Non-homogeneous equations · Differential Equations

Intuition

A forced equation has exactly one solution for each value and slope, just as the homogeneous one does — but the whole family is shifted. Every solution is one particular solution of the forced equation plus the general solution of the homogeneous one. So solving a forced equation is two jobs: the homogeneous job, which the last chapter finished for constant coefficients, and finding just one solution of the forced equation, which the next two lessons do by two different methods.

An address. A particular solution gets you to the right street, and the homogeneous solutions say which house. Any street on the map will do as a start, provided every way of moving along it is then added.

plane diagram: oscillating-about-the-forcing-level

Solutions of $y''+y=1$ . The constant $y_{p}=1$ is one solution, and every other is $1+C_{1}\cos x+C_{2}\sin x$ : the oscillations of the homogeneous equation, centred on the particular solution instead of on zero.

One particular solution, plus all the homogeneous ones

If $y_{p}$ is any one solution of $L[y]=g$ and $y_{1}$ , are independent solutions of , then every solution of is . The constants are found last, from the conditions, after has been added — never before.

y=y_{p}+C_{1}y_{1}+C_{2}y_{2}

Working with it

$y''+y=1$ : $y_{p}=1$ by inspection, so .

Every solution is a particular solution plus a homogeneous one

L[y]=g,\ L[y_{p}]=g\ \Longrightarrow\ y=y_{p}+C_{1}y_{1}+C_{2}y_{2}

Subtract the particular solution from any solution. By linearity the difference solves the homogeneous equation, and the solutions of the homogeneous equation are exactly the combinations of a fundamental pair, by the theorem of the Wronskian lesson. Adding the particular solution back gives the form claimed.

Proof steps

L[y-y_{p}]=L[y]-L[y_{p}]=g-g=0

By linearity the difference of the two solutions solves the homogeneous equation.

y-y_{p}=C_{1}y_{1}+C_{2}y_{2}

Every homogeneous solution is a combination of a fundamental pair.

y=y_{p}+C_{1}y_{1}+C_{2}y_{2}

Adding the particular solution back gives every solution of the forced equation.

Applications

Practice

Particular Plus Homogeneous

Every solution of a forced linear equation is one particular solution plus the general solution of the homogeneous equation.

y=y_{p}+C_{1}y_{1}+C_{2}y_{2}

Try it

$y_{p}=2$ solves $y''-y=-2$ , and $e^{x}$ , $e^{-x}$ solve $y''-y=0$ . What is the general solution of $y''-y=-2$ ?