# Example 18

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 18 Show that the family of curves for which the slope of the tangent at any point 𝑥 , 𝑦 on its 𝑥2+ 𝑦22𝑥𝑦 , is given by 𝑥2− 𝑦2=𝑐𝑥 We know that the slope of the tangent at 𝑥 ,𝑦 of a curve is 𝑑𝑦𝑑𝑥 Given slope of tangent at 𝑥 , 𝑦 is 𝑥2 + 𝑦22𝑥𝑦 Therefore 𝑑𝑦𝑑𝑥= 𝑥2 + 𝑦22𝑥𝑦 Step 1: Find 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥= 𝑥2 + 𝑦22𝑥𝑦 Step 2: Put 𝑑𝑦𝑑𝑥= F 𝑥 𝑦 So, F 𝑥 𝑦= 𝑥2 + 𝑦22𝑥𝑦 Finding F 𝜆𝑥 ,𝜆𝑦 F 𝜆𝑥 ,𝜆𝑦= 𝜆𝑥2 + 𝜆𝑦22 𝜆𝑥 𝜆𝑦 = 𝜆2 𝑥2 + 𝜆2 𝑦22 𝜆2 𝑥𝑦 = 𝜆2 𝑥2 + 𝑦2 𝜆2 2𝑥𝑦 = 𝑥2 + 𝑦2 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 𝑑𝑦𝑑𝑥 by putting 𝑦=𝑣𝑥 𝑑𝑦𝑑𝑥= 𝑥2 + 𝑦22𝑥𝑦 Put 𝑦=𝑣𝑥 Diff. w.r.t. 𝑥 𝑑𝑦𝑑𝑥= 𝑑𝑑𝑥 𝑣𝑥 𝑑𝑦𝑑𝑥=𝑥 𝑑𝑣𝑑𝑥+𝑣 𝑑𝑥𝑑𝑥 𝑑𝑦𝑑𝑥= 𝑥𝑑𝑣𝑑𝑥+𝑣 Putting values 𝑜𝑓 𝑑𝑦𝑑𝑥 and y in (i) v + 𝑥 𝑑𝑣𝑑𝑥= 𝑥2 + (𝑣 𝑥)22𝑥 (𝑣𝑥) v + 𝑥 𝑑𝑣𝑑𝑥= 𝑥2 + 𝑣2 𝑥22 𝑥2 𝑣 − 𝑣 v + 𝑥 𝑑𝑣𝑑𝑥= 𝑥2 + 𝑣2 𝑥2 − 2 𝑥2 𝑣 . 𝑣2 𝑥2 𝑣 v + 𝑥 𝑑𝑣𝑑𝑥= 𝑥2 + 𝑣2 𝑥2 − 2 𝑥2 𝑣22 𝑥2 𝑣 v + 𝑥 𝑑𝑣𝑑𝑥= 𝑥2 + 𝑣2 𝑥2 − 2 𝑣2 𝑥22 𝑥2 𝑣 𝑥 𝑑𝑣𝑑𝑥= 𝑥2 + 𝑣2 𝑥2 − 2 𝑣2 𝑥22 𝑥2 𝑣 𝑥 𝑑𝑣𝑑𝑥= 𝑥2 − 𝑣2 𝑥22 𝑥2 𝑣 𝑥 𝑑𝑣𝑑𝑥= 𝑥2 1 − 𝑣2 𝑥2. 2𝑣 𝑥 𝑑𝑣𝑑𝑥= 1 − 𝑣22𝑣 𝑑𝑣𝑑𝑥= 1 − 𝑣22𝑣 . 1𝑥 2𝑣 𝑑𝑣1 − 𝑣2= 𝑑𝑥𝑥 2𝑣 𝑑𝑣− 𝑣2 − 1= 𝑑𝑥𝑥 2𝑣 𝑑𝑣 𝑣2 − 1= −𝑑𝑥𝑥 Integrating Both Sides 2𝑣 𝑑𝑣 𝑣2−1= −𝑑𝑥𝑥 2𝑣 𝑑𝑣 𝑣2 − 1= −𝑑𝑥𝑥 2𝑣 𝑣2 − 1𝑑𝑣=−𝑙𝑜𝑔 𝑥+𝑐 Solving 𝟐𝒗 𝒗𝟐 − 𝟏𝒅𝒗 Put 𝑣2−1=𝑡 Diff. w.r.t. 𝑣 𝑑 𝑣2 −1𝑑𝑣= 𝑑𝑡𝑑𝑣 2𝑣= 𝑑𝑡𝑑𝑣 𝑑𝑣= 𝑑𝑡2𝑣 2𝑣 𝑣2 −1𝑑𝑣 = 2𝑣𝑡 𝑑𝑡2𝑣 = 𝑑𝑡𝑡 = log 𝑡 Putting t = v2 – 1 = log 𝑣2−1 From (2) 2𝑣 𝑣2 − 1𝑑𝑣=−𝑙𝑜𝑔 𝑥+𝑐 log 𝑣2−1=−𝑙𝑜𝑔 𝑥+𝑐1 Putting 𝑣𝑥=𝑦 or 𝑣= 𝑦𝑥 log 𝑦𝑥2−1=−𝑙𝑜𝑔 𝑥+𝑐1 log 𝑦𝑥2−1=−𝑙𝑜𝑔 𝑥+𝑐1 log 𝑦𝑥2−1+𝑙𝑜𝑔 𝑥=+𝑐1 𝑙𝑜𝑔 𝑦𝑥2−1𝑥=𝑐1 𝑙𝑜𝑔 𝑦2 𝑥2 −1𝑥=𝑐1 Putting 𝑐1= log𝑐 𝑙𝑜𝑔 𝑦2 𝑥2−1𝑥= log 𝑐1 Removing log 𝑦2 𝑥2 −1𝑥=𝐶1 𝑥 𝑦2 𝑥2−𝑥=𝐶1 𝑦2𝑥−𝑥=𝐶1 𝑦2− 𝑥2𝑥=𝐶1 𝑦2− 𝑥2=𝐶1𝑥 𝑥2− 𝑦2=−𝐶1𝑥 Put 𝑐=−𝐶1 𝒙𝟐− 𝒚𝟐=𝒄𝒙 Hence Proved

Example 1
Important

Ex 9.1, 11 Important

Ex 9.1, 12 Important

Example 7 Important

Ex 9.3, 7 Important

Ex 9.3, 10 Important

Example 13 Important

Ex 9.4, 14 Important

Example 17 Important

Example 18 Important You are here

Ex 9.5, 8 Important

Ex 9.5, 15 Important

Example 22 Important

Ex 9.6, 7 Important

Ex 9.6, 13 Important

Ex 9.6, 14 Important

Example 25 Important

Example 27 Important

Example 28 Important

Misc 6 Important

Misc 11 Important

Misc 12 Important

Misc 13 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .