Question

Use reduction of order to solve ty" + 2ty - 2y yi = t is a solution of the differential equation. = 0 if it is known that

Answer 1

Solution: Given differential equation is

$t^2y''+2ty'-2y=0$.......(1)

and

Th = t

assume that a second solution will have the form

$y_2(t)=v(t)y_1(t)$

$\Rightarrow y_2(t)=vt$

=>y(t) = 0 + tu'

$\Rightarrow y_2''(t)=v'+v'+tv''=2v'+tv''$

putting these values in (1), we get

$t^2(2v'+tv'')+2t(v+tv')-2vt=0$

$\Rightarrow 2t^2v'+t^3v''+2tv+2t^2v'-2vt=0$

$\Rightarrow t^3v''+4t^2v'=0$

$\Rightarrow tv''+4v'=0$

consider,

we get

$tw'+4w=0$

$\Rightarrow tw'=-4w$

$\Rightarrow \frac{w'}{w}=-\frac{4}{t}$

By integrating, we get

$log(w)=-4log(t)+log(c)$

$\Rightarrow log(w)=log(t^{-4})+log(c)$

$\Rightarrow log(w)=log\left ( \frac{c}{t^4} \right )$

taking antilog

$\Rightarrow w=\frac{c}{t^4}$

Substituting back $w=v'$ , we get

$v'=\frac{c}{t^4}$

integrate both sides

$\Rightarrow v=-\frac{c}{3t^3}+d$

By choosing , we have

$v=\frac{1}{t^3}$

Thus,

$y_2(t)=\frac{1}{t^3}t$

$\Rightarrow y_2(t)=\frac{1}{t^2}$

Hence the general solution is given by

$y(t)=c_1y_1(t)+c_2y_2(t)$

$\Rightarrow y(t)=c_1t+\frac{c_2}{t^2}$

Which is the required solution.

This complete the solution.