Definition: A differential equation is an equation with an unknown function and it’s derivative(s)
Example 1.0.1: \(D_xy=2x+y \)
Solution: \( y= \int 2x+7dx = x^2+7x+C \)
Example 1.0.2: \(D_xy+y=7 \)
Solution here is a little more complicated
Definition: A linear first order differential equation is one such that it can be written \(\frac {dy}{dx}+P(x)y=Q(x)\)
Example 1.1.1: \(D_xy=y(1-y)=y-y^2\) (the logistic model)
Solution: \(D_xy=\frac {dy}{dx}=y(1-y) \Leftrightarrow \frac {dy}{y(1-y)}=dx \Leftrightarrow \int \frac {dy}{y(1-y)}=\int dx \Leftrightarrow u=1-\frac {1}{y} \Rightarrow -\int \frac {1}{u}du = x+C \Leftrightarrow -\ln |u| = -\ln |1-\frac {1}{y}| = x+C \Leftrightarrow \ln |1-\frac {1}{y}| = -x+C \Leftrightarrow \) Assuming \(1-\frac {1}{y} \geq 0\), \(1-\frac {1}{y}=e^{-x+C} \Leftrightarrow -1=e^{-x+C}y-y=y(e^{-x+C}-1) \Leftrightarrow y=\frac {-1}{e^{-x+C}-1} \)
Solving first order linear D.E.s The simplest general method for solving first order linear D.E.s (\(\frac {dy}{dx}+P(x)y=Q(x)\)) is to add an additional function \(\mu (x)\): \[ \begin {aligned} & D_xy+P(x)y=Q(x) \\ & \Leftrightarrow \mu (x)(D_xy+P(x)y)=D_x(\mu (x)y) \\ & \Leftrightarrow \mu (x) D_xy+ \mu (x) P(x)y=\mu (x)D_xy + D_x\mu (x)y \\ & \Leftrightarrow \mu (x)P(x) = D_x \mu (x)y \\ & P(x)=\frac {\mu _x (x)}{\mu (x)} = \frac {d \mu }{\mu } \Rightarrow \int P(x)dx = \int \frac {d \mu }{\mu } = \ln \mu \Rightarrow \mu (x)=e^{\int P(x)dx} \end {aligned} \] And then \(\mu (x) \) can be plugged back in to find y.
Theorem: If P,Q are continuous on an open interval, \(I\), containing \(x_0\), then the initial value problem (IVP) \(\frac {dy}{dx}+P(x)y=Q(x)\), \(y(x_0)=y_0\) has a unique solution \(y(x)\) on \(I\) given by \(y(x)=e^{-\int P(x)dx} \left (\displaystyle \int e^{\int P(x)dx}Q(x)dx + C \right )\) for an appropriate C
If you have \(\frac {dy}{dx} = f(x,y)\), substitute part of the e.q. with \(v=\alpha (x,y)\), and then plug back into the e.q. at the end.
Usually you want to get a linear D.E. relative to \(v\) and \(D_x v\) (for example \(D_x v + P(x)v = Q(x)\) ) which is easier to solve.
Note 1: sometimes a second order D.E. can be reduced into a first order by substituting \(v=q(D_x y) \Rightarrow D_x v=w(D_x^2 y)\). (e.g. \(x D_x^2 y + D_xy = Q(x)\) sub \(v= D_x y \Rightarrow D_x v = D_x^2 y \Rightarrow x D_x v + v = Q(x)\) is 1st order)
Node 2: you can substitute implicitly (e.g. \(y D_x y + (D_x^2 y)^2=0\) sub \(v=y D_x y \Rightarrow D_x v = (D_x y) ^2 + y D_x^2 y \Rightarrow D_x v = 0 \Rightarrow y D_x y = y \frac {dy}{dx}=v=c \Rightarrow \int ydy= \int cdx + C\))
homogenious equation: An equation of the type: \(D_x y=F(\frac {y}{x})\) \(\Rightarrow \) substitute \(v=\frac {y}{x} \Rightarrow y=vx \Rightarrow D_x y=D_x(v)x+v\) which can be solved by \(v`x+v=F(v) \Rightarrow \frac {D_x v}{F(v)-v} = \frac {1}{x}\)
In general \(f(x,y)dy=g(x,y)dx\) substitute \(y=ux \Rightarrow \frac {dx}{x} = h(u)du \Rightarrow \) integrate.
