Let y = y(x) be the solution of the differential equation (xy–5x21+x2)dx+(1+x2)dy=0, y(0) = 0. Then y(3) is equal to [2025]
(3)
We have, (xy–5x21+x2)dx+(1+x2)dy=0
⇒ dydx=5x21+x2–xy(1+x2)
⇒ dydx+xy(1+x2)=5x21+x2
I.F.=∫ex1+x2dx=1+x2
y×1+x2=∫5x21+x2×1+x2dx+C
y1+x2=5x33+C
∵ y(0)=0
∴ C=0
⇒ y=5x331+x2
∴ y(3)=5×3×331+3=532