Differential Equations Cheatsheet
Jargon
General Solution: a family of functions, has parameters.
Particular Solution: has no arbitrary parameters.
Singular Solution: cannot be obtained from the general solution.
Linear Equations
y(n)(x) + an−1(x)y(n−1)(x) + · · · +a1(x)y0(x) + a0(x)y(x) = f(x)
1st-order
F(y0, y, x)=0 y0+a(x)y=f(x)I.F. = eRa(x)dx Sol:y=Ce−Ra(x)dx
Variable Separable
dy
dx =f(x, y)A(x)dx +B(y)dy = 0
Test:
f(x, y)fxy(x, y ) = fx(x, y)fy(x, y)
Sol: Separate and integrate on both sides.
Exact
M(x, y)dx +N(x, y)dy =0=dg(x, y)
Iff ∂M
∂y =∂N
∂x
Sol: Find g(x, y)by integrating and comparing:
ZMdx and ZN dy
Reduction to Exact via Integrating Factor
I(x, y)[M(x, y)dx +N(x, y)dy ]=0
Case I
If My−Nx
M≡h(y)then I(x, y) = e−Rh(y)dx
Case II
If Nx−My
N≡g(x)then I(x, y) = e−Rg(x)dx
Case III
If M=yf (xy)and N=xg(xy)then I(x, y) =
1
xM−yN
Homogeneous of degree 0
f(tx, ty) = t0f(x, y) = f(x, y)
Sol: Reduce to var.sep. using:
y=xv dy
dx =v+xdv
dx
Bernoulli
y0+p(x)y=q(x)yn
Sol: Change var z=1
yn−1and divide by 1
yn.
Reduction by Translation
y0=Ax +By +C
Dx +Ey +F
Case I: Lines intersect
Sol: Put x=X+hand y=Y+k,
find hand k, solve var.sep. and translate back.
Case II: Parallel Lines (A=B,D =E)
Sol: Put u=Ax +By ,y0=u0−A
Band solve.
Principle of Superposition
If y00+ay0+by =f1(x)has solution y1(x)
y00+ay0+by =f2(x)has solution y2(x)then y00 +ay 0+by =f(x) = f1(x)+ f2(x)
has solution: y(x) = y1(x) + y2(x)
2nd-order Homogeneous
F(y00, y0, y, x) = 0 y00 +a(x)y0+b(x)y= 0 Sol:yh=c1y1(x)+ c2y2(x)
Reduction of Order - Method
If we already know y1, put y2=vy1,
expand in terms of v00, v0, v, and put z=v0
and solve the reduced equation.
Wronskian (Linear Independence)
y1(x)and y2(x)are linearly independent iff
W(y1, y2)(x) =
y1y2
y0
1y0
2
6= 0
Constant Coefficients
A.E. λ2+aλ +b= 0
A. Real roots
Sol:y(x) = C1eλ1x+C2eλ2x
B. Single root
Sol:y(x) = C1eλx +C2xeλx
C. Complex roots
Sol:y(x) = eαx(C1cos βx +C2sin β x)
with α=−a
2and β=√4b−a2
2
Euler-Cauchy Equation
x2y00 +axy0+by = 0 where x6= 0
A.E. :λ(λ−1) + aλ +b= 0
Sol:y(x)of the form xλ
Reduction to Constant Coefficients: Use x=et, t = lnx,
and rewrite in terms of tusing the chain rule.
A. Real roots
Sol:y(x) = C1xλ1+C2xλ2x6= 0
B. Single root
Sol:y(x) = xλ(C1+C2ln |x|)
C. Complex roots (λ1,2=α±iβ)
Sol:y(x) = xα[C1cos(βln |x|) + C2sin(βln |x|)]
2nd-order Non-Homogeneous
F(y00, y0, y, x)=0 y00 +a(x)y0+b(x)y=f(x)Sol:y=yh+yp=C1y1(x) + C2y2(x) + yp(x)
Simple case: y0,y missing
y00 =f(x)
Sol: Integrate twice.
Simple case: y0,x missing
y00 =f(y)
Sol: Change of var: p=y0+ chain rule, then
pdp
dy =f(y)is var.sep.
Solve it, back-replace pand solve again.
Simple case: ymissing
y00 =f(y0, x)
Sol: Change of var: p=y0and then solve twice.
Simple case: xmissing
y00 =f(y0, y)
Sol: Change of var: p=y0+ chain rule, then
pdp
dy =f(p, y)is 1st-order ODE.
Solve it, back-replace pand solve again.
Method of Undetermined Coefficients / “Guesswork”
Sol: Assume y(x)has same form as f(x)with
undetermined constant coefficients.
Validforms:
1. Pn(x)
2. Pn(x)eax
3. eax(Pn(x) cos bx +Qn(x)sinbx
Failurecase: If any term of f(x)is a solution of yh,
multiply ypby xand repeat until it works.
Variation of Parameters (Lagrange Method)
(More general, but you need to know yh)
Sol:yp=v1y1+v2y2+· · · +vnyn
v0
1y1+· · · +v0
nyn= 0
v0
2y0
2+· · · +v0
ny0
n= 0
· · · +· · · +· · · = 0
v0
ny(n−1)
b+· · · +v0
ny(n−1)
n=φ(x)
Solve for all v0
iand integrate.
