Example 13 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Solving homogeneous differential equation
Ex 9.4, 16 (MCQ)
Ex 9.4, 2
Ex 9.4, 15
Ex 9.4, 4 Important
Example 11
Ex 9.4, 14 Important
Ex 9.4, 13
Ex 9.4, 8
Ex 9.4, 7 Important
Example 21 Important
Ex 9.4, 6 Important
Ex 9.4, 5
Example 13 Important You are here
Ex 9.4, 9
Ex 9.4, 12 Important
Ex 9.4, 1 Important
Ex 9.4, 3
Ex 9.4, 11
Example 10 Important
Misc 3 Important
Example 12 Important
Ex 9.4, 10 Important
Misc 8 Important
Misc 9
Solving homogeneous differential equation
Last updated at April 16, 2024 by Teachoo
Example 13 Show that the family of curves for which the slope of the tangent at any point (𝑥 , 𝑦) on its (𝑥^2+𝑦^2)/2𝑥𝑦 , is given by 𝑥^2−𝑦^2=𝑐𝑥 We know that the slope of the tangent at (𝑥 ,𝑦) of a curve is 𝑑𝑦/𝑑𝑥 Given slope of tangent at (𝑥 , 𝑦) is (𝑥^2 +〖 𝑦〗^2)/2𝑥𝑦 Therefore 𝒅𝒚/𝒅𝒙=(𝒙^𝟐 +〖 𝒚〗^𝟐)/𝟐𝒙𝒚 Step 1: Find 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=(𝑥^2 +𝑦^2)/2𝑥𝑦 Step 2: Put 𝑑𝑦/𝑑𝑥= F(𝑥 𝑦) So, F(𝑥 𝑦)=(𝑥^2 + 𝑦^2)/2𝑥𝑦 Finding F(𝜆𝑥 ,𝜆𝑦) F(𝜆𝑥 ,𝜆𝑦)=((𝜆𝑥)^2 + (𝜆𝑦)^2)/2(𝜆𝑥)(𝜆𝑦) =(𝜆^2 𝑥^2 + 𝜆^2 𝑦^2)/(2𝜆^2 𝑥𝑦) =(𝜆^2 (𝑥^2 +〖 𝑦〗^2 ))/(𝜆^2 2𝑥𝑦) = ((𝑥^2 +〖 𝑦〗^2 ))/( 2𝑥𝑦) = F(𝑥 , 𝑦) So , F(𝝀𝒙 ,𝝀𝒚)= F(𝑥 , 𝑦) =𝝀° "F" (𝒙 , 𝒚) So , "F" (𝑥 , 𝑦) is homogeneous function of degree zero, Therefore given equation is a homogeneous differential equation Step 3: Solving 𝑑𝑦/𝑑𝑥 by putting 𝑦=𝑣𝑥 𝑑𝑦/𝑑𝑥=(𝑥^2 + 𝑦^2)/2𝑥𝑦 Put 𝒚=𝒗𝒙 Diff. w.r.t. 