The solution of dydx+P(x)y=0 is:
(3)
Ans. y=Ce-∫P dx
Explanation:
dydx+P(x)y=0
Here, Q=0
∴ I.F.=e∫P dx
∴ Solution of the given equation is
y(I.F.)=∫Q(I.F.) dx+C
⇒ y e∫P dx=0+C
⇒ y=Ce-∫P dx