Example 14 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Solving Linear differential equations - Equation given
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Example 14 You are here
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Solving Linear differential equations - Equation given
Last updated at April 16, 2024 by Teachoo
Example 14 Find the general solution of the differential equation ππ¦/ππ₯βπ¦=cosβ‘π₯ Differential equation is of the form π π/π π+π·π=πΈ where P = β1 & Q = cos x Finding Integrating Factor IF = e^β«1βπππ₯ IF = e^(ββ«1β1ππ₯) IF = π^(βπ) Solution is y(IF) = β«1βγ(πΓπΌπΉ) ππ₯+πγ ππ^(βπ) = β«1βπ^(βπ) ππ¨π¬β‘γπ+πγ Let I = β«1βπ^(βπ) πππβ‘γπ π πγ 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) πππβ‘γπ π πγ 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 = βeβx cos x β [sinβ‘γπ₯ β«1βγπ^(βπ₯) ππ₯γββ«1βγ(cosγβ‘γπ₯ β«1βγπ^(βπ₯) ππ₯γ)γ "dx " 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 [βπ^(βπ) πππβ‘γπ ββ«1βγβπ^(βπ) πππβ‘π π πγ " " 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)/π (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