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Differential Equations Cheatsheet
JMC Year 1, 2017/2018 syllabus
Fawaz Shah
Topics not covered in this summary: phase portraits, similarity transformations.
Contents
1 Definitions 2
2 1st order linear ODEs 2
3 1st order non-linear ODEs 3 3.1 Exact equations............................. 3 3.2 Separable ODEs............................. 3 3.3 Homogenous ODEs........................... 4 3.4 Bernoulli type ODEs.......................... 4
4 2nd order ODEs 4 4.1 Special case - y missing......................... 4 4.2 Special case - x missing......................... 5 4.3 General case - finding the CF..................... 5 4.4 General case - finding the PI...................... 6
5 Solving systems of differential equations 7
1 Definitions
Order (of derivative) An nth^ derivative has order n.
Order (of ODE) The order of the highest derivative present in an ODE.
Degree (of ODE) The highest power to which a term is raised in an ODE (ex- cluding fractional powers).
Linear (ODE) An ODE which has no terms raised to more than the 1st^ power, and with no y, x or other derivative terms multiplied by each other.
System of diff. equations A set of simultaneous equations of derivatives, where derivatives of y, x etc. are given w.r.t. a parameter t
Order (of system) The order of the highest derivative present in the system.
Degree (of system) The highest power to which a term is raised in an ODE (excluding fractional powers).
Linear (system) A system which has no terms raised to more than the 1st^ power, and with no y or other derivative terms multiplied by each other.
Homogeneous (system) A system with no explicit functions of t (i.e. f (t)) present.
2 1st order linear ODEs
Every 1st order linear ODE can be expressed as:
dy dx
These can ALL be solved by the integrating factor method:
- Multiply both sides by exp(
p(x) dx)
- Use the reverse product rule to express the LHS as a single derivative (of a function of y).
- Integrate both sides and rearrange.
3.3 Homogenous ODEs
Homogenous equations can be written in the form:
dy dx
= f (
y x
To solve, set v = yx , so that y = xv. Note that v is still a single-variable function of x, since y is a function of x. Now we can differentiate both sides to get:
dy dx
= v + x
dv dx
We now have simultaneous equations for dydx. Equate and solve for (^) dxdv , and then
solve this 1st order linear ODE in dvdx to find v (and then y).
3.4 Bernoulli type ODEs
A Bernoulli type ODE is of the form:
dy dx
To solve:
- Multiply both sides by (1 − n)y−n
- Let z = y^1 −n^ and substitute into equation, including rewriting one of the terms as dzdx
- The resulting equation is 1st order linear in z, so solve for z (and then y).
4 2nd order ODEs
4.1 Special case - y missing
If we can write the 2nd^ derivative in the form:
d^2 y dx^2
= f (x,
dy dx
(i.e. no y terms present), then we can make a substitution. Let P = dydx. This
means d
(^2) y dx^2 =^
dP dx , the refore we have: dP dx
= f (x, P ) (13)
This is 1st order w.r.t P and can be solved by appropriate 1st order methods.
4.2 Special case - x missing
If we can write the 2nd^ derivative as:
d^2 y dx^2
= f (y,
dy dx
(i.e. no x terms present), then we can make the same substitution. Let P = (^) dxdy.
This means d
(^2) y dx^2 =^
dP dx , therefore we have: dP dx
= f (y, P ) (15)
However, this is not yet a 1st order equation since the derivative is w.r.t. x, but we only have y terms on the RHS.
DIFFERENT TO LAST TIME: we must rewrite dPdx as a derivative with respect to y. Luckily, we can see that:
dP dx
dP dy
dy dx
= P
dP dy
Therefore:
P
dP dy
= f (y, P ) (17)
This is 1st order w.r.t P and can be solved by appropriate 1st order methods.
4.3 General case - finding the CF
The general solution (GS) of a 2nd order ODE can be expressed as the sum of two other functions, called the ’complementary function’ (CF) and a ’particular integral’ (PI). yGS = yCF + yP I (18)
A 2nd order ODE will usually be presented to us in the form:
a
d^2 y dx^2
dy dx
It can be shown that the CF can be calculated from the LHS of the above equation. We write down the auxiliary equation, which is simply the equation:
aλ^2 + bλ + c = 0 (20)
using a, b, c from above. Solving this gives us two values, λ 1 and λ 2.
Case 3: f (x) is a multiple of ebx, ebx^ IS in CF
We now have a clash between the PI and the CF. We can try yP I = Axebx, i.e. sticking an x in the PI to avoid the clash. If this doesn’t work, we can choose yP I = A(x)ebx^ for some real FUNCTION A. Remember to use the CHAIN RULE to differentiate A this time.
At the end remove any clashing terms, i.e. terms of the form Beλx^ where eλx is already present in the CF. Other terms with more x’s included are allowed, e.g. xeλx^ would not count as a clashing term.
Case 4: f (x) = A(x)ebx^ where A(x) is a polynomial
Choose yP I = C(x)ebx^ for some polynomial C(x).
Case 5: f (x) is trigonometric (e.g. sin, cos, sinh etc.)
Look for a pattern in f (x). A good tip for an f (x) with only sines/cosines is to use yP I = A cos(x) + B sin(x) and solve for A and B. A similar story for sinh and cosh. CAUTION: sinh, cosh and tanh are actually exponential functions in disguise, so make sure they do not clash with any eλx^ terms in the CF.
Other cases
If f (x) has a term of the form ex^ cos(x) or ex^ sin(x) then we can rewrite it as the real/imaginary part of a complex function (in this case e(1+i)x^ would be appropri- ate, since it expands to ex(cos(x) + i sin(x)).
If f (x) is more complicated, we may have to be imaginative with the choice of yP I. e.g. for f (x) = Aeax^ + Bebx^ we could choose yP I = Ceax^ + Debx^ for some constants C, D. Again be careful of terms that clash with the CF.
5 Solving systems of differential equations
A homogeneous 1st order system of equations can be written as:
dx dt
= F (x, y) dy dt
= G(x, y)
Let us choose an example coupled system:
dx dt
= ax + by dy dt
= cx + dy
We can rewrite this in matrix form:
d dt
x y
a b c d
x y
The system is now of the form
d dt
v = M v (27)
If we set v = V eλt, where V is a constant vector independent of x, y or t, then we get
λV = M V (M − λIn)V = 0v det(M − λIn) = 0
Predictably, we find two eigenvalues λ 1 , λ 2 and (any) two eigenvectors v 1 , v 2. The solution to the system is given by: ( x y
= A 1 v 1 eλ^1 t^ + A 2 v 2 eλ^2 t^ (29)
The dimension of the eigenvectors will always match the number of variables being dealt with, for example a possible scenario is: ( x y
= A 1
e−^3 t^ + A 2
e^2 t^ (30)
The values of the individual derivatives can be found by reading off the rows of the matrices.
x = 3A 1 e−^3 t^ + 7A 2 e^2 t y = − 5 A 1 e−^3 t^ + − 2 A 2 e^2 t^
Complex eigenvalues
If the eigenvalues turn out to be complex conjugates, the solution can be written as: (^) ( x y
= A 1 v 1 e(a+bi)t^ + A 2 v 2 e(a−bi)t^ (32)
