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Example 12 - Chapter 9 Class 12 Differential Equations
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example 12 Find the equation of the curve passing through the point (1 , 1) whose differential equation is ๐ฅ ๐๐ฆ= (2๐ฅ^2+1)๐๐ฅ(๐ฅโ 0) ๐ฅ ๐๐ฆ = (2x2 + 1)dx dy = "(2x2 + 1)" /๐ฅ dx dy = ("2x2" /๐ฅ+1/๐ฅ) dx dy = (2๐ฅ+1/๐ฅ) dx Integrating both sides. โซ1โ๐๐ฆ = โซ1โ(2๐ฅ+1/๐ฅ) ๐๐ฅ โซ1โ๐๐ฆ = โซ1โใ2๐ฅ ๐๐ฅ+ใ โซ1โใ1/๐ฅ ๐๐ฅใ y = 2 ๐ฅ2/2 + log |๐ฅ| + C y = ๐ฅ2 + log |๐ฅ| + C Since the curve passes through point (1, 1) Putting x = 1, y = 1 is (1) 1 = 12 + log |๐| + C 1 = 1 + 0 + C 1 โ 1 = C โด C = 0 Put C = 0 in (1) i.e y = x2 + log |๐ฅ| + C y = x2 + log |๐ฅ| + 0 y = x2 + log |๐ฅ| Hence, the equation of curve is y = x2 + log |๐|
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Example 4 Not in Syllabus - CBSE Exams 2021
Example 5 Not in Syllabus - CBSE Exams 2021
Example 6 Important Not in Syllabus - CBSE Exams 2021
Example 7 Not in Syllabus - CBSE Exams 2021
Example 8 Not in Syllabus - CBSE Exams 2021
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Example 12 You are here
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Example 21 Not in Syllabus - CBSE Exams 2021
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Example 25 Not in Syllabus - CBSE Exams 2021
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Example 27 Important
Example 28 Important Not in Syllabus - CBSE Exams 2021
Chapter 9 Class 12 Differential Equations
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About the Author
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.