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Question

Mathematics Question on Differential equations

Let 𝑦=𝑦(𝑥)𝑦=𝑦(𝑥) be the solution of the differential equation 𝑥  𝑙𝑜𝑔𝑒⁡𝑥𝑑𝑦𝑑𝑥+𝑦=𝑥2𝑙𝑜𝑔𝑒⁡𝑥,(𝑥>1)𝑥\;𝑙𝑜𝑔_𝑒⁡𝑥\frac{𝑑𝑦}{ 𝑑𝑥} +𝑦=𝑥^2𝑙𝑜𝑔_𝑒⁡𝑥,(𝑥>1). If 𝑦(2)=2𝑦(2)=2, then 𝑦(𝑒)𝑦(𝑒) is equal to

A

2+e22\frac{2+e^2}{2}

B

1+e22\frac{1+e^2}{2}

C

1+e24\frac{1+e^2}{4}

D

4+e24\frac{4+e^2}{4}

Answer

4+e24\frac{4+e^2}{4}

Explanation

Solution

The Correct Option is (D): 4+e24\frac{4+e^2}{4}