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Simora [160]
3 years ago
7

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

4.2, y2 = y1(x) e−∫p(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). 9y'' − 12y' + 4y = 0; y1 = e2x/3
Mathematics
1 answer:
Ber [7]3 years ago
7 0
Given that y_1=e^{2x/3}, we can use reduction of order to find a solution y_2=v(x)y_1=ve^{2x/3}.

\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}
\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}

\implies9y''-12y'+4y=0
\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0
\implies9v''-3v'=0

Let u=v', so that

9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0
e^{-x/3}u'-\dfrac13e^{-x/3}u=0
\left(e^{-x/3}u\right)'=0
e^{-x/3}u=C_1
u=C_1e^{x/3}

\implies v'=C_1e^{x/3}
\implies v=3C_1e^{x/3}+C_2

\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}
\implies y_2=3C_1e^x+C_2e^{2x/3}

Since y_1 already accounts for the e^{2x/3} term, we end up with

y_2=e^x

as the remaining fundamental solution to the ODE.
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