Example 24 - Chapter 9 Class 12 Differential Equations

Transcript

Example 24 Verify that the function 𝑦=𝑐1 𝑒^𝑎𝑥 cos⁡〖𝑏𝑥+𝑐2 𝑒^𝑎𝑥 sin⁡𝑏𝑥 〗 , 𝑤ℎ𝑒𝑟𝑒 𝑐1 , 𝑐2 are arbitrary constants is a solution of the differential equation (𝑑^2 𝑦)/(𝑑𝑥^2 )−2𝑎 𝑑𝑦/𝑑𝑥+(𝑎^2+𝑏^2 )𝑦=0 𝑦=𝑐1 𝑒^𝑎𝑥 cos⁡〖𝑏𝑥+𝑐2 𝑒^𝑎𝑥 sin⁡〖𝑏𝑥, 〗 〗 𝑦 =𝑒^𝑎𝑥 (𝑐1 𝑐𝑜𝑠 𝑏𝑥+ 𝑐2 sin⁡𝑏𝑥) Differentiating w.r.t. x 𝑦^′=(𝑒^𝑎𝑥 )^′ (𝑐_1 cos⁡𝑏𝑥+𝑐_2 sin⁡𝑏𝑥 )+𝑒^𝑎𝑥 (𝑐_1 cos⁡𝑏𝑥+𝑐_2 sin⁡𝑏𝑥 )^′ 𝑦^′=𝑎𝑒^𝑎𝑥 (𝑐_1 cos⁡𝑏𝑥+𝑐_2 sin⁡𝑏𝑥 )+𝑒^𝑎𝑥 (〖−𝑐〗_1 𝑏 sin⁡𝑏𝑥+𝑐_2 𝑏 cos⁡𝑏𝑥 ) 𝑦^′=𝑎𝑒^𝑎𝑥 (𝑐_1 cos⁡𝑏𝑥+𝑐_2 sin⁡𝑏𝑥 )+𝑏𝑒^𝑎𝑥 (〖−𝑐〗_1 sin⁡𝑏𝑥+𝑐_2 cos⁡𝑏𝑥 ) Putting 𝑦 =𝑒^𝑎𝑥 (𝑐1 𝑐𝑜𝑠 𝑏𝑥+ 𝑐2 sin⁡𝑏𝑥) 𝑦^′=𝑎𝑦+𝑏𝑒^𝑎𝑥 (〖−𝑐〗_1 sin⁡𝑏𝑥+𝑐_2 cos⁡𝑏𝑥 ) 𝑦^′−𝑎𝑦=𝑏𝑒^𝑎𝑥 (〖−𝑐〗_1 sin⁡𝑏𝑥+𝑐_2 cos⁡𝑏𝑥 ) Differentiating again w.r.t x 𝑦^′′−𝑎𝑦^′=(𝑏𝑒^𝑎𝑥 )^′ (〖−𝑐〗_1 sin⁡𝑏𝑥+𝑐_2 cos⁡𝑏𝑥 )+𝑏𝑒^𝑎𝑥 (〖−𝑐〗_1 sin⁡𝑏𝑥+𝑐_2 cos⁡𝑏𝑥 )^′ 𝑦^′′−𝑎𝑦^′=𝑎𝑏𝑒^𝑎𝑥 (〖−𝑐〗_1 sin⁡𝑏𝑥+𝑐_2 cos⁡𝑏𝑥 )+𝑏𝑒^𝑎𝑥 (〖−𝑐〗_1 𝑏 cos⁡𝑏𝑥−𝑐_2 𝑏 sin⁡𝑏𝑥 ) 𝑦^′′−𝑎𝑦^′=𝑎𝑏𝑒^𝑎𝑥 (〖−𝑐〗_1 sin⁡𝑏𝑥+𝑐_2 cos⁡𝑏𝑥 )−𝑏^2 𝒆^𝒂𝒙 (𝒄_𝟏 𝒄𝒐𝒔⁡𝒃𝒙+𝒄_𝟐 𝒔𝒊𝒏⁡𝒃𝒙 ) Putting 𝑦 =𝑒^𝑎𝑥 (𝑐1 𝑐𝑜𝑠 𝑏𝑥+ 𝑐2 sin⁡𝑏𝑥) 𝑦^′′−𝑎𝑦^′=𝑎𝒃𝒆^𝒂𝒙 (〖−𝒄〗_𝟏 𝒔𝒊𝒏⁡𝒃𝒙+𝒄_𝟐 𝒄𝒐𝒔⁡𝒃𝒙 )−𝑏^2 𝑦 "Putting" 𝑦^′−𝑎𝑦=𝑏𝑒^𝑎𝑥 (〖−𝑐〗_1 𝑠𝑖𝑛⁡𝑏𝑥+𝑐_2 𝑐𝑜𝑠⁡𝑏𝑥 ) 𝑦^′′−𝑎𝑦^′=𝑎(𝑦^′−𝑎𝑦)−𝑏^2 𝑦 𝑦^′′−𝑎𝑦^′=𝑎𝑦^′−𝑎^2 𝑦−𝑏^2 𝑦 𝑦^′′−𝑎𝑦^′−𝑎𝑦^′+𝑎^2 𝑦+𝑏^2 𝑦=0 𝒚^′′−𝟐𝒂𝒚^′+(𝒂^𝟐+𝒃^𝟐)𝒚=𝟎 Hence verified