# A differential equation and a nontrivial solution f are given below.

banganX 2021-02-05 Answered

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.

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## Expert Answer

lamusesamuset
Answered 2021-02-06 Author has 93 answers

Consider the second linearly independent solution as
Obtain derivatives:
${g}^{\prime }\left(t\right)=6v{e}^{2t}+3{e}^{2t}{v}^{\prime }$
and
${g}^{″}\left(t\right)=12v{e}^{2t}+6{e}^{2t}{v}^{\prime }+6{e}^{2t}{v}^{\prime }+3{e}^{2t}{v}^{″}$
$=12v{e}^{2t}+12{e}^{2t}{v}^{\prime }+3{e}^{2t}{v}^{″}$
Substitute g and its derivative:
$t\left(12v{e}^{2t}+12{e}^{2t}{v}^{\prime }+3{e}^{2t}{v}^{″}\right)-\left(2t+1\right)\left(6v{e}^{2t}+3{e}^{2t}{v}^{\prime }\right)+2\left(3v{e}^{2t}\right)=0$
$12tv{e}^{2t}+12t{e}^{2t}{v}^{\prime }+3t{e}^{2t}{v}^{″}-12tv{e}^{2t}-6t{e}^{2t}{v}^{\prime }-6v{e}^{2t}-3{e}^{2t}{v}^{\prime }+6v{e}^{2t}=0$
$3t{e}^{2t}{v}^{″}+\left(6t-3\right){e}^{2t}{v}^{\prime }=0$
$3t{v}^{″}+\left(6t-3\right){v}^{\prime }=0$
Further follows,
Let $w={v}^{\prime }$, then
$3t{w}^{\prime }+\left(6t-3\right)w=0$
$3t{w}^{\prime }=-\left(6t-3\right)w$
$\frac{dw}{w}=\frac{3-6t}{3t}dt$
$\frac{dw}{w}=\left(\frac{1}{t}-2\right)dt$
Integrate both sides:
$\int \frac{dw}{w}=\int \left(\frac{1}{2}-2\right)dt$
$\mathrm{ln}w=\mathrm{ln}t-2t+C$
$w={e}^{\mathrm{ln}t-2t+C}=At{e}^{-2t}$
Then, we have
As $w={v}^{\prime }$, obtain the function v(t):
${v}^{\prime }=At{e}^{-2t}$
$v\left(t\right)=-\frac{A}{2}t{e}^{-2t}+\int \frac{A}{2}{e}^{-2t}dt$
$v\left(t\right)=-\frac{A}{2}t{e}^{-2t}-\frac{A}{4}{e}^{-2t}+C$
Choose $A=-2$ and $C=0$, thus $v\left(t\right)=t{e}^{-2t}+\frac{1}{2}{e}^{-2t}$
Conclusion:
Then, the function g becomes
$g\left(t\right)=3\left(t{e}^{-2t}+\frac{1}{2}{e}^{-2t}\right){e}^{2t}$
$=3t+\frac{3}{2}$
Therefore, the second linearly independent solution is $g\left(t\right)=3t+\frac{3}{2}$

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