# Example 13 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Examples

Example 1 (i)

Example 1 (ii) Important

Example 1 (iii) Important

Example 2

Example 3 Important

Example 4

Example 5

Example 6

Example 7 Important

Example 8

Example 9 Important

Example 10 Important

Example 11

Example 12 Important

Example 13 Important You are here

Example 14

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19

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Example 21 Important

Example 22 Important

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Question 4 Deleted for CBSE Board 2025 Exams

Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Deleted for CBSE Board 2025 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

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