AP EAMCET · Maths · Differential Equations
If \(y=y(x)\) is the solution of \(x \frac{d y}{d x}=y+x e^{-\left(\frac{y}{x}\right)}, y(1)=\) \(\log \mathrm{e}\), then \(\mathrm{y}(\mathrm{e})=\)
- A \(\log \left(\frac{1}{\mathrm{e}}+1\right)\)
- B e \(\log (1+\mathrm{e})\)
- C \(\mathrm{e} \log \left(\frac{1}{\mathrm{e}}+1\right)\)
- D \(\operatorname{elog}\left(1-\frac{1}{\mathrm{e}}\right)\)
Answer & Solution
Correct Answer
(B) e \(\log (1+\mathrm{e})\)
Step-by-step Solution
Detailed explanation
Which is a homogenous differential equation So let \(\frac{y}{x}=v \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) ...(ii) From \(\mathrm{eq}^{\mathrm{n}}\) (i) \& (ii), we get…
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