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The general solution of the differential equation 𝑦𝑑π‘₯ βˆ’ π‘₯𝑑𝑦 = 0 𝑖𝑠

(a) π‘₯𝑦 = 𝐢 Β 

(b) π‘₯ = Cy 2 Β 

(c) 𝑦 = 𝐢π‘₯ Β 

(d) 𝑦 = Cx 2

Β 

This question is Similar to Misc-16 - Class-12 Miscellaneous

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Question 15 The general solution of the differential equation 𝑦𝑑π‘₯ βˆ’ π‘₯𝑑𝑦 = 0 𝑖𝑠 (a) π‘₯𝑦 = 𝐢 (b) π‘₯ = 〖𝐢𝑦〗^2 (c) 𝑦 = 𝐢π‘₯ (d) 𝑦 = Cπ‘₯^2𝑦 𝑑π‘₯ βˆ’ π‘₯ 𝑑𝑦=0 𝑦 𝑑π‘₯= π‘₯ 𝑑𝑦 𝒅𝒙/𝒙 = π’…π’š/π’š Integrating both sides. ∫1β–’γ€–(𝑑π‘₯ )/π‘₯=(𝑑𝑦 )/(𝑦 )γ€— log x = log y + C log x βˆ’ log y = C log ((𝒙 )/π’š) = C (𝒙 )/π’š = c1 y = π‘₯/𝑐_1 y = Cx So, the correct answer is (c)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.