I2, I4, I6 - Oscillatory solutions

I2, I4, I6 - Oscillatory solutions

Complex auxiliary roots can describe a completely real curve. I4 gives the oscillatory solution form, I6 connects it to a negative discriminant, and I2 uses initial conditions to choose one curve. You will separate the oscillation from its exponential envelope and check both the equation and the starting data.

Complex-conjugate auxiliary roots

Can a curve satisfying y′′+2y′+5y=0y''+2y'+5y=0 start at one, cross zero infinitely often and still tend to zero? Real exponential modes alone cannot explain that repeated crossing. The auxiliary quadratic points towards a different shape.

Here primes denote derivatives with respect to xx. As before, trying y=emxy=e^{mx} gives m2+am+b=0m^2+am+b=0 for y′′+ay′+by=0y''+ay'+by=0. With real constants a,ba,b, a negative discriminant gives a complex-conjugate pair.

Complex auxiliary roots

For m2+am+b=0m^2+am+b=0, with real a,ba,b and a2−4b<0a^2-4b<0, write the roots as

m=α±iβ,α=−a2,β=4b−a22>0m=\alpha\pm\mathrm{i}\beta, \qquad \alpha=-\frac a2, \qquad \beta=\frac{\sqrt{4b-a^2}}2>0

The common real part and the positive imaginary-part magnitude will have different jobs in the real solution.

The real-coefficient condition matters: it is what makes the non-real roots conjugates. A negative discriminant does not mean that the differential equation has no real solutions.

Worked example 1: finding the conjugate pair

Question. Find the auxiliary roots of y′′+2y′+5y=0y''+2y'+5y=0.

Plan. Complete the square in the auxiliary equation, then take both square roots.

m2+2m+5=0form the auxiliary equation(m+1)2=−4complete the squarem+1=±2ii2=−1\begin{aligned} m^2+2m+5&=0 && \text{form the auxiliary equation}\\ (m+1)^2&=-4 && \text{complete the square}\\ m+1&=\pm2\mathrm{i} && \mathrm{i}^2=-1 \end{aligned}

Answer. m=−1±2im=-1\pm2\mathrm{i}, so α=−1\alpha=-1 and β=2\beta=2.

Check. The roots sum to −2-2, and their product is (−1)2+22=5(-1)^2+2^2=5, matching the quadratic coefficients.

Change only the coefficient of yy from 55 to 1010. The real part will stay fixed while the imaginary part changes.

Worked example 2: changing the imaginary part

Question. Find the auxiliary roots of y′′+2y′+10y=0y''+2y'+10y=0.

Plan. Complete the square again, retaining both signs.

m2+2m+10=0(m+1)2=−9m=−1±3i\begin{aligned} m^2+2m+10&=0\\ (m+1)^2&=-9\\ m&=-1\pm3\mathrm{i} \end{aligned}

Answer. α=−1\alpha=-1, β=3\beta=3.

Check. The sum remains −2-2; the product is 1+9=101+9=10. The discriminant is 4−40=−36<04-40=-36<0, as required.

In each case there are two independent real solution modes, even though neither auxiliary root is real. We construct those modes next.

Watch out

Wrong: “The discriminant is negative, so there is no solution.” It means there are no real auxiliary roots. The differential equation still has real sine-and-cosine solutions.

Find the two parameters for a fresh equation.

Exponential envelopes and oscillations

For the opening equation, e−xcos⁡2xe^{-x}\cos2x really does oscillate while shrinking. The cosine changes sign repeatedly; its multiplying exponential tends to zero. Complex roots predict these two features separately.

Real solution for non-real roots

If the real constant-coefficient auxiliary quadratic has roots α±iβ\alpha\pm\mathrm{i}\beta, with β>0\beta>0, then

y=eαx(Acos⁡βx+Bsin⁡βx)y=e^{\alpha x}\bigl(A\cos\beta x+B\sin\beta x\bigr)

where A,BA,B are arbitrary real constants and trigonometric arguments are in radians. Not in the formula booklet: learn it.

To see why, put y=eαxuy=e^{\alpha x}u, with α=−a/2\alpha=-a/2 and β2=b−a2/4>0\beta^2=b-a^2/4>0. Product differentiation gives

y′=eαx(u′+αu)y′′=eαx(u′′+2αu′+α2u)y′′+ay′+by=eαx(u′′+β2u)\begin{aligned} y'&=e^{\alpha x}(u'+\alpha u)\\ y''&=e^{\alpha x}(u''+2\alpha u'+\alpha^2u)\\ y''+ay'+by&=e^{\alpha x}(u''+\beta^2u) \end{aligned}

Both cos⁡βx\cos\beta x and sin⁡βx\sin\beta x satisfy u′′+β2u=0u''+\beta^2u=0. They are independent: at x=0x=0, their values are 1,01,0 and their derivatives are 0,β0,\beta, allowing any initial value and slope. Thus their real linear combination supplies the two constants required.

For a nonzero solution let R=A2+B2R=\sqrt{A^2+B^2}. Since the trigonometric combination lies between −R-R and RR, the curve lies between the envelopes y=±Reαxy=\pm Re^{\alpha x}. As x→∞x\to\infty, a negative α\alpha gives decay, zero gives a constant-amplitude oscillation, and positive α\alpha gives growth of the envelope. In the growing case the oscillating function does not tend to +∞+\infty: it keeps changing sign. The zero solution has R=0R=0 and no oscillation.

Worked example 3: a decaying oscillation

Question. Find the general real solution of y′′+2y′+5y=0y''+2y'+5y=0.

Plan. Use −1-1 as the exponential rate and 22 as the angular frequency.

m2+2m+5=0m=−1±2iy=e−x(Acos⁡2x+Bsin⁡2x)\begin{aligned} m^2+2m+5&=0\\ m&=-1\pm2\mathrm{i}\\ y&=e^{-x}(A\cos2x+B\sin2x) \end{aligned}

Answer. y=e−x(Acos⁡2x+Bsin⁡2x)y=e^{-x}(A\cos2x+B\sin2x). Its envelope decays for increasing xx.

Check. With u=Acos⁡2x+Bsin⁡2xu=A\cos2x+B\sin2x, u′′=−4uu''=-4u. The substitution identity gives y′′+2y′+5y=e−x(u′′+4u)=0y''+2y'+5y=e^{-x}(u''+4u)=0.

Now reverse only the sign of the y′y'-coefficient. This reverses the envelope's trend but leaves the oscillation frequency unchanged.

Worked example 4: the same frequency with growth

Question. Find the general real solution of y′′−2y′+5y=0y''-2y'+5y=0.

Plan. Recalculate the roots before choosing the sign of the exponential.

m2−2m+5=0(m−1)2=−4m=1±2iy=ex(Acos⁡2x+Bsin⁡2x)\begin{aligned} m^2-2m+5&=0\\ (m-1)^2&=-4\\ m&=1\pm2\mathrm{i}\\ y&=e^x(A\cos2x+B\sin2x) \end{aligned}

Answer. y=ex(Acos⁡2x+Bsin⁡2x)y=e^x(A\cos2x+B\sin2x). Its nonzero envelope grows, although the trigonometric factor still oscillates.

Check. For y=exuy=e^xu, direct differentiation gives y′′−2y′+5y=ex(u′′+4u)=0y''-2y'+5y=e^x(u''+4u)=0. The sign of the envelope matches the positive real part of the roots.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account