My question relates to an a second order inhomogeneous equation:


First I need to investigate the homogeneous equation:



Forms Hermite's Equation where $\lambda = 11$

So I need a general solution of the homogeneous equation $Ly(x) = 0$.

To do this I need to linearly independent solutions, $y_1(x)$ and $y_2(x)$ say, and then the general solution of $Ly(x) = 0$.


$$Ay_1(x) + By_2(x)$$

where $A$ and $B$ are arbitrary constants.

I am struggling to find two independent solutions of the hermite equation above:

My attempt

If I take a solution to be if the form $$y=\sum_{n=0}^{\infty}a_{n}x^n$$ Then putting this into the ODE I get the following:


Which I can reduce to:


And I have found that rearranging gives:

$$a_{n+2} = \frac{-2(n+\frac{11}{2})a_{n}}{(n+2)(n+1)}$$

This can be used to develop a recurrence relation...

How can I use this to find my two independent solutions $y_1(x)$ and $y_2(x)$ which I need in order to calculate the inhomogeneous solution?

Maybe there are more efficient ways of calculating this ODE that I am unaware of.


I displayed the equation incorrectly and have since edited it.

Should my solution be in the form of an infinite sum rather than a finite polynomial because I believe that the solution could only be constructed as a series solution which terminates if and only if $\lambda = −2n$ where $n \in \mathbb N$.

In my case $\lambda$ isn't of the above form hence an infinte series is required.


3 Answers 3


We will solve the equation \begin{equation} (a_1x+b_1)f''(x)+(a_2x+b_2)f'(x)+(a_3x+b_3)f(x)=g(x)\tag 1 \end{equation} where $f$, $g\in L_2(\bf R\rm)$ and $a_1$, $a_2$, $a_3$, $b_1$, $b_2$, $b_3$ are constants in $\bf R\rm$.

Let the Fourier Transform of a function $f$ of $L_2(\bf R\rm)$ is $$\widehat{f}(\gamma)=\int^{\infty}_{-\infty}f(t)e^{-i t\gamma}dx$$ the Inverse Fourier Transform is $$f(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}\widehat{f}(\gamma)e^{i \gamma x}d\gamma$$ Then it is known (integration by parts) \begin{equation} \int^{\infty}_{-\infty}f(x)x^ne^{-ix\gamma}dx=i^n(\widehat{f})^{(n)}(\gamma) . \end{equation} \begin{equation} \widehat{(f^{(n)})}(\gamma)=(i\gamma)^n\widehat{f}(\gamma) . \end{equation} $$ \int^{\infty}_{-\infty}f'(x)A(x)e^{-i x\gamma}dx= $$ \begin{equation} =\int^{\infty}_{-\infty}f(x)A'(x)e^{-ix \gamma}dx+(-i\gamma)\int^{\infty}_{-\infty}f(x)A(x)e^{-ix\gamma}dx . \end{equation} $$ \int^{\infty}_{-\infty}f''(x)A(x)e^{-ix\gamma}=\int^{\infty}_{-\infty}f(x)A''(x)e^{-ix\gamma}dx+ $$ \begin{equation} +2(-i\gamma)\int^{\infty}_{-\infty}f(x)A'(x)e^{-ix\gamma}dx+(-i\gamma)^2\int^{\infty}_{-\infty}f(x)A(x)e^{-ix\gamma}dx . \end{equation}

Theorem. When $f$, $g\in L_2(\bf R\rm)$ and $\lim_{|x|\rightarrow \infty}|f(x)x^{2+\epsilon}|=0$, $\epsilon>0$, equation $(1)$ can reduced to \begin{equation} (-ia_1\gamma^2+a_2\gamma+ia_3)\frac{\widehat{f}(\gamma)}{d\gamma}+(-b_1\gamma^2-2ia_1\gamma+ib_2\gamma+a_2+b_3)\widehat{f}(\gamma)=\widehat{g}(\gamma) \end{equation} which is solvable.

