Separation of variables is the technique this course spends the most time on. It works in a simple way. You turn one differential equation in several variables into two or more equations. Each new equation is an ordinary differential equation, one per variable. Then you solve each ODE the same way ODE always worked. Every solved ODE ahead ends back at two ideas this section reviews. First: a homogeneous linear equation's general solution is a combination. That combination is built from independent building-block solutions. Second: a nonhomogeneous equation's solution splits into that same combination. It adds one particular fix, for whatever is forcing the equation.
Think of the general solution to a homogeneous equation as every way to combine two moves. Walk east some, walk north some. Any point you can reach that way counts. Every point you can reach some other way can also be written as a mix of those same two moves. A nonhomogeneous equation adds a fixed detour on top. First, get to wherever the forcing term insists you land. That's y_p. Then the free mixing still applies, on top of that fixed starting point.
For a linear homogeneous equation, take any two solutions that are not constant multiples of each other. They form a fundamental set. Every solution is some combination c_1y_1+c_2y_2 of that pair. No solution is left out, and no combination fails to solve the equation. Independence is what makes the pair span every case. Two dependent solutions describe only one curve, twice.
For a linear nonhomogeneous equation L[y]=g(x), the general solution splits as y = y_c + y_p. Here y_c, the complementary solution, solves the associated homogeneous equation L[y]=0 and supplies the free constants. Meanwhile y_p, one particular solution, absorbs the forcing term g(x) and carries no constants of its own. Swap y_p for a different particular solution, and y_c absorbs the difference. That difference is itself a solution of the homogeneous equation. So the split is not unique in its pieces, only in what it adds up to.
Independence, read off the bottom curve
Drag the point along x. The bottom curve is the Wronskian of the two solutions above it -- watch that it never touches zero, which is exactly what makes them a fundamental set instead of two copies of the same curve.
Loading interactive…
Worked example
A general solution built from two independent pieces
-
1
Find the general solution of y''+4y=0.
-
2
The characteristic equation r^2+4=0 has roots r=\pm2i. So \cos2x and \sin2x both work. They are two independent solutions.
-
3\boxed{y = C_1\cos2x + C_2\sin2x}
A second-order equation has only two independent solutions. There are no others to find. So this combination already covers every solution there could be.
Loading interactive…
Worked example
Complementary plus particular
-
1
Find the general solution of y''+y=2.
-
2
The matching homogeneous equation y''+y=0 has complementary solution y_c = C_1\cos x + C_2\sin x. A constant y_p=2 satisfies the full equation directly. That's because y_p''=0.
-
3\boxed{y = C_1\cos x + C_2\sin x + 2}
Every solution swings around the fixed offset 2. The particular piece sets where that swing is centered. The complementary piece sets everything about the swing itself.
Loading interactive…
Where this goes next
Separation of variables
Splitting a PDE by separation of variables produces exactly one ordinary differential equation per variable. One piece is spatial, and it turns out homogeneous. Often there's also a time or forcing piece that doesn't. Solving the spatial equation means finding its general solution, the same way as here. Solving a version with a source term means finding a particular solution and adding it to that same complementary piece. Nothing about the technique changes once the unknown is X(x) instead of y(x). See ch.05, "Separation of variables and the heat equation," where this split first does real work.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Ordinary Differential Equations
Given fundamental set y1=e^(-x), y2=e^(5x), write the general solution as an expression in x, C1, C2.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
y=C_1e^{-x}+C_2e^{5x}.
Correct answer
Also accepted:
Why
The general solution is C1 y1 + C2 y2.
From Ordinary Differential Equations
For y=C1cos(x)+C2sin(x) with y(0)=0 and y'(0)=3, find C2.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
y'(0)=C_2\cos0-C_1\sin0=C_2=3.
Correct answer
Also accepted:
Why
y(0)=C1; y'(0)=C2.
From Ordinary Differential Equations
For y=C1e^x+C2e^(-x) with y(0)=4, y'(0)=2, find C1.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
C_1+C_2=4, C_1-C_2=2; adding gives 2C_1=6, C_1=3.
Correct answer
Also accepted:
Why
Set up the 2x2 system from y(0) and y'(0).
Quick review
Practice from earlier coursework, pulled in for this topic.
From Ordinary Differential Equations
For y''-y=2, find the constant particular solution y_p=A.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
0-A=2, so A=-2.
Correct answer
Also accepted:
Why
Substitute y_p=A (a constant) and solve for A.
From Ordinary Differential Equations
Given y_c=C1exp(2x)+C2exp(-x) and y_p=3x, evaluate y_c+y_p at x=0 with C1=1, C2=2.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
y_c(0)=1+2=3; y_p(0)=0; sum is 3.
Correct answer
Also accepted:
Why
exp(0)=1 for both exponential terms; y_p contributes 0 at x=0.
From Ordinary Differential Equations
Which best describes the complementary solution y_c?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
That describes y_p, not y_c -- y_c specifically solves the HOMOGENEOUS version.
Initial conditions are applied to the full sum y=y_c+y_p, not to y_c alone.
y_c is defined for every nonhomogeneous equation, by solving its homogeneous counterpart.
Why
y_c solves ay''+by'+cy=0, the homogeneous equation associated with the original.
y_c comes from setting the right-hand side to zero.