# Example 19 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 24, 2018 by

Last updated at Dec. 24, 2018 by

Transcript

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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Example 19 You are here

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Example 28 Important Deleted for CBSE Board 2022 Exams

Chapter 9 Class 12 Differential Equations (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.