FP1 3.1 - Series solutions of differential equations

FP1 3.1 - Series solutions of differential equations

Construct a local Taylor solution from a differential equation and its initial conditions, even when an explicit solution is unavailable (FP1 3.1). You will generate higher derivatives, expand about the correct centre and use Leibniz's theorem to turn repeated differentiation into a recurrence.

The equation supplies the missing derivatives

Can a differential equation tell you the first five terms of its solution without telling you the solution itself? For y′=x+yy'=x+y, knowing only y(0)=1y(0)=1 immediately gives y′(0)=1y'(0)=1. Differentiating the equation gives the next derivative, and the process continues.

The initial condition starts a derivative ladder: each differentiated equation supplies information needed for the next rung. Differentiate while xx is still a variable; substitute the expansion point afterwards. Substituting x=0x=0 too early would erase products whose derivatives matter.

You already know

Taylor coefficients are derivative values divided by factorials: cn=y(n)(a)/n!c_n=y^{(n)}(a)/n! for powers of x−ax-a. A truncation is generally a local approximation; the series must represent the solution where it is used. Leibniz's theorem handles repeated product differentiation.

For an expansion at zero, the standard assembly is supplied; the differential equation's job is to produce its derivative values.

Maclaurin assembly from derivative values

Where the series represents the solution,

y(x)=y(0)+xy′(0)+x22!y′′(0)+x33!y′′′(0)+x44!y(4)(0)+⋯y(x)=y(0)+xy'(0)+\frac{x^2}{2!}y''(0) +\frac{x^3}{3!}y'''(0)+\frac{x^4}{4!}y^{(4)}(0)+\cdots

The coefficients are fixed once the required initial derivatives have been determined.

In the formula booklet.

Keep the derivative values and the coefficients in separate rows if necessary. For instance, y(4)(0)=2y^{(4)}(0)=2 means a coefficient 2/4!=1/122/4!=1/12, not a coefficient 22.

Worked example 1: a first-order derivative ladder

Question. Given y′=x+yy'=x+y and y(0)=1y(0)=1, find the Maclaurin series solution up to and including x4x^4.

Plan. Differentiate the equation three times, evaluate the derivative values at zero, then divide by factorials.

y′=x+ygiven equationy′′=1+y′differentiatey′′′=y′′,y(4)=y′′′differentiate again\begin{aligned} y'&=x+y && \text{given equation}\\ y''&=1+y' && \text{differentiate}\\ y'''&=y'',\qquad y^{(4)}=y''' && \text{differentiate again} \end{aligned}

At zero,

n01234y(n)(0)11222\begin{array}{c|rrrrr} n&0&1&2&3&4\\ \hline y^{(n)}(0)&1&1&2&2&2 \end{array}

Thus

y=1+x+x2+x33+x412+⋯y=1+x+x^2+\frac{x^3}{3}+\frac{x^4}{12}+\cdots

Why this step? The equation provides derivative values in sequence; Taylor's factorials then convert those values into coefficients.

Answer. y=1+x+x2+x3/3+x4/12+⋯y=1+x+x^2+x^3/3+x^4/12+\cdots.

Check. The known function 2ex−x−12\mathrm e^x-x-1 satisfies both the equation and initial condition, and its exponential expansion has the same coefficients.

Change only the initial value to −1-1. The same derivative relations now make every derivative from the second onward vanish.

Worked example 2: a series that terminates

Question. Given y′=x+yy'=x+y and y(0)=−1y(0)=-1, find the series through x4x^4, and determine whether it gives an exact polynomial solution.

Plan. Use the same derivative ladder with the new starting value, then substitute the resulting polynomial into the original equation.

y′(0)=0+(−1)=−1original equationy′′(0)=1−1=0y′′=1+y′y′′′(0)=0,y(4)(0)=0continue the laddery=−1−x+⋯assemble the series\begin{aligned} y'(0)&=0+(-1)=-1 && \text{original equation}\\ y''(0)&=1-1=0 && y''=1+y'\\ y'''(0)&=0,\qquad y^{(4)}(0)=0 && \text{continue the ladder}\\ y&=-1-x+\cdots && \text{assemble the series} \end{aligned}

Why this step? A zero derivative is a value to propagate through the differentiated equations, not a reason to shift subsequent coefficients into lower powers.

Answer. Through x4x^4, the series is −1−x-1-x. In fact, y=−1−xy=-1-x is an exact solution.

Check. For that polynomial, y′=−1y'=-1 and x+y=x−1−x=−1x+y=x-1-x=-1, with y(0)=−1y(0)=-1. Every higher coefficient is zero.

Most finite Taylor polynomials do not satisfy the full equation exactly. This example is special because direct substitution verifies that the polynomial itself is a solution.

Watch out

Wrong: inserting the derivative value y(4)(0)=2y^{(4)}(0)=2 as 2x42x^4. Right: divide by 4!4!, giving x4/12x^4/12. Keep derivative values separate from series coefficients.

