5 L5: FOL and Integrating Factors (2.1-2.2 part b.)
5.1 Definition: First-order linear diffeq
A First-order linear diffeq(FOL) is an equation of the form:
\[a_1(x)y' + a_0(x)y=b(x)\]
\[\sin x\frac{dy}{dx}+e^xy=x^3+1\]
5.2 Definition: Standard form
The standard form of a first-order linear diffeq is
\[y' +P(x)y=Q(x)\]
A first-order linear differential equation is a type of differential equation that involves only first derivatives and is linear in the unknown function and its derivative. The general form of such an equation is characterized by two functions that are dependent on the independent variable, but not on the unknown function or its derivatives.
The standard form of a first-order linear differential equation is:
\(\frac{dy}{dx} + P(x)y = Q(x)\)
Here:
- \(\frac{dy}{dx}\) represents the first derivative of the unknown function \(y\) with respect to the independent variable \(x\).
- \(P(x)\) and \(Q(x)\) are known functions of \(x\). They can be constants or vary with \(x\), but they don’t depend on \(y\) or its derivatives.
- \(y\) is the unknown function of \(x\) that we want to solve for.
The equation is linear because the unknown function \(y\) and its derivative \(\frac{dy}{dx}\) appear to the first power and are not multiplied together or by any other functions of \(y\).
Solving such equations typically involves finding an integrating factor that makes it possible to express the left side of the equation as the derivative of a product of functions, which can then be integrated to find a solution for \(y\).
\[\frac{dy}{dx}+\frac{e^x}{\sin x}y =\frac{x^3+1}{\sin x}\]
5.3 Solving first-order linear diffeq
5.3.1 Integrating factors:
\[y'+P(x)y=Q(x)\]
On the LHS, we have the addition of two terms, one term with \(y'\) and another term with \(y\) …
Let’s multiply by \(u(x)\)
\[\begin{align} u(x)y'+u(x)P(x)y&=u(x)Q(x)\\ \text{mmmm}\\ \text{The left side looks familiar}& \cdots\\ \end{align}\]
Product rule!
\[\begin{align} \frac{d}{dx}(u(x)y)&=u(x)y'+u'(x)y \end{align}\]
\[u'(x)=u(x)P(x)\] Then:
\[\begin{align} \frac{d}{dx}(u(x)y)&=u(x)Q(x)\\ u(x)y&=\int u(x)Q(x)\,dx \end{align}\]
Therefore, under this assumption, the solution is:
\[\boxed{y=\frac{1}{u(x)}\int u(x)Q(x)\,dx}\] where \(\boxed{u'(x)=u(x)P(x)}\) \[\begin{align} \Rightarrow \int \frac{u'(x)}{u(x)}\,dx &=\int P(x)\,dx\\ \ln \left(u(x)\right) &= \int P(x)\,dx \end{align}\]
\[\Rightarrow \boxed{u(x)=e^{\int P(x)\,dx}}\]
Find \(y\)
\[xy'+y=e^x,\qquad x>0\]
Let’s first arrange in the standard form:
\[y' + \frac{1}{x}y =\frac{e^x}{x}\] Therefore: \[P(x)=\frac{1}{x},\qquad Q(x)=\frac{e^x}{x}\] Then let’s integrating first the factor \(u(x)\),
\[\begin{align} u(x)&=e^{\int P(x)\,dx}\\ &=e^{\int \frac{1}{x}\,dx}\\ &=e^{\ln x}\\ &= x \end{align}\]
Now the solution:
\[\begin{align} y&=\frac{1}{u(x)}\int u(x)Q(x)\,dx\\ &=\frac{1}{x} \int x \frac{e^x}{x}\,dx\\ &=\frac{1}{x} \int e^x\,dx\\ \end{align}\]
\[\therefore \boxed{y=\frac{1}{x}\left(e^x+C\right)}\]
5.4 Existence and Uniquenens
Suppose that \(P(x)\) and \(Q(x)\) are continuous over \((\alpha,\beta)\). Then there is one and only one solution \(y=y(x)\) satisfying \[y'+P(x)y=Q(x)\], for all \(x\) on the interval \((\alpha,\beta)\) and for the initial condition \(y(x_0)=y_0\) where \(x_0\in (\alpha,\beta)\)
5.5 Example
Solve \[(1+x)\frac{dy}{dx}+y=\sqrt{x},\qquad x>0\]
\[\begin{align} (1+x)\frac{dy}{dx}+y&=\sqrt{x},\qquad x>0\\ \frac{dy}{dx}+\frac{y}{1+x}&=\frac{\sqrt{x}}{1+x}\\ \Rightarrow \end{align}\]
\[P(x)=\frac{1}{1+x}\qquad Q(x)=\frac{\sqrt{x}}{1+x}\] \(P(x)\) and \(Q(x)\) are continuous for \(x>0\), then I should find a solution!
We can try with integrating factors:
\[u(x)=e^{\int \frac{1}{1+x}\,dx}=e^{\ln(1+x)}=1+x\]
\[\begin{align} y&=\frac{1}{1+x}\int (1+x)\frac{\sqrt{x}}{1+x}\,dx\\ &=\frac{1}{1+x}\left(\frac{2}{3}x^{3/2}+C\right) \end{align}\]
Which will be unique when we set the initial condition.
5.6 Have questions ?
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5.7 References
- Separation of variables https://en.wikipedia.org/wiki/Separation_of_variables
- More Examples https://www.mathsisfun.com/calculus/separation-variables.html
- Video: The Big Theorem of Differential Equations: Existence & Uniqueness https://youtu.be/_WpncZ3RkTg?feature=shared