CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## Find the particular solution of the following differential equation, given that y = 0 when π₯ = π/4 ππ¦/ππ₯+π¦πππ‘π₯ 2/(1 + sinβ‘π₯ )

This question is similar to Example 22 - Chapter 9 Class 12 - Differential EquationsΒ

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### Transcript

Question 8 (Choice 2) Find the particular solution of the following differential equation, given that y = 0 when π₯ = π/4 ππ¦/ππ₯+π¦πππ‘π₯= 2/(1 + sinβ‘π₯ ) ππ¦/ππ₯+π¦πππ‘π₯= 2/(1 + sinβ‘π₯ ) Differential equation is of the form ππ¦/ππ₯+ππ¦=π where P = cot x & Q = π/(π + πππβ‘π ) Now, IF = π^β«1βγπ ππ₯γ IF = π^β«1βγcotβ‘π₯ ππ₯γ IF = γπ^π₯π¨π β‘π¬π’π§β‘π γ^" " IF = sin x Solution is y (IF) =β«1βγ(πΓπΌπΉ) ππ₯+πγ y sin x = β«1βγπ/(π + πππβ‘π )Γ πππ π ππγ + C y sin x = 2β«1βγ(π ππ π₯)/(1 + π ππβ‘π₯ ) ππ₯γ + C y sin x = 2β«1βγ((1 + π ππ π₯ β 1))/(1 + π ππβ‘π₯ ) ππ₯γ + C y sin x = 2β«1βγ((1 + π ππ π₯))/(1 + π ππβ‘π₯ ) ππ₯γβ2β«1βγ1/(1 + π ππβ‘π₯ ) ππ₯γ + C y sin x = πβ«1βππβπβ«1βγπ/(π + πππβ‘π ) ππγ + C y sin x = 2π₯β2β«1βγπ/(π + πππβ‘π ) ππγ + C y sin x = 2π₯β2β«1βγ1/(1 + π ππβ‘π₯ ) Γ(π β π¬π’π§β‘π)/(π β πππβ‘π ) ππ₯γ + C y sin x = 2π₯β2β«1βγ(1 β sinβ‘π₯)/(1 β sin^2β‘π₯ ) ππ₯γ + C y sin x = 2π₯β2β«1βγ(π β πππβ‘π)/γππ¨π¬γ^πβ‘π ππγ + C y sin x = 2π₯β2[β«1βγ1/γπππ γ^2β‘π₯ ππ₯γββ«1βγsinβ‘π₯/γπππ γ^2β‘π₯ ππ₯γ] + C y sin x = 2π₯β2[β«1βγsec^2β‘π₯ ππ₯γββ«1βγsinβ‘π₯/(πππ  π₯) Γ1/cosβ‘π₯ ππ₯γ] + C y sin x = 2π₯β2[β«1βγγπππγ^πβ‘π ππγββ«1βγπ­ππ§β‘π π¬ππβ‘π ππγ] + C y sin x = ππβπ π­ππ§β‘π+π πππ π + C We need to find particular solution when y = 0 when π₯ = π/4 Putting y = 0 and π = π/π 0 Γ sin π/π = 2(π/π)β2 tanβ‘γπ/πγ+2 π ππ π/π + C 0 = π/2β2 Γ 1+2β2 + C 2β2β2βπ/2 = C C = π(πββπ)βπ/π Thus, our particular solution is y sin x = ππβπ π­ππ§β‘π+π πππ π + π(πββπ)βπ/π