We have,
d y d x = ( 2 + 5 y ) ( 5 y − 2 ) {{dy} \over {dx}} = (2 + 5y)(5y - 2) d x d y = ( 2 + 5 y ) ( 5 y − 2 ) ⇒ d y 25 y 2 − 4 = d x \Rightarrow {{dy} \over {25{y^2} - 4}} = dx ⇒ 25 y 2 − 4 d y = d x ⇒ 1 25 ( d y y 2 − 4 25 ) = d x \Rightarrow {1 \over {25}}\left( {{{dy} \over {{y^2} - {4 \over {25}}}}} \right) = dx ⇒ 25 1 ( y 2 − 25 4 d y ) = d x On integrating both sides, we get
1 25 ∫ d y y 2 − ( 2 5 ) 2 = ∫ d x {1 \over {25}}\int {{{dy} \over {{y^2} - {{\left( {{2 \over 5}} \right)}^2}}} = \int {dx} } 25 1 ∫ y 2 − ( 5 2 ) 2 d y = ∫ d x ⇒ 1 25 × 1 2 × 2 / 5 log ∣ y − 2 / 5 y + 2 / 5 ∣ = x + C \Rightarrow {1 \over {25}} \times {1 \over {2 \times 2/5}}\log \left| {{{y - 2/5} \over {y + 2/5}}} \right| = x + C ⇒ 25 1 × 2 × 2/5 1 log y + 2/5 y − 2/5 = x + C ⇒ log ∣ 5 y − 2 5 y + 2 ∣ = 20 ( x + C ) \Rightarrow \log \left| {{{5y - 2} \over {5y + 2}}} \right| = 20(x + C) ⇒ log 5 y + 2 5 y − 2 = 20 ( x + C ) ⇒ ∣ 5 y − 2 5 y + 2 ∣ = A e 20 x \Rightarrow \left| {{{5y - 2} \over {5y + 2}}} \right| = A{e^{20x}} ⇒ 5 y + 2 5 y − 2 = A e 20 x [
∵ \because ∵ e
20C = A]
when x = 0
⇒ \Rightarrow ⇒ y = 0, then A = 1
∴ \therefore ∴ ∣ 5 y − 2 5 y + 2 ∣ = e 20 x \left| {{{5y - 2} \over {5y + 2}}} \right| = {e^{20x}} 5 y + 2 5 y − 2 = e 20 x lim x → − ∞ ∣ 5 f ( x ) − 2 5 f ( x ) + 2 ∣ = lim x → − ∞ e 20 x \mathop {\lim }\limits_{x \to - \infty } \left| {{{5f(x) - 2} \over {5f(x) + 2}}} \right| = \mathop {\lim }\limits_{x \to - \infty } {e^{20x}} x → − ∞ lim 5 f ( x ) + 2 5 f ( x ) − 2 = x → − ∞ lim e 20 x ⇒ lim n → − ∞ ∣ 5 f ( x ) − 2 5 f ( x ) + 2 ∣ = 0 \Rightarrow \mathop {\lim }\limits_{n \to - \infty } \left| {{{5f(x) - 2} \over {5f(x) + 2}}} \right| = 0 ⇒ n → − ∞ lim 5 f ( x ) + 2 5 f ( x ) − 2 = 0 ⇒ lim n → − ∞ 5 f ( x ) − 2 = 0 \Rightarrow \mathop {\lim }\limits_{n \to - \infty } 5f(x) - 2 = 0 ⇒ n → − ∞ lim 5 f ( x ) − 2 = 0 ⇒ lim n → − ∞ f ( x ) = 2 5 = 0.4 \Rightarrow \mathop {\lim }\limits_{n \to - \infty } f(x) = {2 \over 5} = 0.4 ⇒ n → − ∞ lim f ( x ) = 5 2 = 0.4