5. The solutions to the constant coefficient linear ODE Llul = a29\" + an/ + aoy

Question

5. The solutions to the constant coefficient linear ODE Llul = a29" + an/ + aoy = 0 are determined by the roots of the characteristic polynomial, P(m) = a2m2 + aim + a0. Suppose P(m) has a real-valued root, m = r, with multiplicity 2. Then one solution to L[y] 0 is given by 1(x) = erz. To find a second linearly independent solution, assume 2(x) = u(x)erz. (a) Substitute 2(x)-u(x)erz into L[3] 0 to derive an ODE that must be satisfied by u(z) (b) Use the result from (a) to determine u(x) and form the general solution to Llul This solution procedure is known as reduction of order.

Explanation / Answer

Assuming p(x)=u(x)exp(rx)

Then we have p'(x)=u'(x)exp(rx)+ru(x)exp(rx) by chain rule

Differentiating again, we get p''(x)=u''(x)exp(rx)+ru'(x)exp(rx)+ru'(x)exp(rx)+r^2u(x)exp(rx)

That is, p''(x)=u''(x)exp(rx)+2ru'(x)exp(rx)+r^2u(x)exp(rx)

Substituting these expressions into L[y]=a_2y''+a_1y'+a_0y=0 we get

a_2(u''(x)exp(rx)+2ru'(x)exp(rx)+r^2u(x)exp(rx))+a_1(u'(x)exp(rx)+ru(x)exp(rx) )+a_0(u(x)exp(rx))=0

u(x)exp(rx)(r^2a_2+ra_1+a_0)+u'(x)exp(rx)(2ra_2+a_1)+u''(x)exp(rx)(a_2)=0

But as r is a double rooot of the equation P(m)=a_2m^2+a_1m+a_0=0 we must have

P(r)=P'(r)=0

That is, we must have P(r)=a_2r^2+a_1r+a_0=0 and P'(r)=2a_2r+a_1=0

Therefore, u(x)exp(rx)(r^2a_2+ra_1+a_0)+u'(x)exp(rx)(2ra_2+a_1)+u''(x)exp(rx)(a_2)=0 and a_2r^2+a_1r+a_0=2a_2r+a_1=0 imply

u''(x)exp(rx)a_2=0

We know that a_2 is non-zero (otherwise P(m) cannot have 2 roots as it is a polynomial of degree 1) and that exp(rx) is never 0

Hence, we must have u''(x)=0 so that u(x)=ax+b (integrating twice)

Thus, our ODE that must be satisfied by u(x) is u''(x)=0

b) Because u''(x) => u(x)=ax+b, we have the second solution as p(x)=u(x)exp(rx)=(ax+b)(exp(rx))

That is, p(x)=axexp(rx)+bexp(rx) for arbitrary constants a and b

As exp(rx) (and its multiples) are already incorporated in the first solution p_1(x)=exp(rx), we can say that the second linearly independent solution is p_2(x)=xexp(rx)

That is, the linearly independent solutions to our equation are p_1(x)=exp(rx) and p_2(x)=xexp(rx)

And the general solution is p(x)=a*x*exp(rx)+b*exp(rx) for arbitrary constants a and b

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