Ex 9.3, 15 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Ex 9.3
Ex 9.3, 1
Important
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Transcript
Ex 9.3, 15 Find the equation of curve passing through the point (0 , 0) and whose differential equation is π¦^β²=π^π₯ sinβ‘π₯ ππ¦/ππ₯ = π^π₯ sin x ππ¦ = π^π₯ sin x dx Integrating both sides β«1βππ¦ = β«1βγππ₯ sinβ‘γπ₯ ππ₯γ γ y = β«1βγππ πππβ‘γπ π πγ γ Integrating by parts, using formula β«1βγπ (π₯)π(π₯)ππ₯ γ= f(π₯) β«1βγπ(π₯)ππ₯ ββ«1βγ[πβ²(π₯)β«1βπ(π₯)ππ₯] ππ₯ γ γ Taking f(x) = ex and g (x) = sin x y = π^π₯ (βcosβ‘π₯) β β«1βγ(βcosβ‘π₯)γ π^π₯ ππ₯ y = βπ^π₯ πππ β‘π₯+β«1βγπ^π πππβ‘π π πγ Integrating by parts, using formula β«1βγπ (π₯)π(π₯)ππ₯ γ= f(π₯) β«1βγπ(π₯)ππ₯ ββ«1βγ[πβ²(π₯)β«1βπ(π₯)ππ₯] ππ₯ γ γ Taking f(x) = ex and g(x) = cos x y = βπ^π₯ πππ β‘π₯+ π^π₯ β«1βγγπππ π₯γβ‘γ ππ₯γβγ β«1β[(π^π₯) β«1βγcosβ‘π₯ ππ₯ γ]ππ₯ y = βπ^π₯ πππ β‘π₯+π^π₯ π ππβ‘π₯ ββ«1βγπ^π π¬π’π§β‘π π πγ But From (1) y = β«1βγππ₯ π ππβ‘γπ₯ ππ₯γ γ y = βπ^π₯ πππ β‘π₯+π^π₯ π ππβ‘π₯ βπ y + y = βπ^π₯ πππ β‘π₯+π^π₯ π ππβ‘π₯ 2y = π^π₯ (sinβ‘π₯βcosβ‘π₯) y = π^π₯ β«1βγγπ ππ π₯γβ‘γ ππ₯γβγ β«1β[(π^π₯) β«1βγsinβ‘π₯ ππ₯ γ]ππ₯ y = π/π π^π (πππβ‘πβπππβ‘π ) + C Given curve passes through (0, 0) Putting x = 0, y = 0 in equation 0 = 1/2 π^0 (sinβ‘0βcosβ‘0) + C 0 = 1/2 (0β1) + C 0 = (β1)/2 + C C = π/π Putting value of C in (1) y = π/π π^π (πππβ‘πβπππβ‘π ) + π/π y β 1/2 = 1/2 π^π₯ (sinβ‘π₯βcosβ‘π₯ ) (2π¦ β 1)/2 = 1/2 π^π₯ (sinβ‘π₯βcosβ‘π₯ ) 2y β 1 = π^π (πππβ‘πβπππβ‘π )
