# Ex 9.6, 8

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.6, 8For 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 Step 1: Put in form ππ¦/ππ₯ + Py = Q (1 + x2)dy + 2xy.dx = cot x. dx Dividing both sides by (1 + x2) ππ¦/ππ₯ + 2π₯π¦/((1 + π₯2)) ππ₯/ππ₯ = (cotβ‘(π₯))/((1 + π₯^2)). ππ₯/ππ₯ ππ¦/ππ₯ + 2π₯/((1 + π₯2)) y = cotβ‘π₯/((1 + π₯2)) Step 2 : Find P and Q Comparing (1) with ππ¦/ππ₯ + Py = Q where P = 2π₯/((1 + π₯^2)) & Q = cotβ‘π₯/((1 + π₯^2)) Step 3 : Find Integrating factor, I.F I.F. = π^β«1βγπ ππ₯γ I.F. = e^β«1βγ2π₯/((1 + π₯^2 ) ) ππ₯ γ Let t = 1 + x2 dt = 2x dx I.F. = e^β«1βγππ‘/π‘ γ = elog t = t = (1 + x2) So, I.F = 1 + x2 Step 4 : Solution of the equation y Γ I.F = β«1βγπΓπΌ.πΉ.ππ₯+πΆγ Putting values, y.(1 + x2) = β«1βcotβ‘π₯/((1 + π₯2)) Γ (1+π₯2).dx + c y.(1 + x2) = β«1βγcotβ‘π₯ ππ₯γ+π y (1 + x2) = log|sinβ‘π₯ | + C Dividing by (1 + x2) y = (1 + x2)β1 log |π¬π’π§β‘π |+π(π+"x2" )^(βπ) is the general solution of the given equation

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.