y''=dp/dx=dp/dy*dy/dx=pdp/dy

Drag in

y？ pdp/dy+2p=0

Separate the respective integrals of variables.

∫dp=∫-2/y？ dysprosium (Dy)

p= 1/y？ +c

Because y'(0)= 1, y(0)= 1.

Therefore, 1 = 1+C, and C = 0.

So dy/dx= 1/y?

Separate variables and integrate them separately.

∫y？ dy=∫dx

1/3y？ =x+c 1

y(0)= 1

So,

1/3=0+c 1, then c 1= 1/3.

So the equation is: y? =3x+ 1