Question
Asked Jul 19, 2019
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Differential equations

Find the particular solution to the initial value problem:

(x+1)y"-(2-x)y'+y=0,    y(0)=2, y'(0)=-1

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Expert Answer

Step 1

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о (х +1)—п(п-1)с, " - (2-х)| Упс, х"-1 с х" %3D0 E2 $чи-лозрт и р. *Zn(n-1) с, х" In(n-1) с,2 -2) n-2 Сх" %3D0 псх пс 2 ОС ос о Хn(n-1) с, у" - Zn(n-1) c, "3 - 22 псх = 0 пс 1 v=2 0 и-0

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Step 2

Comparing constant terms on each sides,

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C2 2 2c, Co 0 C2 2 2c-Co 2(-1) 2 C2 (y(0)2 Co(0)=-1=c) 2 c2-2

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Step 3

Comparing coefficient of x ...

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2(1)с, +3(2)с, - 2(2)с, - (1)с, ~ с, 30 2с, + бс — 4с, + 2с, %3D0 бс, — 2с, + 2с, %3 0 бсз — 2с, — 2с 2(-2) 2(-) Сз У (0) --1-с) 6 Сз 3

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