Bernoulli euquation: An equation of the type: \(D_x y+P(x)y=Q(x)y^n\) \(\Rightarrow \) substitute \(v=y^{1-n} \Rightarrow D_x v=(1-n)y^{-n} D_xy \Rightarrow D_x v+(1-n)P(x)v=(1-n)Q(x)\)
Remember that n can be negative (for example \(D_x y+P(x)y=Q(x)\frac {1}{y}\))
An equation \(I(x,y)+J(x,y)D_x y=0 \Leftrightarrow I(x,y)dx+J(x,y)dy=0\) is exact iff \(I_y=J_x\). Then \(\exists \psi (x,y)\) s.t. \[ \begin {aligned} & \psi _x (x,y)=I(x,y) \\ & \psi _y (x,y)=J(x,y) \end {aligned} \] and \(\psi (x,y)=c\) is a solution.
For an IVP, given \(f(a)=b\), plugin \(x=a,y=b\) into \(\psi \), and solve for \(c\).
An autonomous DE is one such that the independent variable (e.g. \(x\), \(t\)) is not in the eq. (e.g. \(D_x y=P(y)\)).
An autonomous equation is always seperable
A seperable equation is one of the form \(D_x y = f(x)g(y)\). A seperable equation can be solved as follows: \(D_x y= \frac {dy}{dx} = f(x)g(y) \Rightarrow \int \frac {dy}{g(y)} = \int f(x)dx + C \).
A second order differential equation is a differential equation that includes the second derivative.
A second order constant coefficient homogeneous D.E. is one with the form \(a D_x^2 y + b D_x y + cy=0\).
Theorem (Super Position):
If a second order homogeneous D.E. has 2 solutions \(a,b\), then \(a+b\) is also a solution.
To solve a second order linear constant coefficient homogeneous differential equation \(a D_x^2 y + b D_x y + cy=0\)
let \(y=e^{rx}\). Then by plugging in we get: \[ \begin {aligned} & a(r^2e^{rx}) + b(re^{rx}) + ce^{rx} = 0 \\ & \Leftrightarrow e^{rx}(ar^2 + br + c) = 0 \\ & \Leftrightarrow ar^2 + br + c=0 \end {aligned} \] which can be solved for \(r=r_1,r_2\) \(\Rightarrow \) \(y=Ae^{rx} + Be^{rx}\) \(\forall A,B\) by super position. If
there is a repeated root (\(r_1=r_2\)), than let \(y=Ae^{rx}+Bxe^{rx}\), plug in, and solve.
Let \(D_x^2 y + p(x) D_x y + q(x)y = g(x)\) be the 2nd order non-homogeneous D.E. For simplicity, let \(L[y]= D_x^2 y + p(x) D_x y + q(x)y\) (so the D.E. is \(L[y] = g(x)\) ). Then the solution is of the form \(y=(c_1 y_1 + c_2 y_2 = y_h) + y_p\) where \(y_h\) is the solution to \(L[y]=0\). To find \(y_p\) there is really two methods:
Refer to the following table to find the general equation for \(y_p\) based on \(g(x)\):
| \(g(x)\) | \(y_p\) |
| \(P_n(x)\) | \(t^s Q_n(x)\) |
| \(P_n(x) e^{\alpha x}\) | \(t^s Q_n(x) e^{\alpha x}\) |
| \(P_n(x) e^{\alpha x} (\sin \beta t + \cos \beta t)\) | \(t^s e^{\alpha x} (Q_n(x) \sin \beta x + R_n(x) \cos \beta x) \) |
Where \(s\) is the smallest integer \(\ge 0\) s.t. \(y_p\) is not a solution to \(L[y] = 0\) and \(P_n(x), Q_n(x), R_n(x)\) are polynomials of
degree n.
