# Ex 9.5, 10 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Ex 9.5

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Ex 9.5, 2

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Ex 9.5, 10 You are here

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Ex 9.5, 18 (MCQ)

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Last updated at April 16, 2024 by Teachoo

Ex 9.5, 10 For 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¦ π π/π π + (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 Step 4 : Solution of the equation x Γ I.F. = β«1βγπ1ΓπΌ.πΉ. ππ¦+πΆγ Putting values, x Γ e β y = β«1βγπ¦ Γ π^(βπ¦).γ ππ¦+πΆ Let I = β«1βγπ.π^(βπ) π πγ Using Integration by parts β«1βγπ(π¦)π(π¦)ππ¦=π(π¦) β«1βγπ(π¦)ππ¦βγγ β«1β[πβ²(π¦)β«1βπ(π¦)ππ¦] ππ¦ Taking π(π¦) = y and g (y) = π^(βπ¦) = y β«1βγπ^(βπ) π πγ β β«1β[π /π π π β«1βγπ^(βπ) π πγ] 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 = βππ^(βπ)βπ^(βπ)+πͺ Dividing by π^(βπ¦) x = βy β 1 + Cey x + y + 1 = Cey