A 2nd order linear differential equation is called an \(``\)Euler equation" if it can be expressed in the form \(ax^2 y''+bxy'+cy=f(x)\) where \(a,b,c\) are constants with \(a\neq 0\) and \(f\) is a function. We will only consider these equations for \(x>0\).

Let’s determine the general solution to a homogeneous Euler equation \(ax^2 y''+bxy'+cy=0\) for \(x>0\).

We guess that there might be solutions of the form \(y=x^m\). Then \(y'=mx^{m-1}\) and
\(y''=m(m-1)x^{m-2}\). Then \(ax^2 y''+bxy'+cy=0\) becomes
\(ax^2\,m(m-1)x^{m-2}+bx\,mx^{m-1}+c\,x^m=0\), which simplifies to \((am(m-1)+bm+c)x^m=0\). Since \(x>0\), we can divide both sides by \(x^m\) to get \(am(m-1)+bm+c=0\), which we refer to as the \(``\)Euler-characteristic equation" and call the polynomial
\(am(m-1)+bm+c=am^2+(b-a)m+c\) the \(``\)Euler-characteristic polynomial."

Since it is a quadratic equation, three cases arise. Case \(1\): There are two distinct real solutions \(m_1<m_2\). Case \(2\): There is only one real solution \(m_1\). Case \(3\): There are non-real complex-conjugate solutions \(\alpha \pm \beta i\) where \(\alpha \), \(\beta \) are real numbers and \(\beta \neq 0\).

Case \(1\): We already have two solutions, \(y_1=x^{m_1}\) and \(y_2=x^{m_2}\). Since \(\displaystyle \frac {y_2}{y_1}\) is non-constant it follows that these solutions form a fundamental set of solutions and that the general solution is \(y=c_1x^{m_1}+c_2x^{m_2}\).

Case \(2\): We use that \(y_1=x^{m_1}\) is a solution and
\(am^2+(b-a)m+c=a(m-m_1)^2=a(m^2-2m_1 m+m_1^2)\), the latter implies that \(b-a=-2am_1\). We can build the general solution by using Reduction of Order!

Let \(y=ux^{m_1}\) be a solution to \(ax^2 y''+bxy'+cy=0\) for \(x>0\). Then \(y'=u'x^{m_1}+m_1ux^{m_1 -1}\) and \(y''=u''x^{m_1}+2m_1u'x^{m_1 -1}+m_1(m_1 -1)ux^{m_1 -2}\). So \(ax^2 y''+bxy'+cy=0\) becomes:

So the general solution in this case is \(y=(c_1\ln (x)+c_2)x^{m_1}\).

Case 3: In this case we get two complex-valued solutions, \(\bar {y_1}=x^{\alpha +\beta i}\) and \(\bar {y_2}=x^{\alpha -\beta i}\).

We can re-express these as \(\bar {y_1}=x^{\alpha }x^{\beta i}\) and \(\bar {y_2}=x^{\alpha }x^{-\beta i}\).

We can even re-express these as \(\displaystyle \bar {y_1}=x^{\alpha }e^{ln{\left (x^{\beta i}\right )}}=x^{\alpha }e^{i\beta ln{(x)}}\) and
\(\displaystyle \bar {y_2}=x^{\alpha }e^{\ln {\left (x^{-\beta i}\right )}}=x^{\alpha }e^{-i\beta ln{(x)}}\).

Then using Euler’s formula (\(e^{i\theta }=\cos {(\theta )}+i\sin {(\theta )}\)) we get that

\(\displaystyle \bar {y_1}=x^{\alpha }\left ( \cos {\left (\beta \ln {(x)}\right )}+i\sin {\left (\beta \ln {(x)}\right )} \right )\) and

\(\displaystyle \bar {y_2}=x^{\alpha }\left (\cos {\left (\beta \ln {(x)}\right )}-i\sin {\left (\beta \ln {(x)}\right )}\right )\).

Since any linear combination of solutions to any 2nd order linear homogeneous de is yet another solution we can find the following two real-valued solutions,
\(\displaystyle y_1=\frac {1}{2}\left (\bar {y_1}+\bar {y_2}\right )=x^{\alpha }\cos {\left (\beta \ln {(x)}\right )}\) and \(\displaystyle y_2=\frac {1}{2i}\left (\bar {y_1}-\bar {y_2}\right )=x^{\alpha }\sin {\left (\beta \ln {(x)}\right )}\).

So the general solution in this case is \(y=x^{\alpha }\left [c_1\cos {\left (\beta \ln {(x)}\right )}+c_2\sin {\left (\beta \ln {(x)}\right )}\,\right ]\).

Examples:

(1) Let’s solve the IVP \(2x^2 y'' -xy'+y=0\), \(y(1)=0\), \(y'(1)=1\).

The Euler-characteristic polynomial is \(2m(m-1)-m+1=2m^2-3m+1=(2m-1)(m-1)\). It has roots at \(m=\frac {1}{2}\) and \(m=1\).

So \(y=c_1 \sqrt {x}+c_2 x\). Since \(y(1)=0\) we get \(0=c_1+c_2\) which means \(c_2=-c_1\).

\(\displaystyle y'=\frac {c_1}{2\sqrt {x}}+c_2\). Since \(y'(1)=1\) we get \(\displaystyle 1=\frac {c_1}{2}+c_2\).

Then \(\displaystyle 1= \frac {c_1}{2}-c_1=-\frac {c_1}{2}\) so \(c_1=-2\) and \(c_2=2\).

So the IVP solution is \(y=2x-2\sqrt {x}\).

(2) Let’s solve the IVP \(x^2 y''-xy'+5y=0\), \(y(1)=0\), \(y'(1)=8\).

The Euler-characteristic polynomial is \(m(m-1)-m+5=m^2-2m+5=(m-1)^2+4\). It has roots at \(m=1\pm 2i\). So \(y=x\left [c_1\cos {\left (2\ln {(x)}\right )}+c_2\sin {\left (2\ln {(x)}\right )}\,\right ]\).

Since \(y(1)=0\) we get \(0=c_1\). So \(y=Cx\sin {\left (2\ln {(x)}\right )}\) and \(y'=C\sin {\left (2\ln {(x)}\right )}+2C\cos {\left (2\ln {(x)}\right )}\). Then \(y'(1)=8\) yields \(8=2C\) so \(C=4\).

So the solution to this IVP is \(y=4x\sin {\left (2\ln {(x)}\right )}\).