Then plug into \(L[y_p]=g(t)\) and solve for the coefficients.
Given \(y_h=c_1 y_1 + c_2 y_2\) we wantto find \(u_1, u_2\) s.t. \(D_x(u_1)y_1 + D_x(u_2)y_2 = 0\) and \(D_x(u_1)D_x(y_1)+ D_x(u_2)D_x(y_2)= g(x)\). We can find precise values using the Wronskian.
Definition: The Wronskian of two functions \(w(f,g)\) is defined as \(w(f,g) := \begin {vmatrix} f & g \\ D_x f & D_x g \end {vmatrix}\)
Lemma: If \(w(f,g) \equiv 0\) (\(w(f,g)(x)=0, \forall x\)), then \(f,g\) are linearally dependent. If \(w(f,g) \not \equiv 0\) (\(\exists x\) s.t. \(w(f,g)(x) \neq 0\)), then \(f,g\) are linearally independent.
Then \(D_x u_1 = \frac {-y_2 g}{w(y_1, y_2)}\) and \(D_x u_2 = \frac {y_1 g}{w(y_1, y_2)}\). This tells us that \(y_p = u_1 y_1 + u_2 y_2\) (notice that \(u_1, u_2\) are not derivatives).
Equivelently, this is: \[ \begin {aligned} y_p & = -y_1(x) \int _{x_0}^x \frac {y_2 (s) g(s)}{w(y_1, y_2)(s)} ds + y_2(x) \int _{x_0}^x \frac {y_1 (s) g(s)}{w(y_1, y_2)(s)} ds \\ & = \int _{x_0}^x \frac {y_2(x)y_1(s)-y_1(x)y_2(s)}{w(y_1,y_2)(s)} g(s) ds = 0 \end {aligned} \] Where \(x_0\) is a convenient point int the interval \(I\) in which \(y_1, y_2\) are defined.
A \(n\)th order differential equation is (as the name implies), a differential equation that includes derrivatives of \(n\) orders.
A \(n\)th order constant coefficient homogeneous D.E. is one with the form \(a_0 D_x^n y + a_1 D_x^{n-1} y + \cdots + a_{n-1} D_x y + a_n y=0\).
To solve, by following the same procedure as for the 2nd order parallel, let \(y=e^{rx} \Rightarrow a_0r^n + a_1r^{n-1} + \cdots + a_{n-1}r + a_n = 0\), and solve.
Definition: A system of first order differential equations is defined as such: \(D_t x_i = \sum \limits _{j=1}^n (P_{ij} (t) x_j) + g_i (t)\) for \(1 \le i \le n\) \(\Leftrightarrow \) \(D_t \vec {x} = \begin {bmatrix} P_{ij} (t) \end {bmatrix}_{ij} \vec {x} = \begin {bmatrix} g_1(t) \\ \vdots \\ g_n(t) \end {bmatrix} \)
A system of first order D.E.s is homogeneous if \(g_i(t)=0, \forall t\).
Theorem: If \(\{\vec {e_i}\}\) is a standard basis for \(\Re ^n\), \(\vec {x_i}\) is a solution to the homogeneous system with the initial condidtion \(\vec {x_i} (t_0) = \vec {e_i}\) then \(\{\vec {x_i}\}\) is the fundimental solution set.
For a homogeneous system, you have \(D_x \vec {x}=A \vec {x} \). If you can find the eigenvalues \(\lambda _1, \cdots , \lambda _n\) and eigenvectors \(\vec {v_1}, \cdots , \vec {v_n}\), then \(\vec {x}= \sum \limits _{i=1}^n c_i e^{\lambda _i} \vec {v_i} \)
If there are repeated eigenvalues, solve for the known one like normal: (In this example I’m just showing for a system of two, but it extends) \(\vec {x}=\vec {x_1} + \vec {x_2}\), \(\vec {x_1}=c_1 \vec {v_1} e^{\lambda _1 t} \), let \(\vec {x_2}=\vec {w} t e^t \Rightarrow D_t \vec {x_2} = \vec {w} (e^t + te^t)\). Then plug back into initial D.E., and solve for \(\vec {w}\).
