Solving initial value problems · Differential Equations

Intuition

The method has three steps. Transform the equation, turning derivatives into powers of s and bringing the initial values in; solve the resulting algebraic equation for the transform of the solution; and invert, usually after partial fractions, reading the answer off the table. The same three steps work for every linear equation with constant coefficients, forced or not: the forcing simply adds its own transform to the algebra.

Translating a hard text into a language in which it is easy, working there, and translating the answer back. The translations are the transform and its inverse, and the easy language is algebra.

$y''+3y'+2y=0$ , $y(0)=1$ , $y'(0)=0$ . The transform $Y=\tfrac{s+3}{(s+1)(s+2)}=\tfrac{2}{s+1}-\tfrac{1}{s+2}$ inverts term by term to $y=2e^{-t}-e^{-2t}$ , drawn with its two terms dashed: each partial fraction is one exponential.

Transform, solve, invert

For $ay''+by'+cy=g(t)$ with and given: transform both sides with the rules for derivatives; solve for , which is a ratio of polynomials when the transform of is; split it into partial fractions; and invert each fraction with the table. The denominator of holds the characteristic polynomial, so its roots are the characteristic roots of chapter four, together with those of .

Y(s)=\frac{G(s)+(as+b)\,y(0)+a\,y'(0)}{as^{2}+bs+c}

Worked in three steps

$y''+3y'+2y=0$ , , : , so and .

Every such problem becomes one algebraic equation

\begin{aligned}&(as^{2}+bs+c)\,Y\\&\quad=G(s)+(as+b)\,y(0)+a\,y'(0)\end{aligned}

Transform the equation term by term. Each derivative becomes a power of s times the transform of the solution, less terms made of the initial values. Collecting the terms in the transform leaves the characteristic polynomial times it on one side, and the transform of the forcing with the initial terms on the other; dividing gives the transform of the solution explicitly.

Proof steps

\begin{aligned}&a\bigl(s^{2}Y-sy(0)-y'(0)\bigr)\\&\quad+b\bigl(sY-y(0)\bigr)+cY=G(s)\end{aligned}

Transform each term with the rules for derivatives.

\begin{aligned}&(as^{2}+bs+c)\,Y\\&\quad-(as+b)\,y(0)-a\,y'(0)=G(s)\end{aligned}

Collect the terms in Y.

Y=\frac{G(s)+(as+b)\,y(0)+a\,y'(0)}{as^{2}+bs+c}

Divide by the characteristic polynomial.

Applications

Practice

Using the Three Transform Steps

Transform the equation, bringing in the initial values; solve the algebra for Y; split Y into partial fractions and invert each with the table.

y'-y=1,\ y(0)=0:\quad(s-1)Y=\frac1s

Try it

What is the right order of the three steps of the method?