Miscellaneous

Chapter 9 Class 12 Differential Equations
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Misc 10 Solve the differential equation [π^(β2βπ₯)/βπ₯βπ¦/βπ₯] ππ₯/ππ¦=1(π₯β 0)[π^(β2βπ₯)/βπ₯βπ¦/βπ₯] ππ₯/ππ¦=1 π^(β2βπ₯)/βπ₯β π¦/βπ₯ =ππ¦/ππ₯ ππ/ππ + π/βπ = π^(βπβπ)/βπ Differential equation is of the form ππ¦/ππ₯ + Py = Q where P = π/βπ & Q = π^(βπβπ)/βπ IF. = eβ«1β"pdx" Finding β«1βγπ· ππγ β«1βγπ· ππ=β«1βππ/βπ γ β«1βγπ ππ₯=β«1βγπ₯^((β1)/2) ππ₯γ γ β«1βγπ ππ₯=β«1βγ(π₯ (β1)/2 + 1)/((β1)/2 + 1) ππ₯γ γ β«1βγπ ππ₯=2π₯^(1/2) γ= 2βπ₯ β΄ IF = π^(πβπ) Solution is y (IF) = β«1βγ(πΓπΌπΉ)ππ₯+πγ yπ^(πβπ) = β«1βγ(π^(βπβπ)/βπΓπ^(πβπ) )ππ+πγ yπ^(2βπ₯) = β«1βγππ₯/βπ₯+πγ yπ^(2βπ₯) = β«1βγ1/βπ₯ ππ₯+πγ yπ^(2βπ₯) = β«1βγπ₯^((β1)/2) ππ₯+πγ yπ^(2βπ₯) = β«1βγ(π₯^(β1/2 + 1) )/(β1/2 + 1)+πγ "y" π^(2βπ₯) " = "2π₯^(1/2) + C "y" π^(πβπ) " = "2βπ + C

Davneet Singh

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.