TS EAMCET · Maths · Differential Equations
The solution of \(\frac{d y}{d x}=\frac{x+y}{x-y}\) is
- A \(\tan ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2+y^2}+C\)
- B \(\tan ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2-y^2}+C\)
- C \(\sin ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2+y^2}+C\)
- D \(\cos ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2-y^2}+C\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2+y^2}+C\)
Step-by-step Solution
Detailed explanation
We have, \[ \frac{d y}{d x}=\frac{x+y}{x-y} \] Put \(y=v x\)…
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