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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 (π‘–π‘šπ‘π‘™π‘–π‘π‘–π‘‘ π‘œπ‘Ÿ 𝑒π‘₯𝑝𝑙𝑖𝑐𝑖𝑑) is a solution of the corresponding differential equation . (i) π‘₯𝑦=π‘Ž 𝑒^π‘₯+𝑏 𝑒^(βˆ’π‘₯)+π‘₯^2 : π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 )+2 𝑑𝑦/𝑑π‘₯βˆ’π‘₯𝑦+π‘₯^2βˆ’2=0 π‘₯𝑦=π‘Žπ‘’^π‘₯+𝑏 𝑒^(βˆ’π‘₯)+π‘₯^2 Differentiating w.r.t x (𝑑(π‘₯𝑦))/𝑑π‘₯=𝑑/𝑑π‘₯ [π‘Ž 𝑒^π‘₯+𝑏 𝑒^(βˆ’π‘₯)+π‘₯^2 ] 𝑑π‘₯/𝑑π‘₯ y+𝑑𝑦/𝑑π‘₯ π‘₯ =π‘Žγ€– 𝑒〗^π‘₯+(βˆ’1)𝑏 𝑒^(βˆ’π‘₯)+2π‘₯ y + 𝑑𝑦/𝑑π‘₯ π‘₯ =π‘Žγ€– 𝑒〗^π‘₯βˆ’π‘ 𝑒^(βˆ’π‘₯)+2π‘₯ Differentiating again w.r.t x 𝑑𝑦/𝑑π‘₯+ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) π‘₯+ 𝑑𝑦/𝑑π‘₯×𝑑π‘₯/𝑑π‘₯=𝑑/𝑑π‘₯ [π‘Žπ‘’^π‘₯βˆ’π‘ 𝑒^(βˆ’π‘₯)+2π‘₯" " ] 𝑑𝑦/𝑑π‘₯+ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) π‘₯+ 𝑑𝑦/𝑑π‘₯=π‘Žπ‘’^π‘₯+𝑏𝑒^(βˆ’π‘₯)+2 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) π‘₯+2 𝑑𝑦/𝑑π‘₯=π‘Žπ‘’^π‘₯+𝑏𝑒^(βˆ’π‘₯)+2 Now, we know that π‘₯𝑦=π‘Žπ‘’^π‘₯+𝑏 𝑒^(βˆ’π‘₯)+π‘₯^2 π‘₯π‘¦βˆ’π‘₯^2=π‘Žπ‘’^π‘₯+𝑏 𝑒^(βˆ’π‘₯) π‘Žπ‘’^π‘₯+𝑏 𝑒^(βˆ’π‘₯)=π‘₯π‘¦βˆ’π‘₯^2 Putting (2) in (1) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) π‘₯+2 𝑑𝑦/𝑑π‘₯=𝒂𝒆^𝒙+𝒃𝒆^(βˆ’π’™)+2 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) π‘₯+2 𝑑𝑦/𝑑π‘₯=π’™π’šβˆ’π’™^𝟐+2 (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) 𝒙+𝟐 π’…π’š/π’…π’™βˆ’π’™π’š+𝒙^𝟐=𝟎 ∴ The given function is a solution Misc 2 (Method 2) For each of the exercise given below , verify that the given function (π‘–π‘šπ‘π‘™π‘–π‘π‘–π‘‘ π‘œπ‘Ÿ 𝑒π‘₯𝑝𝑙𝑖𝑐𝑖𝑑) is a solution of the corresponding differential equation . (ii) 𝑦=𝑒^π‘₯ (π‘Ž cos⁑〖π‘₯+𝑏 sin⁑π‘₯ γ€— ) : (𝑑^2 𝑦)/(𝑑π‘₯^2 )βˆ’2 𝑑𝑦/𝑑π‘₯+2𝑦=0 𝑦=𝑒^π‘₯ (π‘Ž cos⁑〖π‘₯+𝑏 sin⁑π‘₯ γ€— ) : 𝑑𝑦/𝑑π‘₯=𝑑/𝑑π‘₯ [𝑒^π‘₯ (acos⁑〖π‘₯+𝑏𝑠𝑖𝑛 π‘₯γ€— )] =(acos⁑〖π‘₯+𝑏𝑠𝑖𝑛 π‘₯γ€— ) . (𝑑𝑒^π‘₯)/𝑑π‘₯+𝑒^π‘₯ 𝑑/𝑑π‘₯ [acos⁑〖π‘₯+𝑏𝑠𝑖𝑛 π‘₯γ€— ] =(acos⁑〖π‘₯+𝑏𝑠𝑖𝑛 π‘₯γ€— ) 𝑒^π‘₯+𝑒^π‘₯ [βˆ’asin⁑〖π‘₯+π‘π‘π‘œπ‘  π‘₯γ€— ] =𝑒^π‘₯ [acos⁑〖π‘₯+𝑏𝑠𝑖𝑛 π‘₯γ€—βˆ’asin⁑〖π‘₯+π‘π‘π‘œπ‘  π‘₯γ€— ] =𝑒^π‘₯ [(π‘Ž+𝑏)cos⁑〖π‘₯+(π‘βˆ’π‘Ž)𝑠𝑖𝑛 π‘₯γ€— ] 𝑑𝑦/𝑑π‘₯=𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) =𝑑/𝑑π‘₯ [𝑒^π‘₯ [(π‘Ž+𝑏)cos⁑〖π‘₯+(π‘βˆ’π‘Ž)𝑠𝑖𝑛 π‘₯γ€— ]] =[(π‘Ž+𝑏)cos⁑〖π‘₯+(π‘βˆ’π‘Ž)𝑠𝑖𝑛 π‘₯γ€— ] . 𝑑/𝑑π‘₯ 𝑒^π‘₯+𝑒^π‘₯ . 𝑑/𝑑π‘₯ [((π‘Ž+𝑏)cos⁑〖π‘₯+(π‘βˆ’π‘Ž)𝑠𝑖𝑛 π‘₯γ€— )] =[(π‘Ž+𝑏)cos⁑〖π‘₯+(π‘βˆ’π‘Ž)𝑠𝑖𝑛 π‘₯γ€— ] 𝑒^π‘₯+𝑒^π‘₯ (βˆ’(π‘Ž+𝑏)sin⁑〖π‘₯+(π‘βˆ’π‘Ž)π‘π‘œπ‘  π‘₯γ€— ) =𝑒^π‘₯ [γ€–(π‘Ž+𝑏+π‘βˆ’π‘Ž) cos〗⁑〖π‘₯+(π‘βˆ’π‘Žβˆ’π‘Žβˆ’π‘) 𝑠𝑖𝑛 π‘₯γ€— ] =𝑒^π‘₯ [γ€–2b cos〗⁑〖π‘₯βˆ’2π‘Žπ‘ π‘–π‘› π‘₯γ€— ] Now we have to Prove (𝑑^2 𝑦)/(𝑑π‘₯^2 )βˆ’2𝑑𝑦/𝑑π‘₯+2𝑦=0 Solving LHS (𝑑^2 𝑦)/(𝑑π‘₯^2 )βˆ’2𝑑𝑦/𝑑π‘₯+2𝑦 = 𝑒^π‘₯ [2𝑏 π‘π‘œπ‘ π‘₯βˆ’2 asin⁑π‘₯ ] βˆ’ 2[𝑒^π‘₯ [(π‘Ž+𝑏) cos⁑π‘₯+(π‘βˆ’π‘Ž) sin⁑π‘₯ ]] + 2𝑒^π‘₯ [π‘Žπ‘π‘œπ‘  π‘₯+𝑏𝑠𝑖𝑛 π‘₯] =𝑒^π‘₯ [2π‘π‘π‘œπ‘  π‘₯βˆ’2(π‘Ž+𝑏) cos⁑〖π‘₯ γ€—+cos⁑〖π‘₯ γ€—+ +2 acos⁑〖π‘₯βˆ’2π‘Žπ‘ π‘–π‘› π‘₯γ€—βˆ’2(π‘βˆ’π‘Ž) sin⁑〖π‘₯+2𝑏 𝑠𝑖𝑛 π‘₯γ€—] =𝑒^π‘₯ . 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 (π‘–π‘šπ‘π‘™π‘–π‘π‘–π‘‘ π‘œπ‘Ÿ 𝑒π‘₯𝑝𝑙𝑖𝑐𝑖𝑑) is a solution of the corresponding differential equation . (iii) 𝑦=π‘₯ sin⁑3π‘₯ : (𝑑^2 𝑦)/(𝑑π‘₯^2 )+9π‘¦βˆ’6 cos⁑〖3π‘₯=0γ€— 𝑦=π‘₯ sin⁑3π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑/𝑑π‘₯ (π‘₯𝑠𝑖𝑛 3π‘₯) =sin⁑3π‘₯. 