Slide7.JPG

Slide8.JPG
Slide9.JPG

Go Ad-free

Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo