Bernoulli differential equation, Cauchy problem

[nuuskur]

Homework Statement

Observe a Cauchy problem \begin{cases}y' + p(x)y =q(x)y^n\\ y(x_0) = y_0\end{cases}
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 $y^{-n}$:
\frac{y'}{y^n}+p(x)\frac{y}{y^n}=q(x)
Substituting $w=y^{1-n}$ then $w' = (1-n)y^{-n}\cdot y'$ Resulting in w' + p(x)w(1-n) = q(x)(1-n)
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?

 

[verty]

Should it be $w' + p(x)w(1 - n) = q(x)(1 - n)$?

 

[nuuskur]

Should it be $w' + p(x)w(1 - n) = q(x)(1 - n)$?

Yes, correct you are. Thanks.

 

[nuuskur]

Edit
NOPE, still not right