Given that the second order differential equation: $$y''+xy'+y=0$$ with initial conditions: $y(0)=1$ & $y'(0)=0$
Find out the value of $y(0.1)$
My try: $$\int y''+\int xy'+\int y=\int 0$$ $$y'+x y+yx=C_1$$ $$y'+2x y=C_1$$ $$\int y'+2y \int x=\int C_1$$ $$y+2y\cdot \frac{x^2}{2}=C_1x+C_2$$ $$y+x^2y=C_1x+C_2$$ $$y=\frac{C_1x+C_2}{1+x^2}$$ setting x=0, y=1 $$1=\frac{C_1\cdot 0+C_2}{1+0^2}\iff C_2=1$$ setting $x=0$ & $y'=0$, $$0+2(0) (1)=C_1\iff C_1=0$$ $$y=\frac{1}{1+x^2}$$ setting $x=0.1$ $$y(0.1)=\frac{1}{1+0.1^2}=0.99$$ But my teacher suggests the correct answer: $exp(-1/200). But I don't know where I am wrong. Someone please help me to get correct answer. Thanks
\begin{align} y''+xy'+y=0 &\iff (y'+xy)'=0\\ &\iff y'+xy=C_1\\ &\iff \exp{(x^2/2)}y'+x\exp{(x^2/2)}y=C_1\exp{(x^2/2)}\\ &\iff (\exp{(x^2/2)}y)'=C_1\exp{(x^2/2)}\\ &\iff \exp{(x^2/2)}y=C_2\text{erfi}{(x/\sqrt{2})}+C_3\\ &\iff y=(C_2\text{erfi}{(x/\sqrt{2})}+C_3)\exp{(-x^2/2)}\\ \end{align} Given the initial conditions we get $$y=\exp{(-x^2/2)}$$ and hence $$y(0.1)=\exp{(-0.005)}$$