𝑥 𝑑𝑦/𝑑𝑥=𝑑/𝑑𝑥 (𝑣𝑥) 𝑑𝑦/𝑑𝑥=𝑥 𝑑𝑣/𝑑𝑥+𝑣 𝑑𝑥/𝑑𝑥 𝒅𝒚/𝒅𝒙=𝒙𝒅𝒗/𝒅𝒙+𝒗 Putting values 𝑜𝑓 𝑑𝑦/𝑑𝑥 and y in (i") " v + 𝒙 𝒅𝒗/𝒅𝒙=(𝒙^𝟐 + (𝒗〖𝒙)〗^𝟐)/(𝟐𝒙 (𝒗𝒙)) " " 𝑥 𝑑𝑣/𝑑𝑥=(𝑥^2 + 𝑣^2 𝑥^2)/(2𝑥^2 𝑣) − 𝑣" " 𝑥 𝑑𝑣/𝑑𝑥=(𝑥^2 + 𝑣^2 𝑥^2 − 2𝑥^2 𝑣 . 𝑣)/(2𝑥^2 𝑣) 𝑥 𝑑𝑣/𝑑𝑥=(𝑥^2 + 𝑣^2 𝑥^2 − 2𝑥^2 𝑣^2)/(2𝑥^2 𝑣) 𝑥 𝑑𝑣/𝑑𝑥=(𝑥^2 − 𝑣^2 𝑥^2)/(2𝑥^2 𝑣) 𝑥 𝑑𝑣/𝑑𝑥=(𝑥^2 (1 − 𝑣^2 ))/(𝑥^2. 2𝑣) 𝑥 𝑑𝑣/𝑑𝑥=(1 − 𝑣^2)/2𝑣 𝑑𝑣/𝑑𝑥=(1 − 𝑣^2)/2𝑣 . 1/𝑥 (2𝑣 𝑑𝑣)/(1 − 𝑣^2 )=𝑑𝑥/𝑥 (2𝑣 𝑑𝑣)/(−(𝑣^2 − 1) )=𝑑𝑥/𝑥 (𝟐𝒗 𝒅𝒗 )/(𝒗^𝟐 − 𝟏)=(−𝒅𝒙)/𝒙 Integrating Both Sides ∫1▒〖(2𝑣 𝑑𝑣 )/(𝑣^2−1)=∫1▒(−𝑑𝑥)/𝑥〗 ∫1▒(𝟐𝒗 )/(𝒗^𝟐 − 𝟏) 𝒅𝒗=−𝒍𝒐𝒈|𝒙|+𝒄 Solving ∫1▒(𝟐𝒗 )/(𝒗^𝟐 − 𝟏) 𝒅𝒗 Put 𝒗^𝟐−𝟏=𝒕 Diff. w.r.t. 𝑣 𝑑(𝑣^2 −1)/𝑑𝑣=𝑑𝑡/𝑑𝑣 2𝑣=𝑑𝑡/𝑑𝑣 " " 𝑑𝑣=𝑑𝑡/2𝑣 " " ∫1▒𝟐𝒗/(𝒗^𝟐 −𝟏) 𝒅𝒗 = ∫1▒〖𝟐𝒗/𝒕 〗 𝒅𝒕/𝟐𝒗 =∫1▒𝑑𝑡/𝑡 = log |𝑡| Putting t = v2 – 1 = log |𝒗^𝟐−𝟏| From (2) ∫1▒(2𝑣 )/(𝑣^2 − 1) 𝑑𝑣=−𝑙𝑜𝑔|𝑥|+𝑐 "log " |𝑣^2−1|=−𝑙𝑜𝑔|𝑥|+𝑐1 Putting 𝑣𝑥=𝑦 or 𝑣=𝑦/𝑥 "log " |(𝒚/𝒙)^𝟐−𝟏|=−𝒍𝒐𝒈|𝒙|+𝒄𝟏 "log " |(𝑦/𝑥)^2−1|+𝑙𝑜𝑔|𝑥|=+𝑐1 "log" |[(𝑦/𝑥)^2−1]𝑥|=𝑐1 "log" |[𝑦^2/𝑥^2 −1]𝑥|=𝑐1 Putting 𝑐1=log𝑐 𝑙𝑜𝑔|(𝑦^2/𝑥^2 −1)𝑥|=log 𝑐1 Removing log (𝒚^𝟐/𝒙^𝟐 −𝟏)𝒙=𝑪𝟏 (𝑥𝑦^2)/𝑥^2 −𝑥=𝐶1 𝑦^2/𝑥−𝑥=𝐶1 (𝑦^2−𝑥^2)/𝑥=𝐶1" " 𝑦^2−𝑥^2=𝐶1𝑥 𝑥^2−𝑦^2=−𝐶1𝑥 Put 𝑐=−𝐶1 𝒙^𝟐−𝒚^𝟐=𝒄𝒙 Hence Proved