# Misc 18 (MCQ) - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Aug. 11, 2021 by Teachoo

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 1 (iii) Important

Misc 2 (i)

Misc 2 (ii) Important

Misc 2 (iii)

Misc 2 (iv) Important

Misc 3 Deleted for CBSE Board 2023 Exams

Misc 4 Important

Misc 5 Important Deleted for CBSE Board 2023 Exams

Misc 6

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10 Important

Misc 11

Misc 12 Important

Misc 13

Misc 14 Important

Misc 15 Important

Misc 16 (MCQ)

Misc 17 (MCQ) Important

Misc 18 (MCQ) You are here

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at Aug. 11, 2021 by Teachoo

Misc 18 The general solution of the differential equation 𝑒^𝑥 𝑑𝑦+(𝑦 𝑒^𝑥+2𝑥)𝑑𝑥=0 is (A) 𝑥 𝑒^𝑦+𝑥^2=𝐶 (B) 𝑥 𝑒^𝑦+𝑦^2=𝐶 (C) 𝑦 𝑒^𝑥+𝑥^2=𝐶 (D) 𝑦 𝑒^𝑦+𝑥^2=𝐶 Given equation 𝑒^𝑥 𝑑𝑦+(𝑦 𝑒^𝑥+2𝑥)𝑑𝑥=0 𝑒^𝑥 𝑑𝑦=−(𝑦 𝑒^𝑥+2𝑥)𝑑𝑥 𝑑𝑦/𝑑𝑥= (−(𝑦𝑒^𝑥 + 2𝑥))/𝑒^𝑥 𝑑𝑦/𝑑𝑥 = (−𝑦𝑒^𝑥)/𝑒^𝑥 −2𝑥/𝑒^𝑥 𝑑𝑦/𝑑𝑥 = −𝑦−2𝑥/𝑒^𝑥 𝑑𝑦/𝑑𝑥 + y = (−2𝑥)/𝑒^𝑥 Differential equation is of the form 𝑑𝑦/𝑑𝑥 + Py = Q where P = 1 & Q = (−2𝑥)/𝑒^𝑥 IF = 𝑒^∫1▒〖𝑃 𝑑𝑥〗 IF = 𝑒^∫1▒〖1 𝑑𝑥〗 IF = 𝑒^𝑥 Solution is y(IF) = ∫1▒〖(𝑄×𝐼𝐹)𝑑𝑥+𝑐〗 yex = ∫1▒〖(−2𝑥)/𝑒^𝑥 𝑒^𝑥 𝑑𝑥+𝑐〗 yex = −∫1▒〖2𝑥 𝑑𝑥+𝑐〗 yex = −2×𝑥^2/2+𝑐 yex = −𝑥^2+𝑐 yex + 𝑥^2=𝑐 ∴ Option (C) is the correct answer