y′′+2y′+y=3e−xx+1
Let's find yc , using auxiliary equation:
k2+2k+1=0(k+1)2=0k1=k2=−1
Therefore yc=C1e−x+xC2e−x
Let
C1=f(x),C2=g(x)
Then
y=fe−x+xge−x⇒y′=f′e−x−fe−x+g(e−x−xe−x)+g′xe−x
Letf′e−x+g′xe−x=0(∗)
Then
y′=−fe−x+g(e−x−xe−x)⇒y′′=−f′e−x+fe−x+g′(e−x−xe−x)+g(−2e−x+xe−x)
Substituting into our equation for y, y', y'', we have:
−f′e−x+fe−x+g′(e−x−xe−x)+g(−2e−x+xe−x)+2(−fe−x+g(e−x−xe−x))++fe−x+xge−x=3e−xx+1 −f′e−x+g′(e−x−xe−x)=3e−xx+1
with(*) we have
{−f′e−x+g′(e−x−xe−x)=3e−xx+1f′e−x+g′xe−x=0
From 2-nd equation:
f′=−g′x
Then
g′x+g′−g′x=3x+1⇒g′=3x+1⇒g=2x+1(x+1)+C
f′=−g′x=−3xx+1⇒f=−52(3x−2)(x+1)x+1+C2y=fe−x+xge−x=(−52(3x−2)(x+1)x+1+C2)e−x+x(2x+1(x+1)+C)e−x
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