Assume that y is the solution of the initial-value problem 5 sin x If y is written as a power series then y = + X + + y - 3y: = + x² 5 y = x = 0 x = 0' ∞ Σ n=0 Cnxn y(0) = 1.

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Assume that y is the solution of the initial-value problem
5 sin x
If y is written as a power series
then
y =
+
X +
x² +
y - 3y
x +
n3
x² +
{{₁
X
x #0
X =
= 0'
∞
y=Σ cnopne
n=0
y(0) = 1.
Note: You do not have to find a general expression for Cn. Just find the coefficients one by one.
Transcribed Image Text:Assume that y is the solution of the initial-value problem 5 sin x If y is written as a power series then y = + X + x² + y - 3y x + n3 x² + {{₁ X x #0 X = = 0' ∞ y=Σ cnopne n=0 y(0) = 1. Note: You do not have to find a general expression for Cn. Just find the coefficients one by one.
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