Question

Questions

=> Let f:R → R be a differentiable function with f(0) = 0 , if y = f(x) satisfies the differential equation dy/dx= (2 + 5y )(5y - 2 ) , Then the value of
$\sf \to \: \displaystyle \lim _{ \rm \: x \to - \infty \: } \rm \: f(x) \: is \: 0.4$
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Answer 1

Given,

$\sf \: \dfrac{dy}{dx} =(2 + 5y)(5y - 2) \\ \\ \implies \sf \: dx= \dfrac{dy}{(2 + 5y)(5y - 2)}$

Integrating on both sides,

$\implies \displaystyle \: \sf \int dx = \int\dfrac{dy}{10y + 25y {}^{2} - 10y - 4} \\ \\ \implies \displaystyle \: \sf x + c= \int\dfrac{dy}{ {25y}^{2} - 4} \\ \\\implies \displaystyle \: \sf x + c = \dfrac{1}{25} \int\dfrac{dy}{ {y}^{2} - ( \frac{2}{5} ) {}^{2} }$

Of the form,

$\star \: \boxed{ \boxed{ \displaystyle \sf \int \dfrac{dx}{ {x}^{2} - {a}^{2} } = \dfrac{1}{2a} log \bigg| \dfrac{x - a}{x + a} \bigg| + c}}$

Thus,

$\implies \sf x + c= \dfrac{1}{25} \times \dfrac{1}{2 \times \frac{2}{5} } log \bigg( \dfrac{5y - 2}{5y + 2} \bigg) \\ \\ \implies \sf x + c = \dfrac{1}{20} log \bigg( \dfrac{5y - 2}{5y + 2} \bigg) \\ \\ \implies \sf 20(x + c) = log \bigg |\dfrac{5y - 2}{5y + 2} \bigg |$

Taking e on both sides,

$\implies \sf e {}^{20x + 20c} = \bigg| \dfrac{5y - 2}{5y + 2} \bigg|$

We know that, f(x) = y and f(0) = 0.

Let C be some constant which is equal to e^20c.

$\implies \sf \: C \times {e}^{0} = \bigg| \dfrac{5(0) - 2}{5(0) + 2} \bigg | \\ \\ \implies \boxed{ \boxed{ \sf C = 1}}$

The equation becomes,

$\implies \sf \:C \times e {}^{20x} = \bigg| \dfrac{5f(x) - 2}{5f(x) + 2} \bigg| \\ \\ \implies \sf e {}^{20x} = \bigg| \dfrac{5f(x) - 2}{5f(x) + 2} \bigg|$

Applying limits on both sides,

$\implies \sf \lim _{ \rm \: x \to - \infty \: }e {}^{20x} = \lim _{ \rm \: x \to - \infty \: } \bigg| \dfrac{5f(x) - 2}{5f(x) + 2} \bigg| \\ \\ \implies \sf \: {e}^{ - \infty } = \lim _{ \rm \: x \to - \infty \: } \bigg| \dfrac{5f(x) - 2}{5f(x) + 2} \bigg| \\ \\ \implies \sf 0 = \lim _{ \rm \: x \to - \infty \: } (5f(x) - 2) \\ \\ \implies \sf 2 = \lim _{ \rm \: x \to - \infty \: } 5f(x) \\ \\ \implies \boxed{ \boxed{ \sf \lim _{ \rm \: x \to - \infty \: } f(x) = \dfrac{2}{5} }}$

Answer 2

[ Kindly check attached picture for the solution ]

Additional Information :-

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$

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