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$ty'- y=t^2-1$

I tried

$ty'' - y'=2t$

$ty''' - y''=2$

$y''=2$

$y' = 2t$

then substitute values but I'm not getting anywhere, I think I'm doing it all wrong. Please help. Cheers!

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    Your solution is wrong. $$y'+ty''-y'=2t$$ – 2017-02-03
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    @Moo should I use integrate factor of $\frac{1}{t}$ – 2017-02-03
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    Thanks @Moo , I think I have solved the problem – 2017-02-03

2 Answers 2

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We have $$ty'-y = t^2-1$$ $$ \Rightarrow y'-\frac {1}{t}y = t - \frac {1}{t} $$

Notice that this is a linear differential equation whose procedure as to how to solve is given here. Hope it helps.

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    @Reboot Please comment as soon as you have finished. – 2017-02-03
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    We have $$ty'-y = t^2-1$$ $$ \Rightarrow y'-\frac {1}{t}y = t - \frac {1}{t} $$ $$I=exp(-\int\frac{1}{t}dt)$$ $$I=\frac{1}{t}$$ $$y'\frac{1}{t}-y=1-\frac{1}{t^2}$$ $$y\frac{1}{t}=\int1-t^{-2}$$ $$y\frac{1}{t}=t+t^{-1}+c$$ $$y=t^{2}+1+c$$ – 2017-02-03
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    @Reboot Perfect!! Just take care of the $c_1t $ factor as Moo says. Hope you understood the method fully. – 2017-02-03
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The general solution to the associated homogeneous equation, $ty'- y= 0$, is $y(t)= Ct$ where $C$ is any constant. Using "variation of parameters" we look for a solution to the entire equation of the form $y(t)= u(t)t$, replacing the constant $C$ by the function $u(t)$ (thus "varying" the "parameter", $C$). Then $y'= t u'(t)+ u(t)$ so that the equation becomes $$t^2u'(t)+ tu(t)- tu(t)= t^2u'(t)= t^2- 1.$$ Dividing both sides by $t^2$, $u'(t)= \frac{t^2- t}{t^2}= 1- \frac{1}{t^2}$. Integrating, $u(t)= t- \frac{1}{t}+ C$ so that $$y(t)= tu(t)= Ct+ t^2- 1.$$

Check: if $y(t)= Ct+ t^2- 1$ then $y'= C+ 2t$ so that $ty'(t)- y= Ct+ 2t^2- Ct- (t^2- 1)= t^2+ 1$.

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    There is a sign error in the integration of $u$, the "check" should you clue in that the right side is not the original right side. – 2018-06-28