Question: If (x<sup>2</sup> –1) $\frac{dy}{dx}$ + 2xy = 1, y (0) = 0 then y(2) =...

Question

If (x² –1) $\frac{dy}{dx}$ + 2xy = 1, y (0) = 0 then y(2) =

Answer 1

$\frac{dy}{dx}$ + $\frac{2xy}{x^{2}–1}$ = $\frac{1}{x^{2}–1}$ ⇒ P = $\frac{2x}{x^{2}–1}$ , Q = $\frac{1}{x^{2}–1}$

(1) I.F. = $e^{\int_{}^{}\frac{2x}{x^{2}–1}.dx}$ = $e^{\log(x^{2}–1)}$ dx = x² – 1

(2) y . (x² – 1) = $\int_{}^{}\frac{1}{x^{2}–1}$ × (x² – 1) dx

⇒ y(x² – 1) = x + c

(3) y (0) = 0 ⇒ put x = y = 0

⇒ (–1) = c ⇒ c = 0

Now curve ⇒ y (x²–1) = x

⇒ y = $\frac{x}{x^{2}–1}$

(4) Now y(2) ⇒ x = 2 ∴ y = $\frac{2}{2^{2}–1}$ = $\frac{2}{3}$

Question: If (x<sup>2</sup> –1) \(\frac{dy}{dx}\) + 2xy = 1, y (0) = 0 then y(2) =...