What is the solution for y^{''}+xy^'-y=0 ?

The solution for y^{''}+xy^'-y=0 is y=x*integral of (e^{(-x^2+c_{1})/2})/(x^2)

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y^{''}+xy^'-y=0

y^{^{''}} + x y^{^{'}} - y = 0

y = x \cdot \int \frac{e ^{\frac{- x ^{2} + c _{1}}{2}}}{x ^{2}} d x

Solution steps

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

Solve using substitution:

y = x \cdot \int \frac{e ^{\frac{- x ^{2} + c _{1}}{2}}}{x ^{2}} d x

y = x \cdot \int \frac{e ^{\frac{- x ^{2} + c _{1}}{2}}}{x ^{2}} d x

\frac{d}{d x} (- \frac{cos ( x )}{x})

\int \frac{5 x ^{3}}{x ^{2} - 1} d x

y^{^{''}} + 3 y^{^{'}} + 2 y = (1 + e^{t})^{- 1}

4 x^{- 3}

x \to - 5 lim (∣ x ∣ - 5)

What is the solution for y^{''}+xy^'-y=0 ?
The solution for y^{''}+xy^'-y=0 is y=x*integral of (e^{(-x^2+c_{1})/2})/(x^2)