Misc 10 - Chapter 9 Class 12 Differential Equations
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Misc 10 Solve the differential equation [π^(β2βπ₯)/βπ₯βπ¦/βπ₯] ππ₯/ππ¦=1(π₯β 0)[π^(β2βπ₯)/βπ₯βπ¦/βπ₯] ππ₯/ππ¦=1 π^(β2βπ₯)/βπ₯β π¦/βπ₯ =ππ¦/ππ₯ π π/π π + π/βπ = π^(βπβπ)/βπ Differential equation is of the form ππ¦/ππ₯ + Py = Q where P = π/βπ & Q = π^(βπβπ)/βπ IF. = eβ«1β"pdx" Finding β«1βγπ· π πγ β«1βγπ· π π=β«1βπ π/βπ γ β«1βγπ ππ₯=β«1βγπ₯^((β1)/2) ππ₯γ γ β«1βγπ ππ₯=β«1βγ(π₯ (β1)/2 + 1)/((β1)/2 + 1) ππ₯γ γ β«1βγπ ππ₯=2π₯^(1/2) γ= 2βπ₯ β΄ IF = π^(πβπ) Solution is y (IF) = β«1βγ(πΓπΌπΉ)ππ₯+πγ yπ^(πβπ) = β«1βγ(π^(βπβπ)/βπΓπ^(πβπ) )π π+πγ yπ^(2βπ₯) = β«1βγππ₯/βπ₯+πγ yπ^(2βπ₯) = β«1βγ1/βπ₯ ππ₯+πγ yπ^(2βπ₯) = β«1βγπ₯^((β1)/2) ππ₯+πγ yπ^(2βπ₯) = β«1βγ(π₯^(β1/2 + 1) )/(β1/2 + 1)+πγ "y" π^(2βπ₯) " = "2π₯^(1/2) + C "y" π^(πβπ) " = "2βπ + C
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo