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Using a suitable substitution, transform the following equation into a first order Linear differential equation:

$y' = f_0(x) +f_1(x)*y + f_2(x) *y^2$ for $f_2(x)\not\equiv 0$

Attempt at a solution: I've found that if $y_1(x)$ is a solution, we can use $y(x) = y_1(x) + \frac{1}{u(x)}$ but I'm unfamiliar with proving stuff for this special class of ODEs. Please help.

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    $\begingroup$ These equations are known as Riccati's equation. Another helpful resource here that i'll leave for your perusal. $\endgroup$ – thesmallprint Nov 5 '17 at 18:39
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For $$y' = f_0(x) +f_1(x)*y + f_2(x) *y^2 \hspace{5mm} \text{for} \, f_2(x) \not\equiv 0$$ then, suppose $q_{0}(x)$ is a solution, let $y(x) = q_{0}(x) + p(x)$ to obtain \begin{align} q_{0}' + p' &= f_{0} + f_{1} q_{0} + f_{1} p + f_{2} (q_{0}^{2} + 2 q_{0} p + p^{2}) \\ p' &= f_{2} p^{2} + (f_{1} + 2 q_{0} f_{2}) p + (- q_{0}^{'} + f_{2} q_{0}^{2} + f_{1} q_{0} + f_{0}) \\ p' &= f_{2} p^{2} + Q_{0} p, \end{align} where $Q_{0} = f_{1} + 2 q_{0} f_{2}$.

Now \begin{align} \frac{p'}{p^{2}} - \frac{Q_{0}}{p} &= f_{2} \\ - \frac{d}{dx} \left(\frac{1}{p}\right) - Q_{0} \, \frac{1}{p} &= f_{2} \end{align} Let $v(x) = 1/p(x)$ to obtain, with use of integrating factor, \begin{align} v' + Q_{0} v &= - f_{2} \\ \frac{d}{dx} \left[ v \, e^{\int^{x} Q_{0} dt} \right] &= - f_{2} \, e^{ \int^{x} Q_{0} dt} \\ v &= e^{- \int^{x} Q_{0} dt} \, \left[c - \int^{x} f_{2}(u) \, e^{\int^{u} Q_{0} dt} \, du \right] \end{align}

Placing everything together provides $$y(x) = q_{0}(x) + e^{\int^{x} (f_{1}(t) + 2 q_{0}(t) f_{2}(t)) dt} \, \left(c + \int^{x} f_{2}(u) \, e^{- \int^{u} (f_{1}(t) + 2 q_{0}(t) f_{2}(t)) dt} \, du \right)^{-1} $$

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