Misc 2 - Class 12

Transcript

Misc 2 For each of the exercise given below , verify that the given function 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑜𝑟 𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡﷯ is a solution of the corresponding differential equation . (i) 𝑦=𝑎 𝑒﷮𝑥﷯+𝑏 𝑒﷮−𝑥﷯+ 𝑥﷮2﷯ : 𝑥 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯+2 𝑑𝑦﷮𝑑𝑥﷯−𝑥𝑦+ 𝑥﷮2﷯−2=0 𝑦=𝑎 𝑒﷮𝑥﷯+𝑏 𝑒﷮−𝑥﷯+ 𝑥﷮2﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑﷮𝑑𝑥﷯ 𝑎 𝑒﷮𝑥﷯+𝑏 𝑒﷮−𝑥﷯+ 𝑥﷮2﷯﷯ =𝑎 𝑒﷮𝑥﷯+ −1﷯𝑏 𝑒﷮−𝑥﷯+2𝑥 =𝑎 𝑒﷮𝑥﷯−𝑏 𝑒﷮−𝑥﷯+2𝑥 𝑑 𝑦﷮2﷯﷮𝑑 𝑥﷮2﷯﷯= 𝑑﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 𝑑﷮𝑑𝑥﷯ 𝑎 𝑒﷮𝑥﷯−𝑏 𝑒﷮−𝑥﷯+2𝑥 ﷯ =𝑎 𝑒﷮𝑥﷯− −1﷯𝑏 𝑒﷮−𝑥﷯+2 =𝑎 𝑒﷮𝑥﷯+𝑏 𝑒﷮−𝑥﷯+2 Now , We have to Prove 𝑥 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯+2 𝑑𝑦﷮𝑑𝑥﷯−𝑥𝑦+ 𝑥﷮2﷯−2=0 Solving L.H.S 𝑥 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯+ 2𝑑𝑦﷮𝑑𝑥﷯−𝑥𝑦+ 𝑥﷮2﷯−2 =𝑥 𝑎 𝑒﷮𝑥﷯+𝑏 𝑒﷮− 𝑥﷯+2﷯+2 𝑎 𝑒﷮𝑥﷯−𝑏 𝑒﷮− 𝑥﷯+2𝑥﷯−𝑥 𝑎 𝑒﷮𝑥﷯+𝑏 𝑒﷮− 𝑥﷯+ 𝑥﷮2﷯﷯ + 𝑥﷮2﷯− 2 =𝑥 𝑎 𝑒﷮𝑥﷯﷯+2 𝑎 𝑒﷮𝑥﷯﷯−𝑥 𝑎 𝑒﷮𝑥﷯﷯+𝑥 𝑏 𝑒﷮− 𝑥﷯﷯−2𝑏 𝑒﷮− 𝑥﷯−𝑥 𝑏 𝑒﷮− 𝑥﷯﷯+2𝑥+4𝑥− 𝑥﷮3﷯+ 𝑥﷮2﷯−2 =𝑎 𝑒﷮𝑥﷯ 𝑥+2−𝑥﷯+𝑏 𝑒﷮−𝑥﷯ 𝑥−2−𝑥﷯+2𝑥+4𝑥− 𝑥﷮3﷯+ 𝑥﷮2﷯−2 =𝑎 𝑒﷮𝑥﷯ 2﷯+𝑏 𝑒﷮−𝑥﷯ −2﷯+6𝑥− 𝑥﷮3﷯+ 𝑥﷮2﷯−2 =2𝑎 𝑒﷮𝑥﷯−2𝑏 𝑒﷮−𝑥﷯− 𝑥﷮3﷯+ 𝑥﷮2﷯+6𝑥−2 ≠ R.H.S ∴ LHS ≠ RHS The given function is not a solution of given differential equation Misc 2 (Method 2) For each of the exercise given below , verify that the given function 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑜𝑟 𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡﷯ is a solution of the corresponding differential equation . (ii) 𝑦= 𝑒﷮𝑥﷯ 𝑎 cos﷮𝑥+𝑏 sin﷮𝑥﷯﷯﷯ : 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯−2 𝑑𝑦﷮𝑑𝑥﷯+2𝑦=0 𝑦= 𝑒﷮𝑥﷯ 𝑎 cos﷮𝑥+𝑏 sin﷮𝑥﷯﷯﷯ : 𝑑𝑦﷮𝑑𝑥﷯= 𝑑﷮𝑑𝑥﷯ 𝑒﷮𝑥﷯ acos﷮𝑥+𝑏𝑠𝑖𝑛 𝑥﷯﷯﷯ = acos﷮𝑥+𝑏𝑠𝑖𝑛 𝑥﷯﷯ . 