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Example 12 - Find equation: (1, 1) , x dy = (2x2 + 1) dx - Examples

Example 12 - Chapter 9 Class 12 Differential Equations - Part 2
Example 12 - Chapter 9 Class 12 Differential Equations - Part 3


Transcript

Example 12 Find the equation of the curve passing through the point (1 , 1) whose differential equation is π‘₯ 𝑑𝑦= (2π‘₯^2+1)𝑑π‘₯(π‘₯β‰ 0) π‘₯ 𝑑𝑦 = (2x2 + 1)dx dy = "(2x2 + 1)" /π‘₯ dx dy = ("2x2" /π‘₯+1/π‘₯) dx dy = (2π‘₯+1/π‘₯) dx Integrating both sides. ∫1▒𝑑𝑦 = ∫1β–’(2π‘₯+1/π‘₯) 𝑑π‘₯ ∫1▒𝑑𝑦 = ∫1β–’γ€–2π‘₯ 𝑑π‘₯+γ€— ∫1β–’γ€–1/π‘₯ 𝑑π‘₯γ€— y = 2 π‘₯2/2 + log |π‘₯| + C y = π‘₯2 + log |π‘₯| + C Since the curve passes through point (1, 1) Putting x = 1, y = 1 is (1) 1 = 12 + log |𝟏| + C 1 = 1 + 0 + C 1 βˆ’ 1 = C ∴ C = 0 Put C = 0 in (1) i.e y = x2 + log |π‘₯| + C y = x2 + log |π‘₯| + 0 y = x2 + log |π‘₯| Hence, the equation of curve is y = x2 + log |𝒙|

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.