Question ID: #649
Let $y=y(x)$ be the solution curve of the differential equation $(1+x^{2})dy+(y-tan^{-1}x)dx=0, y(0)=1$. Then the value of $y(1)$ is:
- (1) $\frac{2}{e^{\frac{\pi}{4}}}+\frac{\pi}{4}-1$
- (2) $\frac{2}{e^{\frac{\pi}{4}}}-\frac{\pi}{4}-1$
- (3) $\frac{4}{e^{\frac{\pi}{4}}}+\frac{\pi}{2}-1$
- (4) $\frac{4}{e^{\frac{\pi}{4}}}-\frac{\pi}{2}-1$
Solution:
Rearrange the differential equation to linear form:
$\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{\tan^{-1}x}{1+x^2}$
Integrating Factor (I.F.) $= e^{\int \frac{1}{1+x^2}dx} = e^{\tan^{-1}x}$.
Solution is given by:
$y \cdot (I.F.) = \int (Q \cdot I.F.) dx + C$
$y e^{\tan^{-1}x} = \int \frac{\tan^{-1}x}{1+x^2} e^{\tan^{-1}x} dx$
Put $\tan^{-1}x = t \Rightarrow \frac{1}{1+x^2}dx = dt$.
RHS $= \int t e^t dt = e^t(t-1) + C$.
Substituting back:
$y e^{\tan^{-1}x} = e^{\tan^{-1}x}(\tan^{-1}x – 1) + C$
Given $y(0) = 1$:
$1 \cdot e^0 = e^0(0 – 1) + C \Rightarrow 1 = -1 + C \Rightarrow C = 2$.
The equation becomes:
$y e^{\tan^{-1}x} = e^{\tan^{-1}x}(\tan^{-1}x – 1) + 2$
$y = (\tan^{-1}x – 1) + 2e^{-\tan^{-1}x}$
Find $y(1)$:
At $x = 1, \tan^{-1}(1) = \frac{\pi}{4}$.
$y(1) = \left(\frac{\pi}{4} – 1\right) + \frac{2}{e^{\pi/4}}$
Ans. (1)
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