# Example 18 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

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Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

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Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Deleted for CBSE Board 2025 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at April 16, 2024 by Teachoo

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₯, π¦) is equal to the sum of the π₯ coordinate (πππ πππ π π) and the product of the coordinate and π¦ coordinate (πππππππ‘π) of that point .Slope of the tangent to the curve at (x, y) = π π/π π Given that Slope of tangent to curve at point (π₯, π¦) is equal to sum of π₯ coordinate and product of the π₯ and π¦ coordinate of that point So, our equation becomes π π/π π = x + xy ππ¦/ππ₯ β xy = x Differential equation is of the form π π/π π + Py = Q where P = βx & Q = x IF = e^β«1βπππ₯ = e^(ββ«1βγπ₯ ππ₯γ) = π^(γβπγ^π/π) Solution is y (IF) = β«1βγ(πΓπΌπΉ)ππ₯+πΆγ y π^((βπ^π)/π) = β«1βγππ^(γβπγ^π/π) π π+πͺγ Putting (βπ^π)/π = t β 2π₯/2 ππ₯=ππ‘ x dx = β dt Thus, our equation becomes ye^((βπ₯^2)/2) = β«1βγβπ^π‘ ππ‘+πγ yπ^((βπ^π)/π) = βπ^π+π Putting back t = (βπ₯^2)/2 ye^((βπ₯^2)/2) = β e^((βπ₯^2)/2) + C π¦/π^(π₯^2/2) = β 1/π^(π₯^2/2) + C y = β1 + Cπ^(π^π/π) Since curve passes through (0, 1) Putting x = 0, y = 1 in (1) y = β1 + Cπ^(π₯^2/2) 1 = β1 + Ce^(0^2/2) 1 = β1 + C. 1 + 1 = C 2 = C C = 2 Putting value of C in (1) y = β1 + Ce^(π₯^2/2) y = β1 + ππ^(π^π/π) β΄ Equation of the curve is y = β1 + ππ^(π^π/π)