For a constant-coefficient case, it is guaranteed that the solutions we get from taking $e^{\lambda x}$ to be the ansatz and finding out the value of $\lambda$ are valid. Generally, for a homogeneous equation


it is equivalent to taking the linear differential operator as $$\mathcal{L}=aD^2 + bD + c$$ where $D = d/dx$, and saying that


So I would say that $e^{\lambda x}$ is a basis for the eigenspace of $\mathcal{L}$ which corresponds to the eigenvalue of $$\Lambda=a\lambda^2 + b\lambda +c$$

henceforth making the equation $\mathcal{L}y=\Lambda y$.

However, this method does not work for a nonconstant-coefficient case. But I don't see why it doesn't work. Instead of talking about general cases, I will use a specific example from now on. Consider the equation


I found that it is entirely valid up to carrying out the characteristic equation.

$$\lambda^2 e^{\lambda x} + x \lambda e^{\lambda x} + e^{\lambda x}=0$$

$$(\lambda^2+x\lambda+1)e^{\lambda x}=0$$

Since $e^{\lambda x}>0$, $$\lambda^2+x\lambda+1 = 0$$

Then we can find $\lambda$ using the quadratic formula (admitting that $\lambda$ is a function of $x$):

$$\lambda_1(x) = \frac{-x + \sqrt{x^2-4}}{2},\ \lambda_2(x) = \frac{-x - \sqrt{x^2-4}}{2}$$

Since the equation is linear, we can guarantee that the span of $\{e^{\lambda_1 x}, e^{\lambda_2 x}\}$ is a solution. Thus,

$$y(x) = e^{-x/2} \times [Ae^{+\frac{\sqrt{x^2-4}}{2}x} + Be^{-\frac{\sqrt{x^2-4}}{2}x}]$$

Except for the fact that the eigenvalue $\lambda$ is a function of $x$ there is nothing peculiar about it. I thought it is fine, because what we are taking as "vectors" is the solution function $y$, not the independent variable $x$ underlying at the back of this function.


2 Answers 2


It doesn't work in your example because when you differentiated $e^{\lambda x}$ to obtain the characteristic equation $\lambda^2+x\lambda+1 = 0$, you implicitly assumed that $\lambda$ is a constant. However, if $\lambda$ is a function of $x$, then $(e^{\lambda x})'=(\lambda'x+\lambda)e^{\lambda x}\neq \lambda e^{\lambda x}.$

You can use the ansatz $e^{\lambda x}$, but in order to determine $\lambda$ you have to solve a differential equation, not an algebraic one.

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    $\begingroup$ Specifically if you assume $y=e^{f(x)}$ then $y'=e^f f',y''=e^f(f'^2+f'')$ so the equation in the question becomes $f''+f'^2+xf'+1=0$. OP is assuming $f'=\lambda$ so that $f''=0$ but this cannot be right in this context. $\endgroup$
    – Ian
    Commented May 8, 2021 at 2:32
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    $\begingroup$ That ode is more complicated than the original one @Ian $\endgroup$ Commented May 8, 2021 at 2:55
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    $\begingroup$ @Aryadeva Yup. But that is what you get out of the ansatz. $\endgroup$
    – Ian
    Commented May 8, 2021 at 13:44

$$\lambda^2 e^{\lambda x} + x \lambda e^{\lambda x} + e^{\lambda x}=0$$ This must be true for all $x \in \mathbb{R}$: $$\lambda^2+x\lambda+1 = 0$$ For $x=0 \implies \lambda=\pm i$

For $x=1 \implies \lambda^2+\lambda +1=0$

So it dosen't work since your solution depends on the variable $x$. The DE is easy to integrate: $$y''+xy'+y=0$$ $$y''+(xy)'=0$$ Reduce the order by integation. $$y'+xy=C_1$$ It's a first ODE you can easily solve.


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