Question

Write a particular solution of the differential equation, (1 + x²) `dy/dx` + 2xy = `1/(1 + x^2)`  when  y = 0, x = 0.

Answer 1

`(1+ x^2)dy/dx + 2xy = 1/(1 + x^2)"when"  y = 0, x = 0`

On dividing by (1 + x²) on both sides

`dy/dx + (2xy)/((1 + x^2)) = 1/((1 + x^2)^2)`

On comparing with `dy/dx + Py = Q`

we get,
P = `"2x"/(1 + x^2) and Q = 1/(1 + x^2)^2`

Find the integrating factor

I.F = e^∫Pdx

= `e^((2x"/"1 + x^2))dx`

Let 1 + x² = t

On differentiating

2xdx = dt

`dx = dt/(2x)`

If = e^{∫(2x/t) (dt/2x)}

= e^∫dt/t

= e^loge^t

If  t = 1 + x²

∵ y × I.F = ∫Q × I.F dx

On putting values

`y(1 + x^2) = ∫ 1/(1 + x^2)^2(1 + x^2)dx`

`y(1 + x^2) = ∫ 1/((1 + x^2))dx`

`y(1 + x^2) = tan^(−1)x + C`    ...[∵ `∫1/((1 + x^2))dx = tan^(−1) x`]

On x = 0 and y = 0, we get

0(1 + 0²) = tan⁻¹ (0) + C

0 = tan⁻¹ (tan 0°) + C

C = 0

so,

y(1 + x²) = tan⁻¹ x