The solution of the Fredholm integral equation \(y(s)=s+2 \int_0^1\left(s t^2+s^2 t\right) y(t) d t\) is

Explanation

Given:

y(s) = s + 2 \(∫_0^1 s t^2 y(t) d t+2 ∫_0^1 s^2 t y(t) d t\)

let \(\int_0^1\)t² y(t) dt = c₁    -- (i)

\(\int_0^1\) t y(t) dt = c₂    -- (ii)

⇒ y(s) = s + 2sc₁ + 2s² c₂ ---- (iii)

By (ii), c2 = \(\int_0^1\)t[t + 2c1 t + 2c2t2]dt

=\(\int_0^1\)(t2 + 2t2 c1 + 2c2 t3)dt

c2​ = \(\frac{t^3}{3}\left[1+\frac{2}{c_1}\right]+\left.\frac{c_2 t^4}{2}\right|_0 ^1=\left(\frac{1}{3}+\frac{2}{3} c_1\right)+\frac{c_2}{2}\)

\(⇒ \frac{3 C_2}{2}=1+\frac{2}{C_1} \)

⇒ 4C1​ - 3c2 = -2     - (iv)

\(C_1 =∫_0^1 t^2\left[t+2 c_1 t+2 t^2 c_2\right] d t \)

\(=∫_0^1\left[t^3\left(1+2 c_1\right)+2 t^4 c_2\right] d t\)

\(c_1 =\left[\left(1+2 c_1\right) \frac{t^4}{4}+\frac{2 c_2 t^5}{5}\right]_0^1=\frac{1+2 c_1}{4}+\frac{2 c_2}{5}\)

⇒ 20c1 = 5 + 10c1 + 8c2

⇒ 10c₁ - 8c₂ = 5 - (v)

Now solving equation (iv) and (v) to find c₁ & c₂.

20c₁ - 15c₂ = - 10 

20c1 - 16c2 = 10

subtracting we get

c2 = - 20

Hence (iv) implies 

 4c1​ + 60 = -2 ⇒ 4c1 = - 62 ⇒ c1 = - 31/2 

thus y(s) = s + 2sc1 + 2s2c2

= s + (-31) s + 2s2(-20)

y(s) = s - 31s - 40s²  = -(30s + 40s²)

⇒ option (3) is correct