Misc 15 (MCQ) - Chapter 9 Class 12 Differential Equations
Last updated at April 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
Misc 15 The general solution of the differential equation π^π₯ ππ¦+(π¦ π^π₯+2π₯)ππ₯=0 is (A) π₯ π^π¦+π₯^2=πΆ (B) π₯ π^π¦+π¦^2=πΆ (C) π¦ π^π₯+π₯^2=πΆ (D) π¦ π^π¦+π₯^2=πΆ Given equation π^π₯ ππ¦+(π¦ π^π₯+2π₯)ππ₯=0 π^π π π=β(π π^π+ππ)π π ππ¦/ππ₯= (β(π¦π^π₯ + 2π₯))/π^π₯ ππ¦/ππ₯ = (βπ¦π^π₯)/π^π₯ β2π₯/π^π₯ ππ¦/ππ₯ = βπ¦β2π₯/π^π₯ π π/π π + y = (βππ)/π^π Differential equation is of the form ππ¦/ππ₯ + Py = Q where P = 1 & Q = (βππ)/π^π Now, IF = π^β«1βγπ ππ₯γ IF = π^β«1βγ1 ππ₯γ IF = π^π Solution is y(IF) = β«1βγ(πΓπΌπΉ)ππ₯+πγ yex = β«1βγ(βππ)/π^π π^π π π+πγ yex = ββ«1βγ2π₯ ππ₯+πγ yex = β2Γπ₯^2/2+π yex = βπ₯^2+π yex + π^π=π So, the correct answer is (c)
