Ex 9.2, 8 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Gen and Particular Solution
Ex 9.2, 12 (MCQ) Important
Question 11 (MCQ) Deleted for CBSE Board 2024 Exams
Question 12 (MCQ) Important Deleted for CBSE Board 2024 Exams
Example 2
Ex 9.2, 1
Ex 9.2, 2
Ex 9.2, 5
Ex 9.2, 3
Example 3 Important
Ex 9.2, 7
Ex 9.2, 10
Ex 9.2, 4 Important
Ex 9.2, 8 Important You are here
Ex 9.2, 9
Ex 9.2, 6 Important
Example 19
Misc 2 (i)
Gen and Particular Solution
Last updated at April 16, 2024 by Teachoo
Ex 9.2, 8 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation : π¦βcosβ‘γπ¦=π₯γ : (π¦ sinβ‘γπ¦+cosβ‘γπ¦+π₯γ γ ) γ π¦γ^β²=π¦ π¦βcosβ‘γπ¦=π₯γ Differentiating both sides w.r.t. π₯ π/ππ₯ [π¦βγcos γβ‘π¦ ]=ππ₯/ππ₯ π(π¦)/ππ₯βπ[cos π¦]/ππ₯=1 ππ¦/ππ₯β(βsin π¦) ππ¦/ππ₯=1 ππ¦/ππ₯+π ππ π¦ ππ¦/ππ₯=1" " ππ¦/ππ₯ [1+sin π¦]=1 ππ¦/ππ₯=1/(1 + sin π¦) Now, we have to verify (π¦π ππ π¦+cos π¦+π₯) π¦^β²=π¦ Taking L.H.S (π¦ sinβ‘γπ¦+cosβ‘γπ¦+π₯γ γ ) γ π¦γ^β² =[π¦π ππ π¦+cos π¦+π¦βcosβ‘π¦ ] π¦^β² =[π¦π πππ¦+π¦] π¦^β² =π¦(1+π πππ¦) π¦^β² =π¦(1+π πππ¦)[1/(1 + sinβ‘π¦ )] =π¦ = R.H.S Hence Verified