# Misc 2 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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 𝑒𝑥+𝑏 𝑒−𝑥+ 𝑥2 : 𝑥 𝑑2𝑦𝑑 𝑥2+2 𝑑𝑦𝑑𝑥−𝑥𝑦+ 𝑥2−2=0 𝑦=𝑎 𝑒𝑥+𝑏 𝑒−𝑥+ 𝑥2 𝑑𝑦𝑑𝑥= 𝑑𝑑𝑥 𝑎 𝑒𝑥+𝑏 𝑒−𝑥+ 𝑥2 =𝑎 𝑒𝑥+ −1𝑏 𝑒−𝑥+2𝑥 =𝑎 𝑒𝑥−𝑏 𝑒−𝑥+2𝑥 𝑑 𝑦2𝑑 𝑥2= 𝑑𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑑𝑥 𝑎 𝑒𝑥−𝑏 𝑒−𝑥+2𝑥 =𝑎 𝑒𝑥− −1𝑏 𝑒−𝑥+2 =𝑎 𝑒𝑥+𝑏 𝑒−𝑥+2 Now , We have to Prove 𝑥 𝑑2𝑦𝑑 𝑥2+2 𝑑𝑦𝑑𝑥−𝑥𝑦+ 𝑥2−2=0 Solving L.H.S 𝑥 𝑑2𝑦𝑑 𝑥2+ 2𝑑𝑦𝑑𝑥−𝑥𝑦+ 𝑥2−2 =𝑥 𝑎 𝑒𝑥+𝑏 𝑒− 𝑥+2+2 𝑎 𝑒𝑥−𝑏 𝑒− 𝑥+2𝑥−𝑥 𝑎 𝑒𝑥+𝑏 𝑒− 𝑥+ 𝑥2 + 𝑥2− 2 =𝑥 𝑎 𝑒𝑥+2 𝑎 𝑒𝑥−𝑥 𝑎 𝑒𝑥+𝑥 𝑏 𝑒− 𝑥−2𝑏 𝑒− 𝑥−𝑥 𝑏 𝑒− 𝑥+2𝑥+4𝑥− 𝑥3+ 𝑥2−2 =𝑎 𝑒𝑥 𝑥+2−𝑥+𝑏 𝑒−𝑥 𝑥−2−𝑥+2𝑥+4𝑥− 𝑥3+ 𝑥2−2 =𝑎 𝑒𝑥 2+𝑏 𝑒−𝑥 −2+6𝑥− 𝑥3+ 𝑥2−2 =2𝑎 𝑒𝑥−2𝑏 𝑒−𝑥− 𝑥3+ 𝑥2+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= 𝑒𝑥 𝑎 cos𝑥+𝑏 sin𝑥 : 𝑑2𝑦𝑑 𝑥2−2 𝑑𝑦𝑑𝑥+2𝑦=0 𝑦= 𝑒𝑥 𝑎 cos𝑥+𝑏 sin𝑥 : 𝑑𝑦𝑑𝑥= 𝑑𝑑𝑥 𝑒𝑥 acos𝑥+𝑏𝑠𝑖𝑛 𝑥 = acos𝑥+𝑏𝑠𝑖𝑛 𝑥 . 𝑑 𝑒𝑥𝑑𝑥+ 𝑒𝑥 𝑑𝑑𝑥 acos𝑥+𝑏𝑠𝑖𝑛 𝑥 = acos𝑥+𝑏𝑠𝑖𝑛 𝑥 𝑒𝑥+ 𝑒𝑥 − asin𝑥+𝑏𝑐𝑜𝑠 𝑥 = 𝑒𝑥 acos𝑥+𝑏𝑠𝑖𝑛 𝑥− asin𝑥+𝑏𝑐𝑜𝑠 𝑥 = 𝑒𝑥 𝑎+𝑏cos𝑥+ 𝑏−𝑎𝑠𝑖𝑛 𝑥 𝑑𝑦𝑑𝑥= 𝑑𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑑𝑥 𝑒𝑥 𝑎+𝑏cos𝑥+ 𝑏−𝑎𝑠𝑖𝑛 𝑥 = 𝑎+𝑏cos𝑥+ 𝑏−𝑎𝑠𝑖𝑛 𝑥 . 𝑑𝑑𝑥 𝑒𝑥+ 𝑒𝑥 . 𝑑𝑑𝑥 𝑎+𝑏cos𝑥+ 𝑏−𝑎𝑠𝑖𝑛 𝑥 = 𝑎+𝑏cos𝑥+ 𝑏−𝑎𝑠𝑖𝑛 𝑥 𝑒𝑥+ 𝑒𝑥 − 𝑎+𝑏sin𝑥+ 𝑏−𝑎𝑐𝑜𝑠 𝑥 = 𝑒𝑥 𝑎+𝑏+𝑏−𝑎 cos𝑥+ 𝑏−𝑎−𝑎−𝑏 𝑠𝑖𝑛 𝑥 = 𝑒𝑥 2b cos𝑥−2𝑎𝑠𝑖𝑛 𝑥 Now we have to Prove 𝑑2𝑦𝑑 𝑥2− 2𝑑𝑦𝑑𝑥+2𝑦=0 Solving LHS 𝑑2𝑦𝑑 𝑥2− 2𝑑𝑦𝑑𝑥+2𝑦 = 𝑒𝑥 2𝑏 𝑐𝑜𝑠𝑥−2 asin𝑥 − 2 𝑒𝑥 𝑎+𝑏 cos𝑥+ 𝑏−𝑎 sin𝑥 + 2 𝑒𝑥 𝑎𝑐𝑜𝑠 𝑥+𝑏𝑠𝑖𝑛 𝑥 = 𝑒𝑥 [2𝑏𝑐𝑜𝑠 𝑥−2 𝑎+𝑏 cos𝑥 + cos𝑥 + +2 acos𝑥−2𝑎𝑠𝑖𝑛 𝑥−2 𝑏−𝑎 sin𝑥+2𝑏 𝑠𝑖𝑛 𝑥] = 𝑒𝑥 . 