Table of Contents
intro
Pretty straightforward for this one.
Adapted from notes for Oregon State Math 256.
My other notes are more fun.
first order linear equations (integrating factors)
Consider a first-order linear differential equation:
\[a_1(x) \frac{dy}{dx} + a_0(x) y = g(x).
\]
By dividing through by \(a_1(x)\), we can rewrite it in the standard form:
\[\frac{dy}{dx} + p(x) y = f(x).
\]
Looking at the LHS, it resembles the product rule applied to \(y(x)\) and some function \(\mu(x)\):
\[\frac{d}{dx}[\mu(x) y] = \mu(x) \frac{dy}{dx} + \mu'(x) y.
\]
Multiplying both sides of the equation by \(\mu(x)\) gives:
\[\mu(x) \frac{dy}{dx} + \mu(x) p(x) y = \mu(x) f(x).
\]
To match the LHS to the form of the product rule for \(\mu(x) y\), we need a function \(\mu(x)\) that satisfies:
\[\mu'(x) = \mu(x) p(x).
\]
Integrating both sides of this equation yields:
\[\ln |\mu(x)| = \int p(x) dx.
\]
Therefore, the possibilities for \(\mu(x)\) are:
\[\mu(x) = \pm e^{\int p(x) dx}.
\]
These functions are known as integrating factors (IF).
In the case of first-order linear differential equations, we only need one IF.
To simplify the process, we choose \(\mu(x) = e^{\int p(x) dx}\) and set the constant of integration to 0.
With this choice, the equation \(\mu(x) \frac{dy}{dx} + \mu(x) p(x) y = \mu(x) f(x)\) becomes \(\left[y e^{\int p(x) dx}\right]' = f(x) e^{\int p(x) dx}\)
This allows us to integrate both sides and solve for \(y\)
Examples:
- \(y'+\frac{3}{x}y=\frac{4}{x^5}\)
We can use an IF of \(\mu(x)=e^{\int \frac{3}{x} dx}=e^{3\ln|x|}=\pm x^3\)
By multiplying both sides of \(y'+2xy=x^3\) by \(\mu=x^3\), we get \(x^3y'+3x^2y=\frac{4}{x^2}\)
This equation can be rewritten as \(\left(x^3y\right)'=\frac{4}{x^2}\)
By integrating both sides, \(x^3y=-\frac{4}{x}+C\)
So \(y=\frac{C}{x^3}-\frac{4}{x^4}\) is the general solution.
- \(xy'-3y=x^4e^x\)
First, we rewrite this in the standard form to get \(y'-\frac{3}{x}y=x^3e^x\)
So an IF is \(\mu(x)=e^{\int-\frac{3}{x} dx}=e^{-3\ln|x|}=\pm x^{-3}\)
Notice the integrand's sign here.
By multiplying both sides of \(y'-\frac{3}{x}y=x^3e^x\) by \(\mu=x^{-3}\), we get \(x^{-3}y'-3x^{-4}y=e^x\)
This equation can be rewritten as \(\left(x^{-3}y\right)'=e^x\)
By integrating both sides, \(x^{-3}y=e^x+C\)
So \(y=x^3e^x+Cx^3\) is the general solution.
\(y'+2xy=x^3\)
An IF is \(\mu=e^{\int 2x dx}=e^{x^2}\)
By multiplying both sides of \(y'+2xy=x^3\) by \(\mu\) we get
\[e^{x^2} y'+2x e^{x^2} y=x^3 e^{x^2}\]
This equation can be rewritten as
\[\left(e^{x^2} y\right)^{\prime}=x^3 e^{x^2}\]
By integration by parts,
\[e^{x^2} y=\frac{1}{2} e^{x^2}\left(x^2-1\right)+C\]
Hence \(y=\frac{1}{2}\left(x^2-1\right)+C e^{-x^2}\) is the general solution.
\(y'+\cos (t) y=\sin (t), y(0)=2\)
First we rewrite this in the standard form to get \(y'+y\tan(t)=\sec(t)\)
So an IF is \(\mu=e^{\int\tan(t) dt}=e^{\ln|\sec(t)|}= \pm \sec(t)\)
By
\[\sec(t) y'+\sec(t)\tan(t) y=\sec^2(t) .
