Matematické Fórum

stuart clark · 28. 07. 2014 09:21

If $\displaystyle y^2 = ax^2+2bx+c\;,$ and $\displaystyle U_{n} = \int \frac{x^n}{y}dx\;,$ Then prove that $(n+1)a.U_{n}+(2n+1)b.U_{n}+c.U_{n-1}=x^n.y$

and deduce that $aU_{1} = y-bU_{0}$ and $\displaystyle 2a^2U_{2} = y(ax-3b)-(ac-3b^2)U_{0}$

Bati · 13. 08. 2014 11:17 — Editoval Bati (13. 08. 2014 16:29)

↑ stuart clark:
There're probably mistakes in $(n+1)a.U_{n}+(2n+1)b.U_{n}+c.U_{n-1}=x^n.y$ , I think it should be $(n+1)a.U_{n+1}+(2n+1)b.U_{n}+nc.U_{n-1}=x^n.y$ (*).
To see this, consider the expression $\int x^{n-1}y\,dx$ and compute it two-way. Firstly, write $\int x^{n-1}y\,dx=\int \frac{x^{n-1}}{y}y^2\,dx=aU_{n+1}+2bU_n+cU_{n-1}$ - this is true for every integer $n$ . Secondly, use integration by parts to obtain $\int x^{n-1}y\,dx=\tfrac1{n}x^ny-\frac1n\int x^n\frac{ax+b}{y}\,dx=\tfrac1nx^ny-\tfrac{a}{n}U_{n+1}-\tfrac{b}{n}U_n$ - that is true for non-zero integers $n$ . By comparing the right-hand sides of last two equations, multiplying $n$ and collecting terms together we get the equation (*). But this isn't enough to simply deduce particular case $aU_{1} = y-bU_{0}$ since what we just proved doesn't hold for $n=0$ . This obstacle, however, can be circumvented with little magic: $\int x^{-1}y\,dx=\int xy\,\frac1{x^2}\,dx$ and we use integration by parts again (differentiate $xy$ ) to get $\int x^{-1}y\,dx=-y+\int x^{-1}y\,dx+aU_1+bU_0$ and this clearly proves the first particular case. The second one is a simple consequence of the first one and (*) in case of $n=1$ .

Matematické Fórum

#1 28. 07. 2014 09:21

Reduction Integral

#2 13. 08. 2014 11:17 — Editoval Bati (13. 08. 2014 16:29)

Re: Reduction Integral

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