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 sin3𝑥 : 𝑑2𝑦𝑑 𝑥2+9𝑦−6 cos3𝑥=0 𝑦=𝑥 sin3𝑥 𝑑𝑦𝑑𝑥= 𝑑𝑑𝑥 𝑥𝑠𝑖𝑛 3𝑥 = sin3𝑥. 𝑑𝑥𝑑𝑥+𝑥 . 𝑑𝑑𝑥 sin3𝑥 = sin3𝑥+3𝑥 𝑐𝑜𝑠3𝑥 𝑑𝑦𝑑𝑥= 𝑑𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑑𝑥 sin3𝑥+3𝑥 cos3𝑥 =3 cos3𝑥+3 cos3𝑥+3.3𝑥 − sin3𝑥 =6 cos3𝑥−9𝑥 sin3𝑥 Now , we have to Prove 𝑑2𝑦𝑑 𝑥2+9𝑦−6 cos3𝑥=0 Solving LHS 𝑑2𝑦𝑑 𝑥2+9𝑦−6 cos3𝑥 =6 cos3𝑥−9𝑥𝑠𝑖𝑛 3𝑥+9 𝑥 sin3𝑥−6 cos3𝑥 =6 cos3𝑥−6 cos3𝑥−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) 𝑥2=2 𝑦2 log𝑦 : 𝑥2+ 𝑦2 𝑑𝑦𝑑𝑥−𝑥𝑦=0 𝑥2=2 𝑦2 log𝑦 Differentiating Both sides w.r.t. x 𝑑𝑑𝑥 𝑥2= 𝑑𝑑𝑥 2 𝑦2 log𝑦 2𝑥= log𝑦 . 𝑑𝑑𝑥 2 𝑦2+2 𝑦2 𝑑( log𝑦)𝑑𝑥 2𝑥= log𝑦 . 2𝑦 . 𝑑𝑦𝑑𝑥 + 2 𝑦2 1𝑦 . 𝑑𝑦𝑑𝑥 2𝑥=4𝑦 . 𝑙𝑜𝑔𝑦 𝑑𝑦𝑑𝑥 + 2𝑦 𝑑𝑑𝑥 y 2𝑥= 4𝑦 . 𝑙𝑜𝑔𝑦+2𝑦 𝑑𝑦𝑑𝑥 2𝑥4𝑦 𝑙𝑜𝑔𝑦 + 2𝑦= 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥= 𝑥2𝑦 𝑙𝑜𝑔𝑦 + 𝑦 Now we Have to Prove 𝑥2+ 𝑦2 𝑑𝑦𝑑𝑥−𝑥𝑦=0 𝑥2+ 𝑦2 𝑑𝑦𝑑𝑥=𝑥𝑦 𝑑𝑦𝑑𝑥= 𝑥𝑦 𝑥2 + 𝑦2 Solving RHS 𝑥𝑦 𝑥2 + 𝑦2 Putting Value of 𝑥2=2 𝑦2𝑏 𝑙𝑜𝑔𝑦 in denominator = 𝑥𝑦2 𝑦2 𝑙𝑜𝑔𝑦 + 𝑦2 = 𝑥𝑦𝑦 2𝑦 𝑙𝑜𝑔𝑦 + 𝑦 = 𝑥2𝑦 𝑙𝑜𝑔𝑦 + 𝑦 Now taking LHS 𝑑𝑦𝑑𝑥= 𝑥2𝑦 𝑙𝑜𝑔𝑦 + 𝑦 = RHS

Chapter 9 Class 12 Differential Equations

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.