Solve all your doubts with Teachoo Black (new monthly pack available now!)
Solving Linear differential equations - Statement given
Solving Linear differential equations - Statement given
Last updated at Dec. 11, 2019 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)