Start the same ladder from another initial value.

Differentiating nonlinear terms before substitution

A nonlinear differential equation can still supply Taylor derivatives; the chain and product rules replace the need for an explicit solution.

For y′′=x+y2y''=x+y^2, differentiation gives

y′′′=1+2yy′,y(4)=2(y′)2+2yy′′y'''=1+2yy', \qquad y^{(4)}=2(y')^2+2yy''

The derivative of yy′yy' has two contributions. Even if y′y' happens to be zero at the expansion point, differentiate the product before using that value. These relations depend on this equation and are not supplied formulae.

Worked example 3: a nonlinear second-order equation

Question. Given y′′=x+y2y''=x+y^2, y(0)=1y(0)=1 and y′(0)=0y'(0)=0, find the series solution through x4x^4.

Plan. Use the original equation for y′′(0)y''(0), differentiate twice for the next values, then assemble the series.

y′′(0)=0+12=1original equationy′′′(0)=1+2(1)(0)=1y′′′=1+2yy′y(4)(0)=2(0)2+2(1)(1)=2product rule result\begin{aligned} y''(0)&=0+1^2=1 && \text{original equation}\\ y'''(0)&=1+2(1)(0)=1 && y'''=1+2yy'\\ y^{(4)}(0)&=2(0)^2+2(1)(1)=2 && \text{product rule result} \end{aligned}

Therefore

y=1+x22+x36+x412+⋯y=1+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\cdots

Why this step? The absent linear term comes from the given zero slope; the non-zero cubic term is still present because the equation itself depends on xx.

Answer. y=1+x2/2+x3/6+x4/12+⋯y=1+x^2/2+x^3/6+x^4/12+\cdots.

Check. Differentiating the polynomial twice gives 1+x+x21+x+x^2, while x+y2x+y^2 has the same terms through x2x^2, the orders determined by a quartic truncation.

Change only the initial slope from zero to one. The squared-slope contribution to the fourth derivative now matters.

Worked example 4: a non-zero initial slope

Question. Given y′′=x+y2y''=x+y^2, y(0)=1y(0)=1 and y′(0)=1y'(0)=1, find the series through x4x^4.

Plan. Keep the differentiated equations unchanged and insert the new slope into each derivative value.

y′′(0)=1original equationy′′′(0)=1+2(1)(1)=3first differentiated equationy(4)(0)=2(1)2+2(1)(1)=4retain both contributions\begin{aligned} y''(0)&=1 && \text{original equation}\\ y'''(0)&=1+2(1)(1)=3 && \text{first differentiated equation}\\ y^{(4)}(0)&=2(1)^2+2(1)(1)=4 && \text{retain both contributions} \end{aligned}

Hence

y=1+x+x22+x32+x46+⋯y=1+x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{6}+\cdots

Why this step? Omitting 2(y′)22(y')^2 would halve the fourth derivative here, even though it caused no visible error in the previous zero-slope case.

Answer. y=1+x+x2/2+x3/2+x4/6+⋯y=1+x+x^2/2+x^3/2+x^4/6+\cdots.

Check. The second derivative starts 1+3x+2x21+3x+2x^2; expanding x+y2x+y^2 gives those same coefficients through x2x^2.

A truncated second-order series loses two powers when differentiated twice. Check the equation only through the orders your computed terms can determine; higher residual terms need higher Taylor coefficients.

Watch out

Wrong: (2yy′)′=2yy′′(2yy')'=2yy''. The product rule also gives 2(y′)22(y')^2. Right: 2(y′)2+2yy′′2(y')^2+2yy'', even if one contribution later vanishes at a particular initial point.

Change the initial height while restoring the zero slope.

Taylor solutions about a non-zero point

For an initial point x=a≠0x=a\ne0, use the same derivative method but evaluate every relation at aa and write powers of x−ax-a.

A factor xx does not disappear when the centre is non-zero. For

y′′+xy′+2y=0y''+xy'+2y=0

the first differentiated relations are

y′′′=−xy′′−3y′,y(4)=−xy′′′−4y′′y'''=-xy''-3y', \qquad y^{(4)}=-xy'''-4y''

The additional multiples arise from differentiating xx in the products. Only after writing these identities should you insert the initial point.

Shifted Taylor assembly

For h=x−ah=x-a, where the Taylor series represents the solution,

y(x)=y(a)+hy′(a)+h22!y′′(a)+h33!y′′′(a)+h44!y(4)(a)+⋯y(x)=y(a)+h y'(a)+\frac{h^2}{2!}y''(a) +\frac{h^3}{3!}y'''(a)+\frac{h^4}{4!}y^{(4)}(a)+\cdots

The booklet's printed Maclaurin version has a=0a=0; this shifted form is its previously learnt extension.

Not in the formula booklet: learn it.

Do not replace y(a)y(a) by the coordinate aa. The expansion centre and the initial function value are different pieces of information.

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