What is the solution for y^{''}+2xy^'=0 ?

The solution for y^{''}+2xy^'=0 is y=c_{1}(sqrt(pi))/2 \erf(x)+c_{2}

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y^{''}+2xy^'=0

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

y = c_{1} \frac{π}{2} erf (x) + c_{2}

Solution steps

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

Solve using substitution:

y = c_{1} \frac{π}{2} erf (x) + c_{2}

y = c_{1} \frac{π}{2} erf (x) + c_{2}

\int sec^{2} (x t a) n^{3} x d x

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

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

\int x^{2} \cdot ab \cdot e^{- ab x + ab c} d x

\frac{\partial}{\partial y} (e^{x^{2}} + 2 x y)

What is the solution for y^{''}+2xy^'=0 ?
The solution for y^{''}+2xy^'=0 is y=c_{1}(sqrt(pi))/2 \erf(x)+c_{2}