Misc 2 - Verify given function is a solution of differential - Gen and Particular Solution

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Misc 2 For each of the exercise given below , verify that the given function is a solution of the corresponding differential equation . (i) = + + 2 : 2 2 +2 + 2 2=0 = + + 2 = + + 2 = + 1 +2 = +2 2 2 = = +2 = 1 +2 = + +2 Now , We have to Prove 2 2 +2 + 2 2=0 Solving L.H.S 2 2 + 2 + 2 2 = + +2 +2 +2 + + 2 + 2 2 = +2 + 2 +2 +4 3 + 2 2 = +2 + 2 +2 +4 3 + 2 2 = 2 + 2 +6 3 + 2 2 =2 2 3 + 2 +6 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 is a solution of the corresponding differential equation . (ii) = cos + sin : 2 2 2 +2 =0 = cos + sin : = acos + = acos + . + acos + = acos + + asin + = acos + asin + = + cos + = = + cos + = + cos + . + . + cos + = + cos + + + sin + = + + cos + = 2b cos 2 Now we have to Prove 2 2 2 +2 =0 Solving LHS 2 2 2 +2 = 2 2 asin 2 + cos + sin + 2 + = [2 2 + cos + cos + +2 acos 2 2 sin +2 ] = . 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 is a solution of the corresponding differential equation . (iii) = sin 3 : 2 2 +9 6 cos 3 =0 = sin 3 = 3 = sin 3 . + . sin 3 = sin 3 +3 3 = = sin 3 +3 cos 3 =3 cos 3 +3 cos 3 +3.3 sin 3 =6 cos 3 9 sin 3 Now , we have to Prove 2 2 +9 6 cos 3 =0 Solving LHS 2 2 +9 6 cos 3 =6 cos 3 9 3 +9 sin 3 6 cos 3 =6 cos 3 6 cos 3 9 3 +9 3 =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 is a solution of the corresponding differential equation . (iv) 2 =2 2 log : 2 + 2 =0 2 =2 2 log Differentiating Both sides w.r.t. x 2 = 2 2 log 2 = log . 2 2 +2 2 ( log ) 2 = log . 2 . + 2 2 1 . 2 =4 . + 2 y 2 = 4 . +2 2 4 + 2 = = 2 + Now we Have to Prove 2 + 2 =0 2 + 2 = = 2 + 2 Solving RHS 2 + 2 Putting Value of 2 =2 2 in denominator = 2 2 + 2 = 2 + = 2 + Now taking LHS = 2 + = RHS

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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.