\]
This equation can be rewritten as
\[(\sec(t) y)^{\prime}=\sec^2(t) .
\]
By integrating both sides, \(\sec(t) y=\tan(t)+C .
\) Then \(y=\sin(t)+C\cos(t)\) is the general solution.
Imposing the initial condition of \(y(0)=2\) yields \(C=2\) and a solution of \(y=\sin(t)+2\cos(t) .
\)
\(y'-2xy=1, y(1)=e\) An IF is \(\mu=e^{\int-2x dx}=e^{-x^2}\)
By multiplying both sides by \(\mu\) we get
\[e^{-x^2} y'-2xe^{-x^2} y=e^{-x^2}\]
This equation can be rewritten as
\[\left(e^{-x^2} y\right)^{\prime}=e^{-x^2}\]
But the integral of \(e^{-x^2}\) does not have an expression in terms of elementary functions.
Looking at the initial condition of \(y(1)=2\) we will use the integral function \(F(x)=\int_1^x e^{-t^2} dt\) as an antiderivative of \(f(x)=e^{-x^2}\)
So we may write that \(e^{-x^2} y=\int_1^x e^{-t^2} dt+C\)
Then the general solution is \(y=e^{x^2} \int_1^x e^{-t^2} dt+C e^{x^2}\) Imposing the initial condition of \(y(1)=e\) yields \(C=1\)
So the solution is \(y=e^{x^2} \int_1^x e^{-t^2} dt+e^{x^2}\) Try doing the homework for this section using the method of IF instead of the method of variation of parameters shown in the text.
Sometimes you can make a substitution to transform a differential equation into something more familiar.
Examples:
\[y'=\frac{x+y-1}{x+y+1}\]
\[u=x+y\] Then
\[y=u-x \Rightarrow y'=u'-1 \Rightarrow u'-1=\frac{u-1}{u+1} \Rightarrow u'=\frac{u-1}{u+1}+1=\frac{2u}{u+1}\]
That’s separable.
\[\begin{aligned} &\Rightarrow \frac{u+1}{2u} du=dx \Rightarrow \int \frac{1}{2}+\frac{1}{2u} du=\int dx \Rightarrow \frac{1}{2}u+\frac{1}{2}\ln|u|=x+c \\ &\Rightarrow \frac{1}{2}(x+y)+\frac{1}{2}\ln|x+y|=x+c \Rightarrow \frac{1}{2}y+\frac{1}{2}\ln|x+y|=x+C \Rightarrow y+\ln|x+y|=2x+2C \end{aligned}\]
Examples:
\(y+x\tan\left(\frac{y}{x}\right)-x\frac{dy}{dx}=0\)
Divide by \(x\) to get \(\frac{y}{x}\) showing up more than once:
\[\frac{y}{x}+\tan\left(\frac{y}{x}\right)-\frac{dy}{dx}=0\]
Then try \[u=\frac{y}{x}\]
\[\Rightarrow y=ux \Rightarrow y'=u+xu' \Rightarrow u+\tan(u)-\left(u+xu'\right)=0 \Rightarrow \tan(u)-xu'=0\]
Separable.
\[\Rightarrow xu'=\tan(u) \Rightarrow \frac{\cos(u)}{\sin(u)}du=\frac{dx}{x} \Rightarrow \int\frac{\cos(u)}{\sin(u)}du=\int\frac{dx}{x}\]
Using \(w=\sin(u) \Rightarrow dw=\cos(u)du\)
\(\Rightarrow \int\frac{dw}{w}=\ln|x|+c \Rightarrow \ln|w|=\ln|x|+c \Rightarrow \ln\left|\sin\left(\frac{y}{x}\right)\right|=\ln|x|+c \Rightarrow \sin\left(\frac{y}{x}\right)=Cx\)
Definition: If a first-order differential equation can be written in the form
\[y' = g\left(\frac{y}{x}\right)\]
then the differential equation is homogeneous.
