# Example 16 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

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Example 16 You are here

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Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Question 4 Deleted for CBSE Board 2025 Exams

Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Deleted for CBSE Board 2025 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at April 16, 2024 by Teachoo

Example 16 Find the general solution of the differential equation π¦ ππ₯β(π₯+2π¦^2 )ππ¦=0 Given equation π¦ ππ₯β(π₯+2π¦^2 )ππ¦=0 π¦ ππ₯=(π₯+2π¦^2 )ππ¦ π π/π π = π/(π + ππ^π ) This is not of the form ππ¦/ππ₯+ππ¦=π β΄ We find π π/π π ππ₯/ππ¦ = (π₯ + 2π¦^2)/π¦ ππ₯/ππ¦ = (π₯ )/π¦ + (2π¦^2)/π¦ π π/π π β (π )/π = 2y Differential equation is of the form ππ₯/ππ¦ + P1 x = Q1 where P1 = (β1)/π¦ & Q1 = 2y Now, IF = π^β«1βγπ_1 ππ¦γ IF = π^β«1βγ(βπ)/π π π" " γ IF = e^(βlogβ‘π¦ ) IF = e^logβ‘(1/π¦) IF = π/π Solution is x(IF) = β«1βγ(πΈΓπ°π)π π+πͺ γ x Γ 1/π¦=β«1βγ2π¦Γ1/π¦γ ππ¦+πΆ π/π = β«1βγππ π+πͺγ π₯/π¦ = 2π¦+πΆ x = y (2y + C) x = 2y2 + Cy