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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Example 23 Find the equation of a curve passing the point (0 , 1). if the slope of the tangent to the curve at any point (π‘₯, 𝑦) is equal to the sum of the π‘₯ coordinate (π‘Žπ‘π‘ π‘π‘–π‘ π‘ π‘Ž) and the product of the coordinate and 𝑦 coordinate (π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’) of that point . Slope of the tangent to the curve at (x, y) = 𝑑𝑦/𝑑π‘₯ Given that Slope of tangent to curve at point (π‘₯, 𝑦) is equal to sum of π‘₯ coordinate and product of the π‘₯ and 𝑦 coordinate of that point So, our equation becomes 𝑑𝑦/𝑑π‘₯ = x + xy 𝑑𝑦/𝑑π‘₯ βˆ’ xy = x Differential equation is of the form 𝑑𝑦/𝑑π‘₯ + Py = Q where P = –x & Q = x IF = e^∫1▒𝑝𝑑π‘₯ = e^(βˆ’βˆ«1β–’γ€–π‘₯ 𝑑π‘₯γ€—) = e^(γ€–βˆ’π‘₯γ€—^2/2) Solution is y (IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑π‘₯+𝐢〗 y 𝑒^((βˆ’π‘₯^2)/2) = ∫1β–’γ€–π‘₯𝑒^(γ€–βˆ’π‘₯γ€—^2/2) 𝑑π‘₯+𝐢〗 Putting (βˆ’π‘₯^2)/2 = t βˆ’ 2π‘₯/2 𝑑π‘₯=𝑑𝑑 x dx = βˆ’ dt y 𝑒^((βˆ’π‘₯^2)/2) = ∫1β–’γ€–π‘₯𝑒^(γ€–βˆ’π‘₯γ€—^2/2) 𝑑π‘₯+𝐢〗 Putting (βˆ’π‘₯^2)/2 = t βˆ’ 2π‘₯/2 𝑑π‘₯=𝑑𝑑 x dx = βˆ’ dt Thus, our equation becomes ye^((βˆ’π‘₯^2)/2) = ∫1β–’γ€–βˆ’π‘’^𝑑 𝑑𝑑+𝑐〗 ye^((βˆ’π‘₯^2)/2) = βˆ’π‘’^𝑑+𝑐 Putting back t = (βˆ’π‘₯^2)/2 ye^((βˆ’π‘₯^2)/2) = βˆ’ e^((βˆ’π‘₯^2)/2) + C 𝑦/𝑒^(π‘₯^2/2) = βˆ’ 1/𝑒^(π‘₯^2/2) + C y = βˆ’1 + Ce^(π‘₯^2/2) Since curve passes through (0, 1) Putting x = 0, y = 1 in (1) y = βˆ’1 + C𝑒^(π‘₯^2/2) 1 = βˆ’1 + Ce^(0^2/2) 1 = βˆ’1 + C. 1 + 1 = C 2 = C C = 2 …(1) Putting value of C in (1) y = βˆ’1 + Ce^(π‘₯^2/2) y = βˆ’1 + 2e^(π‘₯^2/2) ∴ Equation of the curve is y = βˆ’1 + πŸπ’†^(𝒙^𝟐/𝟐)

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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.