# Misc 15 (MCQ) - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 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 Important

Misc 4

Misc 5 Important

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11

Misc 12 Important

Misc 13 (MCQ)

Misc 14 (MCQ) Important

Misc 15 (MCQ) You are here

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Important Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at April 16, 2024 by Teachoo

Misc 15 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 Now, 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 So, the correct answer is (c)