A differential equation and a nontrivial solution f are given below. Find a second linearly independent solution using reduction of order. Assume that all constants of integration are zero. tx''-(2t+1)x+2x=0, t>0, f(t)=3*e^{2t}

A differential equation and a nontrivial solution f are given below. Find a second linearly independent solution using reduction of order. Assume that all constants of integration are zero. tx''-(2t+1)x+2x=0, t>0, f(t)=3*e^{2t}

Question
A differential equation and a nontrivial solution f are given below. Find a second linearly independent solution using reduction of order. Assume that all constants of integration are zero.
\(tx''-(2t+1)x+2x=0,\ t>0,\ f(t)=3*e^{2t}\)

Answers (1)

2021-02-06
Consider the second linearly independent solution as \(g(t)=v(t),\ f(t)=3ve^{2t}
Obtain derivatives:
\(g'(t)=6ve^{2t}+3e^{2t}v'\)
and
\(g''(t)=12ve^{2t}+6e^{2t}v'+6e^{2t}v'+3e^{2t}v''\)
\(=12ve^{2t}+12e^{2t}v'+3e^{2t}v''\)
Substitute g and its derivative:
\(t(12ve^{2t}+12e^{2t}v'+3e^{2t}v'')-(2t+1)(6ve^{2t}+3e^{2t}v')+2(3ve^{2t})=0\)
\(12tve^{2t}+12te^{2t}v'+3te^{2t}v''-12tve^{2t}-6te^{2t}v'-6ve^{2t}-3e^{2t}v'+6ve^{2t}=0\)
\(3te^{2t}v''+(6t-3)e^{2t}v'=0\)
\(3tv''+(6t-3)v'=0\)
Further follows,
Let \(w=v'\), then
\(3tw'+(6t-3)w=0\)
\(3tw'=-(6t-3)w\)
\(\frac{dw}{w}=\frac{3-6t}{3t}dt\)
\(\frac{dw}{w}=(\frac{1}{t}-2)dt\)
Integrate both sides:
\(\int\frac{dw}{w}=\int(\frac{1}{2}-2)dt\)
\(\ln w=\ln t-2t+C\)
\(w=e^{\ln t-2t+C}=Ate^{-2t}\)
Then, we have
As \(w=v'\), obtain the function v(t):
\(v'=Ate^{-2t}\)
\(v(t)=-\frac{A}{2}te^{-2t}+\int\frac{A}{2}e^{-2t}dt\)
\(v(t)=-\frac{A}{2}te^{-2t}-\frac{A}{4}e^{-2t}+C\)
Choose \(A=-2\) and \(C=0\), thus \(v(t)=te^{-2t}+\frac{1}{2}e^{-2t}\)
Conclusion:
Then, the function g becomes
\(g(t)=3(te^{-2t}+\frac{1}{2}e^{-2t})e^{2t}\)
\(=3t+\frac{3}{2}\)
Therefore, the second linearly independent solution is \(g(t)=3t+\frac{3}{2}\)
0

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