Differential Equations

dy

dx

+

x

x2−1

y =

x4=2x

√1−x2

This is a linear differential equation
I.F = e‌∫

x

x2−1

dx = e

1

2

ln|x2−1|=√1−x2
⇒ solution is
y√1−x2 = ∫

x(x3+2)

√1−x2

.√1−x2dx
or y√1−x2 = ∫(x4+2x)‌dx=

x5

5

+x2+c
f(0) = 0 ⇒ c = 0
⇒ f(x)√1−x2=

x5

5

+x2
Now,

√3 2

∫

− √3 2

f(x)‌dx =

√3 2

∫

− √3 2

x2

√1−x2

‌dx (Using property)
= 2‌

√3 2

∫

0

x2

√1−x2

‌dx = 2‌

π 3

∫

0

sin2θ

cosθ

cosθ dθ (Taking x = sinθ)
= 2‌

π 3

∫

0

sin2θdθ = 2[

θ

2

−

sin2θ

4

]0

π

3

= 2(

π

6

)−2(

√3

8

)=

π

3

−

√3

4