[This is not the same as before, \(y' + x^2 y = 0\)
This is a special case of the more general definition: \(f\) is homogeneous of degree \(n\) if \(f(kx, ky) = k^n f(x, y)\) for any constant \(k\)
The substitution \(u = \frac{y}{x}\) will always transform a homogeneous differential equation into a separable one:
\[ \begin{aligned} &y = ux, \quad y' = g\left(\frac{y}{x}\right) \\ &y' = u + xu' \Rightarrow u + xu' = g(u) \Rightarrow \frac{du}{g(u)-u} = \frac{dx}{x}.
\end{aligned} \]
Example: \(y' = \frac{x+y}{x} = 1 + \frac{y}{x} = g\left(\frac{y}{x}\right)\) is homogeneous.
Example: \(y' = \frac{x^2+y}{x} = x + \frac{y}{x} \neq g\left(\frac{y}{x}\right)\) is not homogeneous.
Example: \(xy' = xe^{-y/x} + y, \quad x > 0\)
\(y' = e^{-y/x} + \frac{y}{x}\) is homogeneous.
\[ \begin{aligned} &u = \frac{y}{x} \Rightarrow u + xu' = e^{-u} + u \Rightarrow xu' = e^{-u} \Rightarrow e^u du = \frac{dx}{x} \\ &\Rightarrow e^u = \ln(x) + \overbrace{C}^{\ln(C)} = \ln(Cx) \quad (C > 0) \\ &\Rightarrow \frac{y}{x} = \ln(\ln(Cx)) \Rightarrow y = x\ln(\ ln(Cx)) \end{aligned} \]
A useful class of nonlinear differential equations are "Bernoulli differential equations."
These differential equations can always be made first-order linear by a substitution (so we can solve them) Definition: A differential equation of the form
\[\frac{dy}{dx}+p(x) y=f(x) y^r \quad(r \neq 0,1)\]
A Bernoulli differential equation can be transformed into a first-order linear differential equation using the substitution \(u=y^{1-r}\) or into a separable differential equation using the variation of parameters technique.
\[y'+p(x) y=f(x) y^r \quad(r \neq 0,1)\]
\[\begin{gathered} \xrightarrow[\text {multiply } y^{-r}]{} \quad y^{-r} y'+p(x) y^{1-r}=f(x) \\ u=y^{1-r} \Rightarrow u'= (1-r) y^{-r} y' \Rightarrow \frac{1}{1-r} u'=y^{-r} y' \\ \Rightarrow \underbrace{\frac{1}{1-r} u'+p(x) u=f(x)}_{\text {1st-order linear } } \quad \end{gathered}\]
Examples:
the Bernoulli differential equation \(y'+\frac{1}{x} y=3 y^3\) (set \(r=3\)). Note: \(y=0\) is a solution.
Let’s look for non-trivial solutions.
\[\begin{aligned} & \xrightarrow[\times y^{-3}]{} \quad y^{-3} y'+\frac{1}{x} y^{-2}=3. \\ & \xrightarrow[u\text{-sub} ]{} \quad u=y^{1-3}=y^{-2} \Rightarrow u'= -2 y^{-3} y' \Rightarrow-\frac{1}{2} u'=y^{-3} y' \\ & \xrightarrow[ \tiny{1^{\text{st}}\text{ order linear}} ]{} \quad -\frac{1}{2} u'+\frac{1}{x} u=3 \Rightarrow u'-\frac{2}{x} u=-6 \\ & \Rightarrow \quad \mu=e^{-2 \ln |x|}=x^{-2} \\ & \xrightarrow[\times \mu]{} \quad (x^{-2} u)' =-6 x^{-2} \Rightarrow \int (x^{-2} u)' dx = \int -6 x^{-2} dx \\ & \Rightarrow \quad x^{-2} u=6 x^{-1}+C \Rightarrow u=6 x+C x^2 \Rightarrow y^{-2}=6 x+C x^2 \\ & \Rightarrow \quad y= \pm \frac{1}{\sqrt{6 x+C x^2}}, \quad y=0 \end{aligned}\]