Showing y=0 is a solution to y'=xg(x,y) and positivity preservation

[Fibonacci88]

Given the equation y'= xg(x,y) , suppose that g and (partial) dg/dy are defined and continuous for all (x,y). Show the following:

1) y(x)=0 is a solution

2)if y=y(x), x in (a,b) is a solution and if y(x0)>0, x0 in (a,b), then y(x)>0 for all x in (a,b)

Please i need your help.

 

[HallsofIvy]

Given the equation y'= xg(x,y) , suppose that g and (partial) dg/dy are defined and continuous for all (x,y). Show the following:

1) y(x)=0 is a solution

You can't- it's not true. For example, supose g(x,y)= 1 so the equation is y'= x. Then $y(x)= (1/2)x^2+ C$. y(x)= 0 is not a solution since then y'= 0 and so $y'= 0\ne x$. there may be some other condition that you have left out.

2)if y=y(x), x in (a,b) is a solution and if y(x0)>0, x0 in (a,b), then y(x)>0 for all x in (a,b)

Please i need your help.

 

[Fibonacci88]

sorry for mistake. I am sending the correct version of the problem now.

i need answer for (ii) and (iii)