Let the solution curve y=y(x) of the differential equation 1+e2xdydx+y=1 pass through the point 0,π2. Then, limx→∞ exy(x) is equal to :

1+e2xdydx+y=1⇒dydx+y=11+e2x,which is a L.D.E.∴ I.F. =e∫1dx=ex; solution is y⋅ex=∫ex⋅dx1+e2x+cPut ex=t⇒exdx=dty⋅ex=∫dt1+t2+c⇒yex=tan−1⁡t+c;y=e−xtan−1⁡ex+ce−xCurve passes through0,π2∴π2=1⋅tan−1⁡(1)+c(1)⇒c=π2−π4=π4∴ y=e−xtan−1⁡ex+π4e−xlimx→∞ exy(x)=limx→∞ tan−1⁡ex+π4=π2+π4=3π4

Let the solution curve y=y(x) of the differential equation 1+e2xdydx+y=1 pass through the point 0,π2. Then, limx→∞ exy(x) is equal to :

1+e2xdydx+y=1⇒dydx+y=11+e2x,which is a L.D.E.∴ I.F. =e∫1dx=ex; solution is y⋅ex=∫ex⋅dx1+e2x+cPut ex=t⇒exdx=dty⋅ex=∫dt1+t2+c⇒yex=tan−1⁡t+c;y=e−xtan−1⁡ex+ce−xCurve passes through0,π2∴π2=1⋅tan−1⁡(1)+c(1)⇒c=π2−π4=π4∴ y=e−xtan−1⁡ex+π4e−xlimx→∞ exy(x)=limx→∞ tan−1⁡ex+π4=π2+π4=3π4

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Let the solution curve $y = y (x)$ of the differential equation $(1 + e^{2 x}) (\frac{dy}{dx} + y) = 1$ pass through the point $(0, \frac{π}{2})$ . Then, $lim_{x \to \infty} e^{x} y (x)$ is equal to :

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$\frac{3 π}{4}$

$\frac{π}{2}$

$\frac{3 π}{2}$

$\frac{π}{4}$

answer is B.

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Detailed Solution

$(1 + e^{2 x}) (\frac{dy}{dx} + y) = 1 \Rightarrow \frac{dy}{dx} + y = \frac{1}{1 + e^{2 x}}$ ,
which is a L.D.E.
$∴ I.F. = e^{\int 1 dx} = e^{x}$ ; solution is $y \cdot e^{x} = \int e^{x} \cdot \frac{dx}{1 + e^{2 x}} + c$
Put $e^{x} = t \Rightarrow e^{x} dx = dt$
$y \cdot e^{x} = \int \frac{dt}{1 + t^{2}} + c \Rightarrow {ye}^{x} = \tan^{- 1} t + c; y = e^{- x} \tan^{- 1} e^{x} + {ce}^{- x}$
Curve passes through $(0, \frac{π}{2})$
$\begin{array}{l} ∴ \frac{π}{2} = 1 \cdot \tan^{- 1} (1) + c (1) \Rightarrow c = \frac{π}{2} - \frac{π}{4} = \frac{π}{4} \\ ∴ y = e^{- x} \tan^{- 1} e^{x} + \frac{π}{4} e^{- x} \\ lim_{x \to \infty} e^{x} y (x) = lim_{x \to \infty} (\tan^{- 1} e^{x} + \frac{π}{4}) = \frac{π}{2} + \frac{π}{4} = \frac{3 π}{4} \end{array}$