Ex 9.5, 12 - Chapter 9 Class 12 Differential Equations
Last updated at Dec. 16, 2024 by Teachoo
Solving Linear differential equations - Equation given
Ex 9.5, 18 (MCQ)
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Chapter 9 Class 12 Differential Equations
Concept wise
- Order and Degree
- Gen and Particular Solution
- Formation of Differntial equation when general solution given
- Variable separation - Equation given
- Variable separation - Statement given
- Solving homogeneous differential equation
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Solving Linear differential equations - Equation given
- Solving Linear differential equations - Statement given
Transcript
Ex 9.5, 12 For each of the differential equation find the general solution : (π₯+3π¦^2 ) ππ¦/ππ₯=π¦(π¦>0) Step 1 : Put In form ππ¦/ππ₯ + py = Q or ππ₯/ππ¦ + P1x = Q1 (π₯+3π¦^2 ) ππ¦/ππ₯=π¦ ππ¦/ππ₯ = π¦/(π₯+3π¦^2 ) This is not of the form ππ¦/ππ₯ + Py = Q β΄ We need to find π π/π π ππ₯/ππ¦ = (π₯ + 3π¦^2)/π¦ π π/π π = π/π + (ππ^π)/π Step 2 : Find P1 and Q1 Comparing with ππ¦/ππ₯ + P1x = Q1 where P1 = (βπ)/π & Q1 = 3y Step 3 : Finding Integrating factor IF = π^(β«1βπ_π π π) IF = π^(β«1β(β1)/π¦ ππ¦" " ) IF = eβlog y IF = π^logβ‘γπ¦^(β1) γ IF = yβ1 IF = π/π Step 4 : Solution of the equation Solution is x(IF) = β«1βγ(π1ΓπΌπΉ)ππ¦+πΆγ x(1/π¦)=β«1βγ3π¦Γ1/π¦ ππ¦+πΆγ π₯/π¦ = 3β«1βγππ¦+πΆγ π₯/π¦ = 3π¦+πΆ π = ππ^π+πͺy
