-1
$\begingroup$

4. (Adjoint Form). A 2nd order linear equation $\bbox[3px,border:1px solid black]{p_0(x)y''+p_1(x)y'+p_2(x)y=f(x)}$ is called exact iff it can be expressed as $\frac{d}{dx}\left[Q(x)y'+R(x)y\right]=f(x)$. (Note: Exact equations are nice because they can be solved by two integrations.)

a) Show $x^2y''+x^2y'+(2x-2)y=0$ is exact and solve it by iterated integration. (Your answer will be in terms of integrals.) Show that if an equation is exact then $p_0''-p_1'+p_2=0$.

b) Suppose an "integrating factor" $u(x)$ makes an equation exact. So $p_0uy''+p_1uy'+p_2uy=uf$ is exact. Use part (a) to show that $u(x)$ satisfies $\bbox[3px,border:1px solid black]{p_0u''+(2p_0'-p_1)u'+(p_0''-p_1'+p_2)u=0\ }$. This is called the adjoint equation of the original equation.

c) Suppose an equation can be written as $\frac{d}{dx}\left[p(x)y'\right]+q(x)y=0$. Find the original equation and the adjoint equation and show they are exactly the same. Such an equation is called self-adjoint and this expression is called self-adjoint form. Show that Legendre's equation $(1-x^2)y''-2xy'+\lambda^2y=0$ is self-adjoint and write it in self-adjoint form.


I am really having difficulty with this problem. I understand part a, as I did $\frac{d}{dx}\left[x^2y'+(x^2-2x)y\right]=0$ and then took the integral of both sides to get $\ln y=x+2\ln x+C$. What I don't understand is how the integrating factor makes an equation exact and how to prove that.

$\endgroup$
0
$\begingroup$

You could take a look at the answer to this earlier question. There, the equation in question is made exact by introducing an integrating factor. I hope this will help you to answer b) and c).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.