𝑑 𝑒﷮𝑥﷯﷮𝑑𝑥﷯+ 𝑒﷮𝑥﷯ 𝑑﷮𝑑𝑥﷯ acos﷮𝑥+𝑏𝑠𝑖𝑛 𝑥﷯﷯ = acos﷮𝑥+𝑏𝑠𝑖𝑛 𝑥﷯﷯ 𝑒﷮𝑥﷯+ 𝑒﷮𝑥﷯ − asin﷮𝑥+𝑏𝑐𝑜𝑠 𝑥﷯﷯ = 𝑒﷮𝑥﷯ acos﷮𝑥+𝑏𝑠𝑖𝑛 𝑥﷯− asin﷮𝑥+𝑏𝑐𝑜𝑠 𝑥﷯﷯ = 𝑒﷮𝑥﷯ 𝑎+𝑏﷯cos﷮𝑥+ 𝑏−𝑎﷯𝑠𝑖𝑛 𝑥﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 𝑑﷮𝑑𝑥﷯ 𝑒﷮𝑥﷯ 𝑎+𝑏﷯cos﷮𝑥+ 𝑏−𝑎﷯𝑠𝑖𝑛 𝑥﷯﷯﷯ = 𝑎+𝑏﷯cos﷮𝑥+ 𝑏−𝑎﷯𝑠𝑖𝑛 𝑥﷯﷯ . 𝑑﷮𝑑𝑥﷯ 𝑒﷮𝑥﷯+ 𝑒﷮𝑥﷯ . 𝑑﷮𝑑𝑥﷯ 𝑎+𝑏﷯cos﷮𝑥+ 𝑏−𝑎﷯𝑠𝑖𝑛 𝑥﷯﷯﷯ = 𝑎+𝑏﷯cos﷮𝑥+ 𝑏−𝑎﷯𝑠𝑖𝑛 𝑥﷯﷯ 𝑒﷮𝑥﷯+ 𝑒﷮𝑥﷯ − 𝑎+𝑏﷯sin﷮𝑥+ 𝑏−𝑎﷯𝑐𝑜𝑠 𝑥﷯﷯ = 𝑒﷮𝑥﷯ 𝑎+𝑏+𝑏−𝑎﷯ cos﷮𝑥+ 𝑏−𝑎−𝑎−𝑏﷯ 𝑠𝑖𝑛 𝑥﷯﷯ = 𝑒﷮𝑥﷯ 2b cos﷮𝑥−2𝑎𝑠𝑖𝑛 𝑥﷯﷯ Now we have to Prove 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯− 2𝑑𝑦﷮𝑑𝑥﷯+2𝑦=0 Solving LHS 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯− 2𝑑𝑦﷮𝑑𝑥﷯+2𝑦 = 𝑒﷮𝑥﷯ 2𝑏 𝑐𝑜𝑠𝑥−2 asin﷮𝑥﷯﷯ − 2 𝑒﷮𝑥﷯ 𝑎+𝑏﷯ cos﷮𝑥﷯+ 𝑏−𝑎﷯ sin﷮𝑥﷯﷯﷯ + 2 𝑒﷮𝑥﷯ 𝑎𝑐𝑜𝑠 𝑥+𝑏𝑠𝑖𝑛 𝑥﷯ = 𝑒﷮𝑥﷯ [2𝑏𝑐𝑜𝑠 𝑥−2 𝑎+𝑏﷯ cos﷮𝑥 ﷯+ cos﷮𝑥 ﷯+ +2 acos﷮𝑥−2𝑎𝑠𝑖𝑛 𝑥﷯−2 𝑏−𝑎﷯ sin﷮𝑥+2𝑏 𝑠𝑖𝑛 𝑥﷯] = 𝑒﷮𝑥﷯ . 0 =0 = R.H.S ∴ LHS = RHS So, The Given Function is a solution of the Differential Equation Misc 2 For each of the exercise given below , verify that the given function 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑜𝑟 𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡﷯ is a solution of the corresponding differential equation . (iii) 𝑦=𝑥 sin﷮3𝑥﷯ : 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯+9𝑦−6 cos﷮3𝑥=0﷯ 𝑦=𝑥 sin﷮3𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑﷮𝑑𝑥﷯ 𝑥𝑠𝑖𝑛 3𝑥﷯ = sin﷮3𝑥﷯. 𝑑𝑥﷮𝑑𝑥﷯+𝑥 . 