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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.