2} + 8 x, x^{2} - 4 x

\frac{d}{d x} (∣ x ∣^{∣ c o s (x) ∣})

\frac{\partial}{\partial x} (0.5 ln (x) + 0.5 ln (y))

What is the solution for x^2y^{''}+7xy^'+5y=0,y(1)=0,y^'(1)=2 ?
The solution for x^2y^{''}+7xy^'+5y=0,y(1)=0,y^'(1)=2 is y= 1/(2x)-1/(2x^5)