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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 18 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) = π’…π’š/𝒅𝒙 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 π’…π’š/𝒅𝒙 = x + xy 𝑑𝑦/𝑑π‘₯ βˆ’ xy = x Differential equation is of the form π’…π’š/𝒅𝒙 + 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) = ∫1▒〖𝒙𝒆^(γ€–βˆ’π’™γ€—^𝟐/𝟐) 𝒅𝒙+π‘ͺγ€— 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’†^𝒕+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^(𝒙^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 Putting value of C in (1) y = βˆ’1 + Ce^(π‘₯^2/2) y = βˆ’1 + πŸπ’†^(𝒙^𝟐/𝟐) ∴ Equation of the curve is y = βˆ’1 + πŸπ’†^(𝒙^𝟐/𝟐)

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.