Q.

Let y = y(x) be the solution of the differential equation dydx+2x(1+x2)2y=xe1(1+x2); y(0) = 0. Then the area enclosed by the curve f(x)=y(x)e–1(1+x2) and the line y – x = 4 is __________.          [2024]


Ans.

(18)

We have, dydx+2x(1+x2)2y=xe1(1+x2)

IF=e∫2x(1+x2)2dx=e–11+x2

∴  General solution of given differential equation is,

y·e–11+x2=∫xe11+x2·e–11+x2dx

⇒ ye–11+x2=x22+c

Since y(0) = 0

⇒ 0=0+c

⇒ c=0

∴  y(x)=x22e11+x2

⇒ f(x)=x22

So, intersection of curve y=f(x)=x22 and y = x + 4 is given by x22=x+4

⇒ x2–2x–8=0 ⇒ (x–4)(x+2)=0

⇒ x=4, –2 ⇒ y=8, 2

∴  Required area = ∫–24((x+4)–x22)dx

= [x22+4x–x36]–24=[8+16–323–2+8–86]

= 30 – 12 = 18.