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

The solution for x^2y^{''}-x(x+2)y^'+(x+2)y=0 is y=c_{1}e^xx+c_{2}x

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

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

y = c_{1} e^{x} x + c_{2} x

Solution steps

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

Solve using substitution:

y = c_{1} e^{x} x + c_{2} x

y = c_{1} e^{x} x + c_{2} x

x + 7 x - 7

x \to \infty lim (3 + x)

9 y^{^{''}} + 6 y^{^{'}} + y = e^{- \frac{1}{3} t}

\frac{d}{d x} (\frac{1}{x ^{2} x})

\int \frac{1}{( 1 - x ^{2} ) ^{0.5}} d x

What is the solution for x^2y^{''}-x(x+2)y^'+(x+2)y=0 ?
The solution for x^2y^{''}-x(x+2)y^'+(x+2)y=0 is y=c_{1}e^xx+c_{2}x