Using the above theorem in your equation we have $$ y''-2xy'-11y=X_{[0,\infty)}(x)e^{-a x}\tag{eq} $$ We get $$ -(13+s^2)Y(s)-2sY'(s)=g(s)\textrm{, }g(s)=\frac{1}{a-is} $$ Solving this equation we get $$ Y(s)=-e^{1/2(-s^2/2-13\log s)}\int^{s}_{1}\frac{e^{1/2(t^2/2+13\log t)}}{2(a-it)t}dt $$ Hence $$ y(x)=-\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{1/2(-s^2/2-13\log s)}\left(\int^{s}_{1}\frac{e^{1/2(t^2/2+13\log t)}}{2(a-it)t}dt\right)e^{isx}ds $$ The equation $$ y''-2xy'-11y=0 $$ Have general solution $$ y(x)=C_1H_{-11/2}(x)+C_2\cdot {}_1F_1\left(\frac{11}{2};\frac{1}{2};x^2\right) $$ Hence equation $(eq)$ have general solution $$ y(x)=C_1H_{-11/2}(x)+C_2\cdot {}_1F_1\left(\frac{11}{2};\frac{1}{2};x^2\right)- $$ $$ -\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{1/2(-s^2/2-13\log s)}\left(\int^{s}_{1}\frac{e^{1/2(t^2/2+13\log t)}}{2(a-it)t}dt\right)e^{isx}ds, $$ where $H_n(x)$ is the $n-$th Hermite function and ${}_1F_1(a;b;x)$ is the ${}_1F_{1}$ hypergeometric function.

For more details in this kind of equations see here .


  1. The degree of the term $x^my^{(n)}$ is $\nu=m-n$. We gather together all the terms of the DE of $\nu$ degree. In this way we can split a differential equation $A(x)y''+B(x)y'(x)+C(x)=0:(DE)$ into $N$ distinct groups of terms with $\nu_i$, $i=1,2,\ldots,n$ degree. The number $N$ is called degree of the DE.

  2. If the degree $N$ is 2, we call the DE 2-degree. i.e $$ (1-x^2)y''-2xy'+\lambda y=0\textrm{ (Legendre) } $$ $$ y''-2xy'''+\lambda y=0\textrm{ (Hermite) } $$ $$ x^2y''+xy'+(x^2-\nu^2)y=0\textrm{ (Bessel) } $$ Every 2-degree DE have two parts: The part with the largest degree ($L_{max}$ part degree) and the part of the smaller degree ($L_{min}$ part-degree).

i) If the part of largest degree have the term $y''$ we call it DE of the first kind.

ii) If the part of largest degree have the term $y'$ we call it DE of the second kind.

iii) If the part of the largest degree have the term $y$ we call it DE of the third kind.

  1. Step $l$ of a 2-degree (DE) is the difference of the degree of the larger part minus the degree of the smaler part.

I) If the 2-degree (DE) is of the first kind, then its solution is $$ y(x)=x^{\mu}\Phi(a,b;c;\lambda x^{l}), $$ where $$ \Phi(a,b,c;x)=c_1\cdot {}_2F_1(a,b;c;x)+c_2\cdot x^{1-c}{}_2F_1(a+1-c,b+1-c;2-c;x), $$ where ${}_2F_1(a,b;c;,x)$ is the well known Gauss hypergeometric series.

The asymptotic behavior $y=x^s$ of (DE) around $x=0$ lead us to the starting powers $s_1,s_2$. Then $x^{s_1}=x^{\mu}(x^l)^0\Rightarrow s_1=\mu$, $x^{s_2}=x^{\mu}(x^l)^{1-c}\Rightarrow s_2=\mu+l(1-c)$.

The asymptotic behavior in infinty $y=x^{k}$, lead us to $x^{k_1}=x^{\mu}(x^l)^{-a}\Rightarrow k_1=\mu-la$, $x^{k_2}=x^{\mu}(x^l)^{-b}\Rightarrow k_2=\mu-lb$

The parameter $\lambda$ is evaluated demmanding that change of variable $t=\lambda x^l$ leaves the finite singular points of (DE) at $t=1$.

II) If the 2-degree (DE) is of the second kind, then its solution is $$ y(x)=x^{\mu}\Phi(a;c;\lambda x^{l}), $$ where $$ \Phi(a;c;x)=c_1\cdot {}_1F_1(a;c;x)+c_2\cdot x^{1-c}\cdot {}_1F_1(a+1-c;2-c;x). $$ In $x=0$, we have $x^{s_1}=x^{\mu}(x^l)^0\Rightarrow s_1=\mu$ and $x^{s_2}=x^{\mu}(x^l)^{1-c}\Rightarrow s_2=\mu+l(1-c)$ and in $x=\infty$, we have $x^{k_1}=x^{\mu}(x^l)^{-a}\Rightarrow k_1=\mu-la$. The $\lambda$ is evaluated from the asymptotic subtitution $y_{\infty}(x)\approx e^{\lambda x^l}$ in the (DE).

