please help to solve and i will fan and give a medal Y" + x y' + y= 0 when y(0)=1 and y'(0)=0

Question

anonymous · Answer

then just plug in the variables, and its just system of equations
so when y(0), it becomes $$y"+xy'+y=0$$
For y(0),
$$y''+xy'+0=1$$
which means $$y''+xy'=1$$
then for y'(0)
$$y''+x(0)+y=0$$
Anything multiplied by 0 is 0, so that means
$$y''+0+y=0$$ 
put y on the other side,
$$y''=-y$$
and you can go plug that in for the other equation, y(0)=1
$$(-y)+xy'+(0)=1$$
$$(0)+xy'+(0)=1$$
$$xy'=1$$
Since when you put 0 in for y', it becomes 0, that means x=1

anonymous · Answer

does that mean this is another process of solving differential equation ivp ?