# Ex 9.6, 10

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.6, 10For each of the differential equation given in Exercises 1 to 12, find the general solution : (π₯+π¦) ππ¦/ππ₯=1 Step 1: Put in form ππ¦/ππ₯ + Py = Q or ππ₯/ππ¦ + P1x = Q1 (x + y) ππ¦/ππ₯ = 1 Dividing by (x + y), ππ¦/ππ₯ = 1/((π₯ + π¦)) ππ₯/ππ¦ = (π₯+π¦) ππ₯/ππ¦ β x = π¦ ππ₯/ππ¦ + (β1) x = π¦ Step 2: Find P1 and Q1 Comparing (1) with ππ₯/ππ¦ + P1x = Q1 P1 = β1 and Q1 = y Step 3: Find Integrating factor, I.F. I.F. = π^β«1βγπ_1 ππ¦γ = e^β«1βγ(β1) ππ¦γ = eβy So, I.F. = eβy Step 4 : Solution of the equation x Γ I.F. = β«1βγπ1ΓπΌ.πΉ. ππ¦+πΆγ Putting values, x Γ e β y = β«1βγπ¦Γπ^(βπ¦).γ ππ¦+πΆ Let I = β«1βγπ¦.π^(βπ¦) ππ¦γ = y β«1βγπ^(βπ¦) ππ¦γ β β«1β[π/ππ¦ π¦ β«1βγπ^(βπ¦) ππ¦γ] dy = y π^(βπ¦)/(β1) β β«1βγ1. γ π^(βπ¦)/(β1) dy. = βπ¦.π^(βπ¦) + β«1βγπ^(βπ¦) ππ¦γ = βπ¦.π^(βπ¦) + π^(βπ¦)/(β1) = βπ¦.π^(βπ¦) β π^(βπ¦) Putting value of I in (2) x e βy = β«1βγπ¦Γπ^(βπ¦).γ ππ¦+πΆ x e βy = βπ¦π^(βπ¦)βπ^(βπ¦)+πΆ Dividing by π^(βπ¦) x = βy β 1 + Cey x + y + 1 = Cey

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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.