JEE Mains · Maths · STD 12 - 9. differential equations
Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be the solution of the differential equation \(\left((x+2) e^{\left(\frac{y+1}{x+2}\right)}+(y+1)\right) d x=(x+2) d y, y(1)=1\) If the domain of \(y=y(x)\) is an open interval \((\alpha, \beta)\), then \(|\alpha+\beta|\) is equal to \(......\)
- A \(3\)
- B \(4\)
- C \(5\)
- D \(9\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(y+1=Y \Rightarrow d y=d Y\) \(x+2=X \Rightarrow d x=d X\) \(\Rightarrow\left(X e^{\frac{Y}{x}}+Y\right) d X=X d Y\) \(\Rightarrow X d Y-Y d X=X e^{Y / x} d X\) \(\Rightarrow d\left(\frac{Y}{X}\right) e^{-\frac{Y}{x}}=\frac{d X}{X}\) \(-e^{-y / x}=\ell n|x|+c\)…
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