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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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