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JEE Mains · Maths · STD 12 - 9. differential equations

माना \( y=y(x) \) अवकल समीकरण \( (1+x^{2})dy+(y-\tan^{-1}x)dx=0 \) का हल वक्र है, जबकि \( y(0)=1 \) है। तब \( y(1) \) का मान ........... है।

  1. A \( \frac{2}{e^{\frac{\pi}{4}}}+\frac{\pi}{4}-1 \)
  2. B \( \frac{2}{e^{\frac{\pi}{4}}}-\frac{\pi}{4}-1 \)
  3. C \( \frac{4}{e^{\frac{\pi}{4}}}+\frac{\pi}{2}-1 \)
  4. D \( \frac{4}{e^{\frac{\pi}{4}}}-\frac{\pi}{2}-1 \)
Verified Solution

Answer & Solution

Correct Answer

(A) \( \frac{2}{e^{\frac{\pi}{4}}}+\frac{\pi}{4}-1 \)

Step-by-step Solution

Detailed explanation

\( \frac{dy}{dx}+\frac{y}{x^{2}+1}=\frac{\tan^{-1}x}{x^{2}+1} \) \( I.f.=e^{\tan^{-1}x} \) \( y\times e^{\tan^{-1}x}=\int e^{\tan^{-1}x}.\frac{\tan^{-1}x}{1+x^{2}}dx \) \( y\times e^{\tan^{-1}x}=\tan^{-1}x(e^{\tan^{-1}x})-e^{\tan^{-1}x}+c \) \( y(0)=1\Rightarrow c=2 \)…
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