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Misc 6 Find the equation of the curve passing through the point (0 , πœ‹/4) whose differential equation is sin π‘₯ cos⁑〖𝑦 𝑑π‘₯+cos⁑〖π‘₯ sin⁑〖𝑦 𝑑𝑦=0γ€— γ€— γ€— sin x cos y dx + cos x sin y dy = 0 sin x cos y dx = βˆ’ cos x sin y dy 〖𝐬𝐒𝐧 〗⁑𝒙/πœπ¨π¬β‘π’™ 𝒅𝒙 = βˆ’ π’”π’Šπ’β‘π’š/π’„π’π’”β‘π’š π’…π’š Put cos x = u Diff w.r.t. x 𝑑/𝑑π‘₯ cos⁑〖π‘₯=𝑑𝑒/𝑑π‘₯ γ€— βˆ’sin x = 𝑑𝑒/𝑑π‘₯ dx = (βˆ’π’…π’–)/π’”π’Šπ’β‘π’™ Put cos y = v Diff w.r.t. y 𝑑/𝑑𝑦 cos⁑〖𝑦=𝑑𝑣/𝑑𝑦 γ€— βˆ’sin y = 𝑑𝑣/𝑑𝑦 dy = (βˆ’π’…π’—)/π’”π’Šπ’β‘π’š Integrating both sides ∫1β–’sin⁑〖π‘₯ 𝑑π‘₯γ€—/cos⁑π‘₯ = ∫1β–’γ€–βˆ’sin〗⁑〖𝑦 𝑑𝑦〗/cos⁑𝑦 ∴ ∫1β–’γ€–sin⁑π‘₯/𝑒 Γ— (βˆ’π‘‘π‘’)/sin⁑π‘₯ γ€—= ∫1β–’γ€–(βˆ’sin⁑𝑦)/𝑣 Γ— (βˆ’π‘‘π‘£)/sin⁑𝑦 γ€— ∫1▒𝒅𝒗/𝒖=βˆ’βˆ«1▒𝒅𝒖/𝒗 log |𝑒|=βˆ’π‘™π‘œπ‘”|𝑣|+𝑐 log |𝑒| + log |𝑣| = c log |𝑒.𝑣|=𝑐 Putting back values of u and v. log |πœπ¨π¬β‘γ€–π’™.πœπ¨π¬β‘π’š γ€— |= βˆ’c Since the curve passes through (𝟎, 𝝅/πŸ’) Putting x = 0 and y = πœ‹/4 in (1) log |cos⁑(0).cos⁑(πœ‹/4)|=𝑐 log |1. 1/√2|=𝑐 ∴ c = log 𝟏/√𝟐 Substitute value of C in (2) log |cos⁑〖π‘₯ cos⁑𝑦 γ€— |=𝑐 log |πœπ¨π¬β‘γ€–π’™.πœπ¨π¬β‘π’š γ€— |=π₯𝐨𝐠⁑|𝟏/√𝟐| ∴ cos x. cos y = 𝟏/√𝟐 cos y = 1/(√2 cos⁑π‘₯ ) cos y = (𝒔𝒆𝒄 𝒙)/√𝟐

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Davneet Singh

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.