𝑑π‘₯/𝑑π‘₯+π‘₯ . 𝑑/𝑑π‘₯ (sin⁑3π‘₯ ) =sin⁑3π‘₯+3π‘₯ π‘π‘œπ‘ 3π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) =𝑑/𝑑π‘₯ (sin⁑〖3π‘₯+3π‘₯ cos⁑3π‘₯ γ€— ) =3 cos⁑〖3π‘₯+3 cos⁑〖3π‘₯+3.3π‘₯(βˆ’sin⁑3π‘₯ )γ€— γ€— =6 cos⁑〖3π‘₯βˆ’9π‘₯ sin⁑3π‘₯ γ€— Now , we have to Prove (𝑑^2 𝑦)/(𝑑π‘₯^2 )+9π‘¦βˆ’6 cos⁑〖3π‘₯=0γ€— Solving LHS (𝑑^2 𝑦)/(𝑑π‘₯^2 )+9π‘¦βˆ’6 cos⁑3π‘₯ =6 cos⁑3π‘₯βˆ’9π‘₯𝑠𝑖𝑛 3π‘₯+9(π‘₯ sin⁑3π‘₯ )βˆ’6 cos⁑3π‘₯ =6 cos⁑3π‘₯βˆ’6 cos⁑3π‘₯βˆ’9π‘₯𝑠𝑖𝑛 3π‘₯+9π‘₯𝑠𝑖𝑛 3π‘₯ =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 (π‘–π‘šπ‘π‘™π‘–π‘π‘–π‘‘ π‘œπ‘Ÿ 𝑒π‘₯𝑝𝑙𝑖𝑐𝑖𝑑) is a solution of the corresponding differential equation . (iv) π‘₯^2=2𝑦^2 log⁑𝑦 : (π‘₯^2+𝑦^2 ) 𝑑𝑦/𝑑π‘₯βˆ’π‘₯𝑦=0 π‘₯^2=2𝑦^2 log⁑𝑦 Differentiating Both sides w.r.t. x 𝑑/𝑑π‘₯ (π‘₯^2 )=𝑑/𝑑π‘₯ (2𝑦^2 log⁑𝑦 ) 2π‘₯=log⁑𝑦 . 𝑑/𝑑π‘₯ (2𝑦^2 )+2𝑦^2 (𝑑(log⁑〖𝑦)γ€—)/𝑑π‘₯ 2π‘₯=log⁑𝑦 . 2𝑦 . 𝑑𝑦/𝑑π‘₯ + 2𝑦^2 1/𝑦 . 𝑑𝑦/𝑑π‘₯ 2π‘₯=4𝑦 . π‘™π‘œπ‘”π‘¦ 𝑑𝑦/𝑑π‘₯ + 2𝑦 𝑑/𝑑π‘₯ y 2π‘₯=(4𝑦 . π‘™π‘œπ‘”π‘¦+2𝑦) 𝑑𝑦/𝑑π‘₯ 2π‘₯/(4𝑦 π‘™π‘œπ‘”π‘¦ + 2𝑦)=𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=π‘₯/(2𝑦 π‘™π‘œπ‘”π‘¦ + 𝑦) Now we Have to Prove (π‘₯^2+𝑦^2 ) 𝑑𝑦/𝑑π‘₯βˆ’π‘₯𝑦=0 (π‘₯^2+𝑦^2 ) 𝑑𝑦/𝑑π‘₯=π‘₯𝑦 𝑑𝑦/𝑑π‘₯=π‘₯𝑦/(π‘₯^2 +γ€– 𝑦〗^2 ) Solving RHS π‘₯𝑦/(π‘₯^2 + 𝑦^2 ) Putting Value of π‘₯^2=2𝑦^2 𝑏 π‘™π‘œπ‘”π‘¦ in denominator =π‘₯𝑦/(2𝑦^2 π‘™π‘œπ‘”π‘¦ +γ€– 𝑦〗^2 ) =π‘₯𝑦/𝑦(2𝑦 π‘™π‘œπ‘”π‘¦ + 𝑦) =π‘₯/(2𝑦 π‘™π‘œπ‘”π‘¦ + 𝑦) Now taking LHS 𝑑𝑦/𝑑π‘₯=π‘₯/(2𝑦 π‘™π‘œπ‘”π‘¦ + 𝑦) = RHS ∴ LHS = RHS So, The Given Function is a solution of the Differential Equation

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.