III) The last case is when the 2-degree (DE) is if third kind. Then $$ y(x)=x^{\mu}Z_{k}(\lambda x^{l/2}), $$ where $$ Z_{k}=c_1J_{k}(x)+c_2Y_{k}(x)\textrm{, where }J_k(x)\textrm{ and }Y_k(x)\textrm{ are the Bessel... } $$ The the asymptotic behavior in $x=0$ give us $x^{s_1}=x^{\mu}(x^{l/2})^k\Rightarrow s_1=\mu+\frac{lk}{2}$, $x^{s_2}=x^{\mu}(x^{l/2})^{-k}\Rightarrow s_2=\mu-\frac{lk}{2}$. The $\lambda$ is evaluated from the asyptotic behavior in $x=\infty$, $y_{\infty}(x)\approx \exp\left(\pm i\lambda x^{l/2}\right)$

Example 1. $$ y''-xy=0\tag 1 $$ Obviously (1) is of 2-degree with step $l=1-(-2)=3$. We have $$ L_{mim}=D^2\textrm{ and }L_{max}=-x. $$ Since the largest term has no differentials this equation is a 2-degree of the third kind with step $l=3$. Hence its general solution is $$ y(x)=x^{\mu}Z_{k}(\lambda x^{3/2}).\tag 2 $$ Setting $y=x^{s}$ in (1), we get $s(s-1)x^{s-2}=0\Leftrightarrow s_1=0$, $s_2=1$, we get from (2): $x^1=x^{\mu}(x^{3/2})^{-k}\Rightarrow 1=\mu+\frac{3}{2}k$. Also $x^{0}=x^{\mu}(x^{3/2})^{-k}\Rightarrow 0=\mu-\frac{3}{2}k$. Hence $\mu=1/2$, $k=1/3$. For the evaluation of $\lambda$, we set $y_{\infty}(x)\approx \exp(\pm i \lambda x^{3/2})$, (using the asymptotic formula: $(e^{S})''\approx (S')^2e^{S}$, where $S=\lambda x^{\rho}$, $\rho>0$, $x>>1$), we get after inserting this into (1): $\lambda=\pm i\frac{2}{3}$. Hence the ecxact solution of (1) (Airy equation) is $$ y(x)=x^{1/2}Z_{1/3}\left(i\frac{2}{3}x^{3/2}\right) $$

Example 2. $$ xy''+(2-x^2)y'-2xy=0\tag 2 $$ We rewrite (2) in the form $(xy''+2y')+(-x^2y'-2xy)=0$. Hence $L_{min}=xD^2+2D$, $L_{max}=-x^2D-2x$. Hence (2) is a two degree DE of second kind with step $l=2$. Hence the solution is of the form $$ y(x)=x^{\mu}\Phi(a;c;\lambda x^{2}) $$ The asymptotic behavior at $x=0$ is: The starting powers are (set $y=x^s$ in $L_{min}y=0$ to get) $s_1=0$, $s_2=-1$. $x^0=x^{\mu}(x^2)^{0}\Rightarrow \mu=0$ and $x^{-1}=x^{\mu}(x^2)x^{1-c}\Rightarrow c=\frac{3}{2}$.

The asymptotic behavior at $x=\infty$ is: Solve $L_{max}x^s=0\Leftrightarrow -x^{2}sx^{s-1}-2xx^s=0\Leftrightarrow s=-2$. Hence $k_1=-2$. Hence $x^{-2}=x^{\mu}(x^2)^{-a}\Leftrightarrow a=1$. Also if we set $y_{\infty}(x)\approx\exp(\lambda x^2)$ in (2) we get $\left(e^{\lambda x^2}\right)'=2\lambda x e^{\lambda x^2}$, $\left(e^{\lambda x^2}\right)''=4\lambda^2 x^2 e^{\lambda x^2}$. Hence seting these in (2), we get $4\lambda^2x^3e^{\lambda x^2}-2\lambda x^3e^{\lambda x^2}=0\Rightarrow 4\lambda^2=2\lambda\Rightarrow \lambda=1/2$. Hence the ecxact solution of (2) is $$ y(x)=\Phi\left(1;\frac{3}{2};\frac{x^2}{2}\right) $$
Hence $$ y(x)=c_1\cdot {}_1F_1\left(1;\frac{3}{2};\frac{x^2}{2}\right)+c_2 \cdot x^{-1/2}\cdot{}_1F_1\left(\frac{1}{2};\frac{1}{2};\frac{x^2}{2}\right) $$

