# 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:

1. $$y_1(x)>y_2(x)$$ for all $$x\in\mathbb{R}$$
2. 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 $$y_1(x_1) for some $$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?

• you are correct,because in general solution of ODE,every solutions differ from each other by constanst ,$y\left( x \right) ={ e }^{ -\int { P\left( x \right) dx } }\left( \int { q\left( x \right) { e }^{ \int { P\left( x \right) dx } } } +C \right)$ Jul 31, 2016 at 10:22

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 $x \in I$.

• So I was right about the conclusion (1. True 2. False) but I was wrong on the explanation? Thank you. Jul 31, 2016 at 10:44
• @user342907, the Lipschitz requirement in Picard's theorem are on dependance on y. I linear ODE as above satisfies the Liptschitz condition (locally, since the cosefficients are continuous, hence loclly bounded). Jul 31, 2016 at 21:41
• @user342907 (@VictorZurkowski) I'm sorry, of course the Picard theorem does apply for this case even though one doesn't really need it as this is a first order linear equation. Sorry for misleading you. Aug 1, 2016 at 23:00

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.