# 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.
$$tx''-(2t+1)x+2x=0,\ t>0,\ f(t)=3*e^{2t}$$

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}$$

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