TS EAMCET · Maths · Differential Equations
The general solution of \(y^2 d x+\left(x^2-x y+y^2\right) d y=0\) is :
- A \(\tan ^{-1}\left(\frac{y}{x}\right)=\log y+C\)
- B \(2 \tan ^{-1}\left(\frac{x}{y}\right)+\log x+C=0\)
- C \(\log \left(y+\sqrt{x^2+y^2}\right)+\log y+C=0\)
- D \(\sinh ^{-1}\left(\frac{x}{y}\right)+\log y+C=0\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}\left(\frac{y}{x}\right)=\log y+C\)
Step-by-step Solution
Detailed explanation
We have, \(y^2 d x+\left(x^2-x y+y^2\right) d y=0\) \(\Rightarrow \quad \frac{d y}{d x}=\frac{-y^2}{x^2-x y+y^2}\) It is a homogeneous linear differential equation Put \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)…
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