Let $\{y_i\}_{i=0,...,d}$ be a sequences of functions depending on $x$ and satisfies $y'_i(x)=-\frac{(i+1)y_i(x)}{\sum_{i=0}^dy_i(x)}$ for $0\le i\le d$ with original conditions $y_d(0)=1$ and $y_i(0)=0$ for $i
How to solve such ordinary differential equations?
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ordinary-differential-equations
1 Answers
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Since $y_i(0) = 0$, and $y_i'$ is proportional to $y_i$, one concludes that $y_i(x) = $ for all $x$. Which leaves $y_d' = - \frac{(d+1)y_d}{y_d} = -(d+1)$.
For a general case, you may observe that (as long as neither $y_i$ nor the denominator vanish) that
$$\frac{y_i'}{y_i} = (i+1) \frac{y_0'}{y_0}$$ therefore
$$y_i = a_i|y_0|^{i+1}$$
where $a_i$ shall be determined from the initial conditions. Substituting it back to the first equation gives
$$y_0' = -\frac{1}{\sum_0^d a_i |y_0|^i}$$
and, as long as $y_0 > 0$, one has
$$\sum_0^d\frac{a_i}{i+1}y_0^{i+1}(x) = C - x$$