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Last updated at Dec. 24, 2018 by Teachoo
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Example 19 Find the general solution of the differential equation ππ¦/ππ₯βπ¦=cosβ‘π₯ Differential equation is of the form ππ¦/ππ₯+ππ¦=π where P = β1 & Q = cos x IF = e^β«1βπππ₯ IF = e^(ββ«1β1ππ₯) IF = π^(βπ₯) Solution is y(IF) = β«1βγ(πΓπΌπΉ) ππ₯+πγ π¦π^(βπ₯) = β«1βπ^(βπ₯) cosβ‘γπ₯+πγ Let I = β«1βπ^(βπ₯) cosβ‘γπ₯ ππ₯γ Integrating by parts with β«1βγπ(π₯) π(π₯) ππ₯=π(π₯) β«1βγπ(π₯) ππ₯ ββ«1βγ[π^β² (π₯) β«1βγπ(π₯) ππ₯] ππ₯γγγγ Take f (x) = cos x & g (x) = π^"βx" I = cos x β«1βγπ^(βπ₯) ππ₯γβ β«1β[βsinβ‘γπ₯β«1βγπ^(βπ₯) ππ₯γγ ]ππ₯ I = γβπγ^(βπ₯)cos x ββ«1βγβsinβ‘γπ₯ (βπ^(βπ₯) γ)γ ππ₯ I = βeβx cos x β β«1βγπ^(βπ₯) sinβ‘γπ₯ ππ₯γ γ. Integrating by parts with β«1βγπ(π₯) π(π₯) ππ₯=π(π₯) β«1βγπ(π₯) ππ₯ ββ«1βγ[π^β² (π₯) β«1βγπ(π₯) ππ₯] ππ₯γγγγ Take f (x) = sin x g (x) = eβx I = βeβx cos x β [sinβ‘γπ₯ β«1βγπ^(βπ₯) ππ₯γββ«1βγ(cosγβ‘γπ₯ β«1βγπ^(βπ₯) ππ₯γ)γ "dx " γ ] I = βeβx cos x β [βπ^(βπ₯) sinβ‘γπ₯ ββ«1βγβπ^(βπ₯) cosβ‘π₯ ππ₯γ " " γ ] I = βeβx cos x β [βπ^(βπ₯) sinβ‘γπ₯+β«1βγπ^(βπ₯) cosβ‘π₯ ππ₯γ " " γ ] I = βeβx cos x + π^(βπ₯) sinβ‘γπ₯ββ«1βγπ^(βπ) πππβ‘π π πγ " " γ I = eβx (sin x β cos x) β I 2I = eβx (sin x β cos x) I = π^(βπ₯)/2 (sin x β cos x) From (1) y π^(βπ₯) = β«1βγπ^(βπ₯) cosβ‘γπ₯+πγ γ y π^(βπ₯) = π^(βπ₯)/2 (sin x β cos x) + c y = π/π (sin x β cos x) + cπ^π
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