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﷮𝑑 𝑥﷮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 𝑐𝑜𝑠𝑏 𝑥+ 𝑐2 sin﷮𝑏𝑥﷯) 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮𝑎𝑥﷯. 𝑎 𝑐1 cos﷮𝑏𝑥+𝑐2 sin﷮𝑏𝑥, ﷯﷯﷯ + 𝑒﷮𝑎𝑥﷯ 𝑐1,𝑏 sin﷮𝑏 𝑥+𝑐2 b cos﷮𝑏𝑥﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮𝑎𝑥﷯𝑎 𝑐1 cos﷮𝑏𝑥+𝑐2 sin﷮𝑏𝑥, ﷯﷯﷯ + 𝑏𝑒﷮𝑎𝑥﷯ 𝑐2 cos﷮𝑏 𝑥−𝑐1 sin﷮𝑏𝑥﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮𝑎𝑥﷯( 𝑎𝑐﷮1﷯ cos﷮𝑏𝑥+𝑎𝑐2 sin 𝑏𝑥+𝑏𝑐2 cos﷮𝑏𝑥−﷯𝑏𝑐1 sin﷮𝑏𝑥﷯﷯) 𝑑𝑦﷮𝑑𝑥﷯ = a 𝑒﷮𝑎𝑥﷯( 𝑐﷮1﷯ cos﷮𝑏𝑥+𝑐2 sin 𝑏𝑥﷯) +𝑏 𝑒﷮𝑎𝑥﷯ ( 𝐶﷮2﷯cos﷮𝑏𝑥−﷯𝑐1 sin﷮𝑏𝑥﷯) 𝑑𝑦﷮𝑑𝑥﷯ = 𝑎𝑦 +﷮𝑏 𝑒﷮𝑎𝑥﷯ ( 𝑐﷮2﷯cos﷮𝑏𝑥−﷯𝑐1 sin﷮𝑏𝑥﷯﷯) Differentiating with respect to x 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯=𝑎 𝑑𝑦﷮𝑑𝑥﷯+ 𝑎𝑏𝑒﷮𝑎𝑥﷯ 𝑐﷮2﷯ cos﷮𝑏𝑥− 𝑐﷮1﷯ sin﷮𝑏𝑥﷯﷯﷯+ 𝑏𝑒﷮𝑎𝑥﷯(− 𝑐﷮2﷯ sin﷮𝑏𝑥×𝑏− 𝑐﷮1﷯ cos﷮𝑏𝑥﷯ ×𝑏﷯) 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯=𝑎 𝑑𝑦﷮𝑑𝑥﷯+ 𝑎𝑏𝑒﷮𝑎𝑥﷯ 𝑐﷮2﷯ cos﷮𝑏𝑥− 𝑐﷮1﷯ sin﷮𝑏𝑥﷯﷯﷯− 𝑏﷮2﷯ 𝑒﷮𝑎𝑥﷯( 𝑐﷮1﷯ cos﷮𝑏𝑥+ 𝑐﷮2﷯ sin﷮𝑏𝑥﷯﷯) 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯ =𝑎 𝑑𝑦﷮𝑑𝑥﷯+ 𝑎𝑏 𝑒﷮𝑎𝑥﷯ 𝑐﷮2﷯ cos﷮𝑏𝑥− 𝑐﷮1﷯ sin﷮𝑏𝑥﷯﷯﷯− 𝑏﷮2﷯𝑦 From (2) 𝑑𝑦﷮𝑑𝑥﷯=𝑎𝑦+ 𝑏 𝑒﷮𝑎𝑥﷯ 𝑐﷮2﷯ cos﷮𝑏𝑥− 𝑐﷮1﷯ sin﷮𝑏𝑥﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯−𝑎𝑦= 𝑏𝑒﷮𝑎𝑥﷯ 𝑐﷮2﷯ cos﷮𝑏𝑥− 𝑐﷮1﷯ sin﷮𝑏𝑥﷯﷯﷯ 𝑏𝑒﷮𝑎𝑥﷯ 𝑐﷮2﷯ cos﷮𝑏𝑥− 𝑐﷮1﷯ sin﷮𝑏𝑥﷯﷯﷯ = 𝑑𝑦﷮𝑑𝑥﷯−𝑎𝑦 Putting value in (3) 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯=𝑎 𝑑𝑦﷮𝑑𝑥﷯+ 𝑎𝒃 𝒆﷮𝒂𝒙﷯ 𝒄﷮𝟐﷯ 𝒄𝒐𝒔﷮𝒃𝒙− 𝒄﷮𝟏﷯ 𝒔𝒊𝒏﷮𝒃𝒙﷯﷯﷯− 𝑏﷮2﷯𝑦 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯= 𝑎 𝑑𝑦﷮𝑑𝑥﷯+𝑎 𝒅𝒚﷮𝒅𝒙﷯−𝒂𝒚﷯− 𝑏﷮2﷯𝑦 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯ = 𝑎 𝑑𝑦﷮𝑑𝑥﷯+𝑎 𝑑𝑦﷮𝑑𝑥﷯− 𝑎﷮2﷯𝑦− 𝑏﷮2﷯𝑦 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯ = 2𝑎 𝑑𝑦﷮𝑑𝑥﷯−𝑦 ( 𝑎﷮2﷯+ 𝑏﷮2﷯) 𝒅﷮𝟐﷯𝒚﷮ 𝒅𝒙﷮𝟐﷯﷯−𝟐𝒂 𝒅𝒚﷮𝒅𝒙﷯+𝒚 𝒂﷮𝟐﷯+ 𝒃﷮𝟐﷯﷯=𝟎 Hence, Proved.