Get live Maths 1-on-1 Classs - Class 6 to 12

Examples

Example 1 (i)

Example 1 (ii) Important

Example 1 (iii) Important

Example 2

Example 3 Important

Example 4 Deleted for CBSE Board 2023 Exams

Example 5 Deleted for CBSE Board 2023 Exams

Example 6 Important Deleted for CBSE Board 2023 Exams

Example 7 Deleted for CBSE Board 2023 Exams

Example 8 Deleted for CBSE Board 2023 Exams

Example 9

Example 10

Example 11

Example 12 Important

Example 13

Example 14 Important

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19

Example 20 Important

Example 21

Example 22 Important

Example 23 Important You are here

Example 24

Example 25 Deleted for CBSE Board 2023 Exams

Example 26

Example 27 Important

Example 28 Important

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at March 16, 2023 by Teachoo

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₯, 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) = ππ¦/ππ₯ 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 ππ¦/ππ₯ = x + xy ππ¦/ππ₯ β xy = x Differential equation is of the form ππ¦/ππ₯ + Py = Q where P = βx & Q = x IF = e^β«1βπππ₯ = e^(ββ«1βγπ₯ ππ₯γ) = e^(γβπ₯γ^2/2) Solution is y (IF) = β«1βγ(πΓπΌπΉ)ππ₯+πΆγ y π^((βπ₯^2)/2) = β«1βγπ₯π^(γβπ₯γ^2/2) ππ₯+πΆγ Putting (βπ₯^2)/2 = t β 2π₯/2 ππ₯=ππ‘ x dx = β dt y π^((βπ₯^2)/2) = β«1βγπ₯π^(γβπ₯γ^2/2) ππ₯+πΆγ Putting (βπ₯^2)/2 = t β 2π₯/2 ππ₯=ππ‘ x dx = β dt Thus, our equation becomes ye^((βπ₯^2)/2) = β«1βγβπ^π‘ ππ‘+πγ ye^((βπ₯^2)/2) = 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 + Ce^(π₯^2/2) 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 β¦(1) Putting value of C in (1) y = β1 + Ce^(π₯^2/2) y = β1 + 2e^(π₯^2/2) β΄ Equation of the curve is y = β1 + ππ^(π^π/π)