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Example 3 Verify that the function 𝑦=π‘Ž cos⁑〖π‘₯+𝑏 sin⁑〖π‘₯, γ€— γ€— where , π‘Ž, π‘βˆˆπ‘ is a solution of the differential equation (𝑑^2 𝑦)/(𝑑π‘₯^2 )+𝑦=0 𝑦=π‘Ž cos⁑〖π‘₯+𝑏 sin⁑〖π‘₯ γ€— γ€— π’…π’š/𝒅𝒙=𝑑/𝑑π‘₯ (π‘Ž cos⁑〖π‘₯+𝑏 sin⁑〖π‘₯ γ€— γ€— ) =π‘Ž 𝑑(cos⁑π‘₯ )/𝑑π‘₯+𝑏 𝑑(sin⁑π‘₯ )/𝑑π‘₯ =π‘Ž(γ€–βˆ’sin〗⁑π‘₯ )+𝑏(cos⁑π‘₯ ) =βˆ’π’‚ π’”π’Šπ’π’™+𝒃 𝒄𝒐𝒔𝒙 Now, (𝑑^2 𝑦)/(𝑑π‘₯^2 )=𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) =𝒅/𝒅𝒙 (βˆ’π’‚ π’”π’Šπ’π’™+𝒃 𝒄𝒐𝒔𝒙) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) =βˆ’π‘Ž 𝑑(sin⁑π‘₯ )/𝑑π‘₯+𝑏 (𝑑(cos⁑π‘₯))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) =βˆ’π‘Ž(cos⁑π‘₯ )+𝑏(βˆ’sin⁑π‘₯) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) =βˆ’π‘Ž π‘π‘œπ‘ β‘γ€–π‘₯βˆ’π‘ 𝑠𝑖𝑛⁑π‘₯ γ€— (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) =βˆ’(𝒂 𝒄𝒐𝒔⁑〖𝒙+𝒃 π’”π’Šπ’β‘π’™ γ€— ) Putting y = π‘Ž cos⁑〖π‘₯+𝑏 sin⁑〖π‘₯ γ€— γ€— (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 )=βˆ’π’š (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 )+π’š=𝟎 ∴ Hence Verified

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo