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

The solution for 2x^2y^{''}+6xy^'+2y=0,y(1)=2,y^'(1)=-1 is y= 2/x+(ln(x))/x

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

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

y = \frac{2}{x} + \frac{ln ( x )}{x}

Solution steps

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

Solve linear ODE:

y = \frac{2}{x} + \frac{ln ( x )}{x}

y = \frac{2}{x} + \frac{ln ( x )}{x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 t e^{3 t}

x \to 2 - lim (\frac{4}{( x - 2 ) ^{5}})

y^{^{'}} - y = 4 sin (3 t)

\frac{1}{s - 1}

n = 4 \sum \infty \frac{1}{( n ) ( n + 1 )}

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