Two statements about solutions of linear ODE 
Let $y'+p(x)y=g(x)$ where $p(x),g(x)$ are continuous on $\mathbb{R}$
and suppose $y_1,y_2$ are two solutions to the given ODE. Let
$x_0\in\mathbb{R}$ such that $y_1(x_0)>y_2(x_0)$. Determine whether the following statements are true or false:

*

*$y_1(x)>y_2(x)$ for all $x\in\mathbb{R}$

*It is possible that there exists $x_1\in\mathbb{R}$ such that $y_1(x_1)=y_2(x_1)$

My initial thought was that the first statement is true and the second is false. The second is false due to uniqueness theorem (if $y_1(x_1)=y_2(x_1)$ then $y_1=y_2$ which contradicts that $y_1(x_0)>y_2(x_0)$). Then the first statement is true because otherwise (that is if $y_1(x_1)<y_2(x_1)$ for some $x_1\in\mathbb{R}$) the functions $y_1,y_2$ must intersect (I suppose they are continuous?). But then I'm not sure that the uniqueness theorem works here (we only know that $p(x),g(x)$ are continuous). Any suggestions?
 A: Uniqueness holds for solutions of ODE with continuous coefficients when the coefficient of the highest derivative is never 0, as in this case.
Your reasoning is correct. 
A: Since this is a linear equation, you have an explicit formula for the possible solutions so you don't need to rely on a general existence and uniqueness theorem. If $y_1,y_2$ are solutions to your equation then $y(x) = y_1(x) - y_2(x)$ is a solution to the linear homogeneous equation 
$$ y'(x) + p(x)y(x) = 0. $$
If $y(x)$ satisfies the equation above on some interval $I$ containing $x_1 \in \mathbb{R}$, then multiplying the equation by the integrating factor $\exp \left( \int_{x_1}^x p(t) \, dt \right)$ we obtain
$$ y'(x)\exp \left( \int_{x_1}^x p(t) \, dt \right) + p(x)y(x)\exp \left( \int_{x_1}^x p(t) \, dt \right) =
\\
\left( y(x) \exp \left( \int_{x_1}^x p(t) \, dt \right) \right)' = 0. $$
Then, we must have
$$ y(x) = y(x_1) \exp \left( -\int_a^x p(t) \, dt \right). $$
The formula above shows that if $y_1(x_1) = y_2(x_1)$ then $y(x_1) = 0$ and so $y(x) \equiv 0$ on $I$ which implies that $y_1(x) \equiv y_2(x)$ on $I$. It also shows that if $y_1(x_1) > y_2(x_1)$ then $y(x) > 0$ for all $x \in I$ and so $y_1(x) > y_2(x)$ for all $x \in I$.
