# Misc 10 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Miscellaneous

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Misc 1 (ii)

Misc 1 (iii) Important

Misc 2 (i)

Misc 2 (ii) Important

Misc 2 (iii)

Misc 2 (iv) Important

Misc 3 Important

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Misc 8 Important

Misc 9

Misc 10 Important You are here

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Misc 12 Important

Misc 13 (MCQ)

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Misc 15 (MCQ)

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at April 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