\(
y'=−y^2(t)+y(t)−t,\\
u\left(t\right)=e^{-\int{-y\left(t\right)\mbox{d}t}}\\
u\left(t\right)=e^{\int{y\left(t\right)\mbox{d}t}}\\
u'\left(t\right)=y\left(t\right)e^{\int{y\left(t\right)\mbox{d}t}}\\
u'\left(t\right)=y\left(t\right)u\left(t\right)\\
u''\left(t\right)=y'\left(t\right)u\left(t\right)+y\left(t\right)u'\left(t\right)\\
u''\left(t\right)=y'\left(t\right)u\left(t\right)+\frac{u'\left(t\right)}{u\left(t\right)}u'\left(t\right)\\
u''\left(t\right)u\left(t\right)=y'\left(t\right)u^2\left(t\right)+\left(u'\left(t\right)\right)^2\\
y'\left(t\right)u^2\left(t\right)=u''\left(t\right)u\left(t\right)-\left(u'\left(t\right)\right)^2\\
y'\left(t\right)=\frac{u''\left(t\right)u\left(t\right)-\left(u'\left(t\right)\right)^2}{u^2\left(t\right)}\\
y'\left(t\right)=\frac{u''\left(t\right)}{u\left(t\right)}-\left(\frac{u'\left(t\right)}{u\left(t\right)}\right)^2\\
\frac{u''\left(t\right)}{u\left(t\right)}-\left(\frac{u'\left(t\right)}{u\left(t\right)}\right)^2=-\left(\frac{u'\left(t\right)}{u\left(t\right)}\right)^2+\frac{u'\left(t\right)}{u\left(t\right)}-t\\
\frac{u''\left(t\right)}{u\left(t\right)}=\frac{u'\left(t\right)}{u\left(t\right)}-t\\
u''\left(t\right)=u'\left(t\right)-tu\left(t\right)\\
u''\left(t\right)-u'\left(t\right)+tu\left(t\right)=0
\)
\(
u\left(t\right)=\sum_{n=0}^{\infty}{c_{n}t^{n}}\\
\sum_{n=0}^{\infty}{\left(n+2\right)\left(n+1\right)c_{n+2}t^{n}}+\sum_{n=0}^{\infty}{-\left(n+1\right)c_{n+1}t^n}+\sum_{n=0}^{\infty}{c_{n}t^{n+1}}=0\\
2c_{2}+\sum_{n=1}^{\infty}{\left(n+2\right)\left(n+1\right)c_{n+2}t^{n}}-c_{1}-\sum_{n=1}^{\infty}{-\left(n+1\right)c_{n+1}t^n}+\sum_{n=0}^{\infty}{c_{n}t^{n+1}}=0\\
2c_{2}-c_{1}+\sum_{n=0}^{\infty}{\left(n+3\right)\left(n+2\right)c_{n+3}t^{n+1}}-\sum_{n=0}^{\infty}{-\left(n+2\right)c_{n+2}t^{n+1}}+\sum_{n=0}^{\infty}{c_{n}t^{n+1}}=0\\
2c_{2}-c_{1}+\sum_{n=0}^{\infty}{\left[\left(n+3\right)\left(n+2\right)c_{n+3}-\left(n+2\right)c_{n+2}+c_{n}\right]t^{n+1}}\\
\)
To daje nam równanie rekurencyjne
\(
\begin{cases}c_{0}\in\mathbb{R}\\c_{1}\in\mathbb{R}\\c_{2}=\frac{1}{2}c_{1}\\c_{n+3}=\frac{\left(n+2\right)c_{n+2} - c_{n}}{\left(n+3\right)\left(n+2\right)}\end{cases}\\
\)
Przyjmijmy że \(a_{n}=n!c_{n}\)
mamy wtedy
\(\begin{cases}a_{0}=c_{0}\\a_{1}=c_{1}\\a_{2}=c_{1}\\\frac{a_{n+3}}{\left(n+3\right)!}=\frac{\left(n+2\right)a_{n+2}}{\left(n+3\right)\left(n+2\right)\left(n+2\right)!}-\frac{a_{n}}{\left(n+3\right)\left(n+2\right)n!}\end{cases}\\
\begin{cases}a_{0}=c_{0}\\a_{1}=c_{1}\\a_{2}=c_{1}\\\frac{a_{n+3}}{\left(n+3\right)!}=\frac{a_{n+2}}{\left(n+3\right)\left(n+2\right)!}-\frac{a_{n}}{\left(n+3\right)\left(n+2\right)n!}\end{cases}\\
\begin{cases}a_{0}=c_{0}\\a_{1}=c_{1}\\a_{2}=c_{1}\\\frac{a_{n+3}}{\left(n+3\right)!}=\frac{a_{n+2}}{\left(n+3\right)!}-\frac{a_{n}}{\left(n+3\right)\left(n+2\right)n!}\end{cases}\\
\begin{cases}a_{0}=c_{0}\\a_{1}=c_{1}\\a_{2}=c_{1}\\\frac{a_{n+3}}{\left(n+3\right)!}=\frac{a_{n+2}}{\left(n+3\right)!}-\frac{\left(n+1\right)a_{n}}{\left(n+3\right)\left(n+2\right)\left(n+1\right)n!}\end{cases}\\
\begin{cases}a_{0}=c_{0}\\a_{1}=c_{1}\\a_{2}=c_{1}\\\frac{a_{n+3}}{\left(n+3\right)!}=\frac{a_{n+2}}{\left(n+3\right)!}-\frac{\left(n+1\right)a_{n}}{\left(n+3\right)!}\end{cases}\\
\begin{cases}a_{0}=c_{0}\\a_{1}=c_{1}\\a_{2}=c_{1}\\a_{n+3}=a_{n+2}-\left(n+1\right)a_{n}\end{cases}\\
\)
Mamy zatem takie równanie rekurencyjne
\(
\begin{cases}a_{0}\in\mathbb{R}\\a_{1}\in\mathbb{R}\\a_{2}=a_{1}\\a_{n+3}=a_{n+2}-\left(n+1\right)a_{n}\end{cases}\\
\)
i jak znaleźć wzór jawny dla tego równania rekurencyjnego ?
Równanie Riccatiego
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