Examples

Chapter 9 Class 12 Differential Equations
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Example 16 Find the general solution of the differential equation π¦ ππ₯β(π₯+2π¦^2 )ππ¦=0 Given equation π¦ ππ₯β(π₯+2π¦^2 )ππ¦=0 π¦ ππ₯=(π₯+2π¦^2 )ππ¦ ππ¦/ππ₯ = π¦/(π₯ + 2π¦^2 ) This is not of the form ππ¦/ππ₯+ππ¦=π β΄ We find ππ₯/ππ¦ ππ₯/ππ¦ = (π₯ + 2π¦^2)/π¦ ππ₯/ππ¦ = (π₯ )/π¦ + (2π¦^2)/π¦ ππ₯/ππ¦ β (π₯ )/π¦ = 2y Differential equation is of the form ππ₯/ππ¦ + P1 x = Q1 where P1 = (β1)/π¦ & Q1 = 2y Now, IF = π^β«1βγπ_1 ππ¦γ IF = e^β«1βγ(β1)/π¦ ππ¦" " γ IF = e^(βlogβ‘π¦ ) IF = e^logβ‘(1/π¦) IF = 1/π¦ IF = e^logβ‘(1/π¦) IF = 1/π¦ Solution is x(IF) = β«1βγ(πΓπΌπΉ)ππ¦+πΆ γ x Γ 1/π¦=β«1βγ2π¦Γ1/π¦γ ππ¦+πΆ π₯/π¦ = β«1βγ2ππ¦+πΆγ π₯/π¦ = 2π¦+πΆ x = y (2y + C) x = 2y2 + Cy (As a log x = log xπ) (As π^πππβ‘π₯ = x)