𝑑﷮𝑑𝑥﷯ sin﷮3𝑥﷯﷯ = sin﷮3𝑥﷯+3𝑥 𝑐𝑜𝑠3𝑥 𝑑𝑦﷮𝑑𝑥﷯= 𝑑﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 𝑑﷮𝑑𝑥﷯ sin﷮3𝑥+3𝑥 cos﷮3𝑥﷯﷯﷯ =3 cos﷮3𝑥+3 cos﷮3𝑥+3.3𝑥 − sin﷮3𝑥﷯﷯﷯﷯ =6 cos﷮3𝑥−9𝑥 sin﷮3𝑥﷯﷯ Now , we have to Prove 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯+9𝑦−6 cos﷮3𝑥=0﷯ Solving LHS 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯+9𝑦−6 cos﷮3𝑥﷯ =6 cos﷮3𝑥﷯−9𝑥𝑠𝑖𝑛 3𝑥+9 𝑥 sin﷮3𝑥﷯﷯−6 cos﷮3𝑥﷯ =6 cos﷮3𝑥﷯−6 cos﷮3𝑥﷯−9𝑥𝑠𝑖𝑛 3𝑥+9𝑥𝑠𝑖𝑛 3𝑥 =0 = RHS ∴ LHS = RHS So, The Given Function is a solution of the Differential Equation Misc 2 For each of the exercise given below , verify that the given function 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑜𝑟 𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡﷯ is a solution of the corresponding differential equation . (iv) 𝑥﷮2﷯=2 𝑦﷮2﷯ log﷮𝑦﷯ : 𝑥﷮2﷯+ 𝑦﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯−𝑥𝑦=0 𝑥﷮2﷯=2 𝑦﷮2﷯ log﷮𝑦﷯ Differentiating Both sides w.r.t. x 𝑑﷮𝑑𝑥﷯ 𝑥﷮2﷯﷯= 𝑑﷮𝑑𝑥﷯ 2 𝑦﷮2﷯ log﷮𝑦﷯﷯ 2𝑥= log﷮𝑦﷯ . 𝑑﷮𝑑𝑥﷯ 2 𝑦﷮2﷯﷯+2 𝑦﷮2﷯ 𝑑( log﷮𝑦)﷯﷮𝑑𝑥﷯ 2𝑥= log﷮𝑦﷯ . 2𝑦 . 𝑑𝑦﷮𝑑𝑥﷯ + 2 𝑦﷮2﷯ 1﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 2𝑥=4𝑦 . 𝑙𝑜𝑔𝑦 𝑑𝑦﷮𝑑𝑥﷯ + 2𝑦 𝑑﷮𝑑𝑥﷯ y 2𝑥= 4𝑦 . 𝑙𝑜𝑔𝑦+2𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯ 2𝑥﷮4𝑦 𝑙𝑜𝑔𝑦 + 2𝑦﷯= 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑥﷮2𝑦 𝑙𝑜𝑔𝑦 + 𝑦﷯ Now we Have to Prove 𝑥﷮2﷯+ 𝑦﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯−𝑥𝑦=0 𝑥﷮2﷯+ 𝑦﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯=𝑥𝑦 𝑑𝑦﷮𝑑𝑥﷯= 𝑥𝑦﷮ 𝑥﷮2﷯ + 𝑦﷮2﷯﷯ Solving RHS 𝑥𝑦﷮ 𝑥﷮2﷯ + 𝑦﷮2﷯﷯ Putting Value of 𝑥﷮2﷯=2 𝑦﷮2﷯𝑏 𝑙𝑜𝑔𝑦 in denominator = 𝑥𝑦﷮2 𝑦﷮2﷯ 𝑙𝑜𝑔𝑦 + 𝑦﷮2﷯﷯ = 𝑥𝑦﷮𝑦 2𝑦 𝑙𝑜𝑔𝑦 + 𝑦﷯﷯ = 𝑥﷮2𝑦 𝑙𝑜𝑔𝑦 + 𝑦﷯ Now taking LHS 𝑑𝑦﷮𝑑𝑥﷯= 𝑥﷮2𝑦 𝑙𝑜𝑔𝑦 + 𝑦﷯ = RHS