# Exact Equations and Integrating Factor

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.