# theorem of existence and uniqueness for first order linear differential equation

The theorem of existence and uniqueness is: Let $y'+p(x)y=g(x)$, $y(x_{0})=y_{0}$ be a first order linear differential equation such that $p(x)$ and $g(x)$ are both continuous for $a<x<b$. Then there is a unique solution that satisfies it.

When a differential equation has no solution that satisfies $y(x_{0})=y_{0}$, what does this mean?? Can the theorem be verified??

• When you say "When a differential equation has no solution that satisfies $y(x_0)=y_0$", do you mean one that looks like $y'+py=g$? – Git Gud Nov 9 '13 at 0:49
• The general solution of the differential equation is $y(x)=x(c+\int(s(x))dx)$, $s(x)=\int(\frac{sinx}{x})$ $s(0)=1$and the initial condition is $y(0)=1$. But for no c, the initial value can be verified. – Mary Star Nov 9 '13 at 1:08
• What differential equation is that? – Git Gud Nov 9 '13 at 1:10
• $xy'(x)-y(x)=xsin(x)$ – Mary Star Nov 9 '13 at 1:11
• That is not equivalent to $y'(x)-\dfrac{y(x)}x=\sin(x)$, unless you take suitable intervals. – Git Gud Nov 9 '13 at 1:21

The existence and uniqueness theorem for first-order linear differential equations can be stated as follows. Suppose that $P$ and $Q$ are continuous on the open interval $I$. If $a$ and $b$ are any real numbers, then there is a unique function $y = f(x)$ satisfying the initial-value problem $y' + P(x)y = Q(x)$ with $f(a) = b$ on the interval $I$. With regard to your question, the important point is that $a$ and $b$ are arbitrary real numbers and the unique solution $f$ to the differential equation satisfies $f(a) = b$ for every choice of $a$ and $b$. Since every first-order linear differential equation satisfying the constraints of the theorem has a solution satisfying $f(a) = b$, there is no case in which such an equation has no solution satisfying $f(a) = b$.
If we look at the simpler case of homogeneous first-order linear differential equations of the form $y' + P(x)y = 0$, where $P$ is continuous on the open interval $I$, we can directly verify that for every choice of $a$ and $b$, the function $f(x) = be^{-A(x)}$ where $A(x) = \int_{a}^x P(t) dt$ is a solution to $y' + P(x)y = 0$. Now letting $g$ be an arbitrary solution of $y' + P(x)y = 0$, we establish uniqueness by showing that $g(x)e^{A(x)} = b$. Differentiating, we see that $h(x) = g(x)e^{A(x)}$ is constant on the interval $I$. But $h(a) = b$, so we must have $h(x) = b$. This demonstrates that $g = f$. Notice that the choice of $a$ and $b$ does not affect the existence or uniqueness of solutions. Can you verify this in the case of non-homogeneous first-order linear differential equations of the form $y' + P(x)y = Q(x)$?
• In my case the differential equation is $y'(x)-\frac{1}{x}y(x)=sin(x)$ with $f(0)=1$. The general solution of this equation is: $y(x)=x[c+\int{s(x)}dx]$, where $s(x)=\frac{sin(x)}{x}$. For $x=0$ $y(0)=0$ for any c but $f(0)=1$.So there is no solution that satisfies f(0)=1. Am I wrong? – Mary Star Nov 9 '13 at 1:47
• Since the function $P(x) = -1/x$ is not continuous at $0$, $P$ is not continuous on any open interval containing $0$. This means that there is not necessarily a unique solution of the indicated form to the differential equation. It appears that no solution satisfies f(0) = 1, but this doesn't interfere with the theorem because one of its hypotheses isn't satisfied. – danportin Nov 9 '13 at 1:50
• When you transform your differential equation into the form required by the theorem, you violate one of the hypotheses of the theorem - namely, that $P$ be continuous on the relevant interval. Since the $P$ in question isn't continuous on that interval, the theorem doesn't apply. This doesn't mean that there isn't a unique solution to the differential equation, just that the existence-uniqueness theorem for first-order linear differential equations won't provide the answer. – danportin Nov 12 '13 at 7:36