Find all solutions of a given initial value problem

Let the function $v = g(x)$ be the unique solution of the initial-value problem

$v' + kP(x)v = kQ(x)$

with $g(a) = b$ where $k \neq 0$ is a constant, $P(x)$ and $Q(x)$ are continuous functions on an interval $I$ , and $a \in I$ with $b$ any real number. If $n \neq 1$ and $k = n-1$ prove that the function $y = f(x)$ for which $f(x) \neq 0$ for all $x \in I$ is a solution of

$y' + P(x) y = Q(x) y^n \qquad \text{on } I, \qquad \text{with } f(a)^k = b$

if and only if $(f(x))^k = g(x)$ for all $x \in I$ .

Proof. $(\Rightarrow)$ Assume $y = f(x)$ is a solution of $y' + P(x)y = Q(x) y^n$ . Let $v = y^{1-n} = y^k$ . Then,

$v' = ky^{k-1} y' \quad \implies \quad y' = \frac{v'}{ky^{k-1}}.$

(This division is allowed since $y \neq 0$ on $I$ implies $y^{k-1} \neq 0$ on $I$ .) Therefore,

$\begin{align*} y' + P(x) y = Q(x) y^n && \implies && \frac{v'}{ky^{k-1}} + P(x) y &= Q(x) y^n \\[9pt] && \implies && \frac{v'}{ky^{k-1+n}} + P(x) y^{1-n} &= Q(x) \\[9pt] && \implies && \frac{v'}{k} + P(x) y^k &= Q(x) &(k = 1-n) \\[9pt] && \implies && v' + kP(x) v = kQ(x). \end{align*}$

Thus, if $y = f(x)$ is a solution of $y' + P(x)y = Q(x)y^n$ then $g(x) = v = y^k$ is the unique solution of

$v' + kP(x)v = kQ(x) \qquad \text{with} \qquad g(a) = (f(a))^k = b.$

$(\Leftarrow)$ Conversely, if $(f(x))^k = g(x)$ on $I$ then

$g'(x) = k(f(x))^{k-1} f'(x) \quad \implies \quad f'(x) = \frac{1}{k} g'(x) (f(x))^{1-k}.$

Therefore,

$f'(x) + P(x) f(x) = \frac{1}{k} g'(x)(f(x))^{1-k} + P(x) f(x)$

Then, since we know by hypothesis that $v = g(x)$ is the unique solution of $v' + kP(x) v = kQ(x)$ we have

$g'(x) = kQ(x) - kP(x) g(x).$

Substituting this into the above equation (and noting that our assumption that $(f(x))^k = g(x)$ implies $(f(x))^{1-k} = \frac{f(x)}{g(x)}$ ) we have

$\begin{align*} f'(x) + P(x) f(x) &= \frac{1}{k} (kQ(x) - kP(x)g(x)) (f(x))^{1-k} + P(x) f(x) \\[9pt] &= Q(x) (f(x))^{1-k} - P(x) g(x) (f(x))^{1-k} + P(x)f(x) \\[9pt] &= Q(x) (f(x))^{1-k} - P(x) g(x) \frac{f(x)}{g(x)} + P(x) f(x) \\[9pt] &= Q(x) (f(x))^{1-k} - P(x) f(x) + P(x) f(x) \\[9pt] &= Q(x) (f(x))^{1-k} \\ &= Q(x) (f(x))^n. \end{align*}$

Therefore $y = f(x)$ is a solution of $y' + P(x)y = Q(x)y^n$ with $(f(a))^k = g(a) = b. \qquad \blacksquare$

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Find all solutions of a given initial value problem

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