Given a second order, often you can convert to a system of first orders, or vice versa. This is done by assigning the variables in the system to be different level derivatives.
Example 4.1.2.1: \(a D_x^2 x + bx = 0 \Rightarrow \) let \(x_1=x, x_2=D_x x \Rightarrow D_x x_2 = \frac {-b x_1}{a}, D_x x_1 = x_2\)
You can also go from multiple of a higher order to lower orders:
Example 4.1.2.2: For two second orders we define \(y_1=x_1,\) \(y_2=D_x x_1,\) \(y_3 = x_2,\) \(y_4 = D_x x_2\)
Consider \(D_t X = AX\) \(\forall X \in M_{n \times n}(S)\). If \(\vec {x} = c_1 \vec {x_1} + c_2 \vec {x_2} + \cdots + c_n \vec {x_n}\) (means \(\vec {x_i}=\vec {v_i}e^{\lambda _i t}\) most likely) is a solution to \(D_t \vec {x} = A \vec {x}\), then there is a solution \(X=\Phi (t)=\begin {bmatrix} \vec {x_1} & \vec {x_2} & \cdots & \vec {x_n} \end {bmatrix}\) where \(\vec {x_i}\) are the columns of the matrix. If there is an initial condition \(\vec {x} (0) = \vec {x_0}\) then \(\Phi (t) \vec {c} = \vec {x_0} \Leftrightarrow \vec {c} = \Phi ^ {-1} (t) \vec {x}\)
Then to solve \(D_t X = AX,\) \(\vec {x}(0) = \vec {x_0}\), we have \(\vec {x}(t) = \Phi (t) \Phi ^{-1}(0) \vec {x_0}\) To solve \(D_t X = AX\), we really want \(X=e^{At}\). We know \(e^x = \sum \limits _{n=0}^{\infty } \frac {x^n}{n!} \Rightarrow e^A = \sum \limits _{n=0}^{\infty } \frac {A^n}{n!} \Rightarrow e^{At} = \sum \limits _{n=0}^{\infty } \frac {A^n t^n}{n!}\)
Note 4.1.3.1: If \(AB=BA\), \(e^{A+B}=e^A e^B; (e^A)^{-1} = e^{-A}; e^{0 \in M_{n \times n} (S)} = I\) If \(A=diag(a_1, a_2, \cdots , a_n)\) then \(e^A=diag(e^{a_1}, e^{a_2}, \cdots , e^{a_n})\) If \(A=SDS^{-1}\) for a diagonal \(D\) then \(e^A=S e^D S^{-1}\). if \(A\) is non-diagonalizable, then cehck if there is \(n\) s.t. \(A^n=0\). If so, than a polynomial can be constructed from \(e^A=\sum \limits _{i=0}^n \frac {A^i}{i!}\) \(\left (=\sum \limits _{n=0}^{\infty } \frac {A^n}{n!}\right )\)
Example 4.1.3.2 For \(A=\begin {bmatrix}0&1&2 \\ 0&0&3 \\ 0&0&0 \end {bmatrix}\), \(A^2 \neq 0\), \(A^3=0 \Rightarrow e^A=\sum \limits _{i=0}^2 \frac {A^i}{i!}=(A^0=I)+A+\frac {1}{2}A^2\)
Note 4.1.3.3: If \(AB=BA\), \(C=A+B\) then \(e^{Ct}=e^{At}e^{Bt}\). Note that if \(A=nI\), \(AB=nIB=nB=Bn=BnI=BA\).
Thus the solution to \(D_t \vec {x}=a \vec {x},\) \(\vec {x}(0) = \vec {x_0}\) is \(\vec {x}(t)=e^{At} \vec {x_0 }\) \((=\Phi (t) \Phi ^{-1} (0) \vec {x_0} \Rightarrow e^{At} = \Phi (t) \Phi ^{-1} (0) \vec {x_0} )\)