# Ex 9.5, 8 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Ex 9.5

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Ex 9.5, 8 Important You are here

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Ex 9.5, 18 (MCQ)

Ex 9.5, 19 (MCQ) Important

Last updated at April 16, 2024 by Teachoo

Ex 9.5, 8 For each of the differential equation given in Exercises 1 to 12, find the general solution : (1+π₯^2 )ππ¦+2π₯π¦ ππ₯=cotβ‘γπ₯ ππ₯(π₯β 0)γ Given equation (1 + x2)dy + 2xy dx = cot x dx Dividing both sides by dx (1 + x2)ππ¦/ππ₯ + 2xy ππ₯/ππ₯ = cot x ππ₯/ππ₯ (1 + x2)ππ¦/ππ₯ + 2xy = cot x Dividing both sides by (1 + x2) π π/π π + ππ/((π + ππ)) y = πππβ‘π/((π + ππ)) Comparing (1) with ππ¦/ππ₯ + Py = Q where P = ππ/((π + π^π)) & Q = πππβ‘π/((π + π^π)) Finding Integrating factor, I.F I.F. = π^β«1βγπ ππ₯γ I.F. = π^β«1βγππ/((π + π^π ) ) π π γ Let t = 1 + x2 dt = 2x dx I.F. = e^β«1βγππ‘/π‘ γ = elog |t| = t Putting back t = (1 + x2) = (1 + x2) Solution of the equation is y Γ I.F = β«1βγπΓπΌ.πΉ.ππ₯+πΆγ Putting values, y.(1 + x2) = β«1βπππβ‘π/((π + ππ)) Γ (π+ππ).dx + c y.(1 + x2) = β«1βγcotβ‘π₯ ππ₯γ+πΆ y (1 + x2) = log |sinβ‘π₯ | + C Dividing by (1 + x2) y = (1 + x2)β1 log |π¬π’π§β‘π |+πͺ(π+"x2" )^(βπ) is the general solution of the given equation Note: This answer does not match with the answer of the book. If we have made any mistake, please comment