$$y''+\frac{y'}{x-1}=x(x-1)$$
I use the substitution $u=y'$ and then $y''=uu'$, but then I do not know what to do next.
A differential equation of order 2.
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ordinary-differential-equations
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0This is a first order linear differential equation in $y'$ – 2017-01-23
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0So may I write it like it was y''=u'? – 2017-01-23
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0OK. $y''=u'$ and $y'=u$. – 2017-01-23
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0OK. Is there any time I just have to write that $y''=uu'$? – 2017-01-23
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0Yes. when in your equation there is not any $x$, you take $y''=uu'$ and $y'=u$. In this case $y$ will be independent variable. – 2017-01-23
1 Answers
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$$y''+\frac{y'}{x-1}=x(x-1)$$ Let $y'=u$. Then, $y''=u'$. $$u'+\frac{1}{x-1}u=x(x-1)$$
This is a linear first order differential equation of the form $$u'+uP(x)=Q(x)$$ which can be solved by the usual method.