Example 3. Solve $$ (1-x^4)y''+\left(n(n+1)x^2-\frac{m(m+1)}{x^2}\right)y=0 $$ Answer $$ y(x)=x^{m+1}\Phi\left(\frac{m-n}{4},\frac{m+n+1}{4};\frac{2m+5}{4};x^4\right) $$

Example 4. The Hermite equation is $$ y''-2xy'+2\nu y=0 $$ This equation is 2-degree of the second kind and step $l=0-(-2)=2$...etc $$ y(x)=\Phi\left(\frac{-\nu}{2};\frac{1}{2};x^2\right) $$


You have correctly found the recurrent relation which gives $a_{n+2}$ as function of $a_{n}$ Hence, if you start with $a_0$ you get all the even coefficients and if you start with $a_1$ you get all the odd coefficients. Hence we get $$y_1(x)=\sum_{k=0}^\infty a_{2k}x^{2k}\quad \text{and}\quad y_2(x)=\sum_{k=0}^\infty a_{2k+1}x^{2k+1}.$$ This solutions are defined up to a constant ($a_0$ and $a_1$ respectevely) that you can arbitrarly choose (but not null, otherwise we get the trivial solution).

Using this ansatz we are implicitly supposing that the solution is analytic in some interval. Hence you can check that $y_1-c y_2 \equiv 0$ means that each coefficient of the series is $0$ and hence both $y_1$ and $y_2$ should be zero. Hence if we choose $a_0,a_1 \neq 0$ we get independent solutions.

We could use the 2 homogeneus solutions with the method of variation of parameters to get the inhomogeneous solution.

Otherwise we could expand $e^{-a x}=\sum_{k=0}^\infty \frac{(-a)^k x^k}{k!}$ and we find the coefficients $a_i$ terms by terms.

  • $\begingroup$ Firstly, thanks for your feedback. Secondly, by variation of parameters do you mean using a Wronskian method to find a particular solution $\endgroup$
    – Student146
    Nov 3, 2020 at 14:44
  • $\begingroup$ Yes, we are talking about the same method. $\endgroup$
    – Jonas
    Nov 3, 2020 at 14:47
  • $\begingroup$ Thanks, I will try this. Would you predict that the wronksian would be difficult to calculate given it contains infinite sums? $\endgroup$
    – Student146
    Nov 3, 2020 at 14:49
  • $\begingroup$ I would expand $e^{- a x}$ and try to solve it term-by-term. $\endgroup$
    – Jonas
    Nov 3, 2020 at 15:04
  • $\begingroup$ Having used what you provided I obtained the following particular solution: $y_{p}(x) =\frac{e^{-ax}}{a^2}$ which gives me a general solution of $y(x) = \sum_{k=0}^\infty a_{2k}x^{2k}+\sum_{k=0}^\infty a_{2k+1}x^{2k+1} + \frac{e^{-ax}}{a^2}$. $\endgroup$
    – Student146
    Nov 3, 2020 at 15:12

Approach $1$: intrgral solution

First consider $y_c''-2xy_c'-11y_c=0$ :

Similar to Help on solving an apparently simple differential equation,

Let $y_c=\int_Ce^{xs}K(s)~ds$ ,

Then $(\int_Ce^{xs}K(s)~ds)''-2x(\int_Ce^{xs}K(s)~ds)'-11\int_Ce^{xs}K(s)~ds=0$











$\ln K(s)=\dfrac{9\ln s}{2}-\dfrac{s^2}{4}+c_1$


$\therefore y_c=\int_Ccs^\frac{9}{2}e^{-\frac{s^2}{4}+xs}~ds$

But since the above procedure in fact suitable for any complex number $s$ ,

$\therefore y_{c,n}=\int_{a_n}^{b_n}c_n(m_nt)^\frac{9}{2}e^{-\frac{(m_nt)^2}{4}+xm_nt}~d(m_nt)={m_n}^\frac{9}{2}c_n\int_{a_n}^{b_n}t^\frac{9}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}~dt$

For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $m_n$ such that:

$\lim\limits_{t\to a_n}t^\frac{11}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}=\lim\limits_{t\to b_n}t^\frac{11}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}$

$\int_{a_n}^{b_n}t^\frac{9}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}~dt$ converges

For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm1$

$\therefore y_c=C_1\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\cosh xt~dt$ or $C_1\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\sinh xt~dt$

Hence $y_c=C_1\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\sinh xt~dt+C_2\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\cosh xt~dt$


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