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Question 14 What is the general solution of the differential equation 𝑒^(𝑦^β€² )=π‘₯ ? (A) 𝑦=π‘₯logπ‘₯+𝑐 (B) 𝑦=π‘₯logπ‘₯βˆ’π‘₯+𝑐 (C) 𝑦=π‘₯logπ‘₯+π‘₯+𝑐 (D) 𝑦=π‘₯+𝑐Given differential equation 𝑒^(𝑦^β€² )=π‘₯ Taking log both sides log(𝑒^(𝑦^β€² ))=log⁑π‘₯ 𝑦^β€² Γ— log e=log⁑π‘₯ Since log e = 1 𝑦^β€²=π‘™π‘œπ‘”β‘π‘₯ 𝑑𝑦/𝑑π‘₯ =π‘™π‘œπ‘”β‘π‘₯ 𝑑𝑦/𝑑π‘₯ =π‘™π‘œπ‘”β‘π‘₯ 𝑑𝑦=π‘™π‘œπ‘”β‘π‘₯ 𝑑π‘₯ Integrating both sides ∫1▒𝑑𝑦=∫1β–’γ€–π‘™π‘œπ‘”β‘π‘₯ 𝑑π‘₯γ€— π’š=∫1β–’γ€–π’π’π’ˆβ‘π’™ 𝒅𝒙〗 Putting ∫1β–’γ€–π‘™π‘œπ‘”β‘π‘₯ 𝑑π‘₯γ€—=π‘₯ log⁑π‘₯βˆ’π‘₯+𝐢 π’š=𝒙 π’π’π’ˆβ‘π’™βˆ’π’™+π‘ͺ So, the correct answer is (B)

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