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Homework Statement
Observe a Cauchy problem [tex]\begin{cases}y' + p(x)y =q(x)y^n\\ y(x_0) = y_0\end{cases}[/tex]
Assume ##p(x), q(x)## are continuous for some ##(a,b)\subseteq\mathbb{R}##
Verify the equation has a solution and determine the condition for there to be exactly one solution.
Homework Equations
Theorem:
Assume ##y' = f(x,y)## and ##y(x_0) = y_0##, where ##x_0\in (a,b)##. If ##f## is continuous in some ##D\subseteq\mathbb{R}^2##, there exists continuous ##f_y := \frac{\partial f}{\partial y}## in ##D## and ##(x_0,y_0)\in D## then the Cauchy problem has exactly one solution.
The Attempt at a Solution
Solve the equation. If ##y^n\neq 0## then multiplying both sides by ##y^{-n}##:
[tex]\frac{y'}{y^n}+p(x)\frac{y}{y^n}=q(x)[/tex]
Substituting ##w=y^{1-n}## then ##w' = (1-n)y^{-n}\cdot y'## Resulting in [tex]w' + p(x)w(1-n) = q(x)(1-n)[/tex]
From Lagrange's method we can say ##w = C(x)e^{-\int p(x)dx}##
Apply the theorem:
##w' = f(x,w)## and we have ##f(x,w) = q(x)(1-n) - p(x)w(1-n)##. From which we can say ##\frac{\partial f}{\partial w} = -p(x)(1-n)##. As ##p(x)## is assumed to be continuous then ##f_w## is defined and continous. ##(x_0, y_0)## is guaranteed to be in ##D## because ##D## is only bound by ##a## and ##b## on the X-axis. There are no limitations on the Y-axis.
Problem is how do I justify that ##f(x,w)## is necessarely continuous in ##D##. I know that ##f_w## is continous, if somehow ##f_x## would be continuous then ##f## itself is continous. From analysis I remember that if ##f## is an elementary function, then it and its differentials are continous, but ##p(x) , q(x)## don't have to be elementary functions, do they?
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