



CBSE Class 12 Sample Paper for 2026 Boards
CBSE Class 12 Sample Paper for 2026 Boards
Last updated at Sept. 2, 2025 by Teachoo
Transcript
Question 34 (A) Solve the differential equation: π¦+π/ππ₯(π₯π¦)=π₯(sin π₯+π₯)Now, our equation is π¦+π/ππ₯(π₯π¦)=π₯(sin π₯+π₯) Using product formula π¦+(π(π₯))/ππ₯ π¦+π₯ ππ¦/ππ₯=π₯(sin π₯+π₯) π¦+π¦+π₯ ππ¦/ππ₯=π₯(sin π₯+π₯) π₯ ππ¦/ππ₯+2π¦=π₯(sin π₯+π₯) Diving both sides by x π₯/π₯ \ Γ ππ¦/ππ₯+2π¦/π₯=π₯(sin π₯+π₯)/π₯ ππ¦/ππ₯+2π¦/π₯=(sin π₯+π₯) Comparing with ππ¦/ππ₯ + Py = Q P = π/π & Q = (πππ π+π) Finding Integrating factor (IF) IF = e^β«1βπππ₯ = π^β«1βγπ/π π πγ = e^(2β«1βγ1/π₯ ππ₯γ) = π^(π πππβ‘|π| ) = e^logβ‘γπ₯^2 γ = π^π Solution of differential equation is y Γ IF = β«1βγπ.πΌπΉ ππ₯γ Putting values y Γ x2 = β«1βγ(πππ π+π) π^π π π γ yx2 = β«1βγπ ππβ‘π₯ Γ π₯^2 γ ππ₯+β«1βπ^π π π yx2 = β«1βγπ ππβ‘π₯ Γ π₯^2 γ ππ₯+π₯^4/4+πΆ yx2 = β«1βγπ^π πππβ‘π γ π π+π^π/π+πͺ Evaluating β«1βγπ^π πππβ‘π γ π π separately β«1βγπ₯^2 sinβ‘π₯ γ ππ₯ We know that β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = x2 and g(x) = sin x β«1βγπ₯^2 sinβ‘π₯ γ ππ₯=π^π β«1βπ¬π’π§β‘π π πββ«1β(π (π^π )/π π β«1βγπππβ‘π π πγ) π π = β π₯^2 cosβ‘π₯ β β«1βγ2π₯ Γ βcosβ‘γπ₯ ππ₯γ γ = β π₯^2 cosβ‘π₯+2 β«1βγπ πππβ‘π γ π π+πΆ Applying by parts again in β«1βγπ πππβ‘π γ = β π₯^2 cosβ‘π₯+2 [πβ«1βπππβ‘π π πββ«1β(π π/π π β«1βγπππβ‘π π πγ) π π]+πΆ = β π₯^2 cosβ‘π₯+2 [π πππ πββ«1βγπ Γ πππ πγ π π]+πΆ = β π₯^2 cosβ‘π₯+2 [π₯ π ππ π₯ββ«1βγπππ πγ π π]+πΆ = β π₯^2 cosβ‘π₯+2 [π₯ π ππ π₯β(βπππ π₯) ]+πΆ = β π^π πππβ‘π+π [π πππ π+ππ¨π¬β‘π ]+πͺ Putting value of β«1βγπ₯^2 sinβ‘π₯ γ ππ₯ in (1) yx2 = β«1βγπ₯^2 π ππβ‘π₯ γ ππ₯+π₯^4/4+πΆ yx2 = β π^π πππβ‘π+π [π πππ π+ππ¨π¬β‘π ]+π^π/π+πͺ Which is the required solution