# Solving $f'(x) = f(x) + x f(x)$ with power series

I've been working on a problem for quite some time now. I want to solve the differential equation $$f'(x) = f(x) + x f(x)$$ with condition $$f(0) = 0$$ with the power series method. I already know the answer which is $$f(x) = x e^x$$ but I can't seem to find it using that particular method. Here's my attempt :

We assume a solution of the form $$f(x) = \sum_{k \geq 0} a_k x^k$$. The derivative have the form $$f'(x) = \sum_{k \geq 1} k \cdot a_k x^{k-1}$$ and we can rewrite the differential equation as :

$$\sum_{k \geq 1} k \cdot a_k x^{k-1} = \sum_{k \geq 0} a_k x^k + \sum_{k \geq 0} a_k x^{k+1}$$

I then make an index change in the from k to k+1 in the first sum and k to k-1 in the last one to match all the $$x^k$$ :

$$\sum_{k \geq 0} (k+1) \cdot a_{k+1} x^{k} = \sum_{k \geq 0} a_k x^k + \sum_{k \geq 1} a_{k-1} x^{k}$$

and I rewrite :

$$a_1 + \sum_{k \geq 1} (k+1) \cdot a_{k+1} x^{k} = a_0 + \sum_{k \geq 1} a_k x^k + \sum_{k \geq 1} a_{k-1} x^{k}$$

Because of the initial condition , we know that $$a_0 = 0$$ and so we have

$$a_1 + \sum_{k \geq 1} [(k+1) \cdot a_{k+1} -a_k - a_{k-1} ] x^{k} =0$$

Here's my problem. I normally don't have that $$a_1$$ standing there when I solve that type of problem. So I can't say that the coefficient $$(k+1) \cdot a_{k+1} -a_k - a_{k-1} = 0$$ as I normally would to find the recurrence relation... Is there anything I should do differently to progress?

Thanks!

• It will be more useful (and easier on your hands) to use to learn MathJax to write mathematics. – Oliver Diaz Apr 15 at 23:46
• Your last equation implies $a_1=0$. Are you sure $f(x)=xe^x$ is a solution? – Karl Apr 15 at 23:57
• There is the small problem that $xe^x$ doesn't actually solve the differential equation. The solution to any ODE of the form $f'=g\cdot f$ is $$f=C\exp\left(\int g\right)$$ – Ninad Munshi Apr 15 at 23:57
• You are all completely correct. My teacher left this as an exercise at the end of the class. He must have been in a hurry because there's a mistake in the statement ! The ODE should have been f'(x) =f/x + f which is way easier to solve! Thanks everyone and sorry for bothering with that!! – Riemann9471 Apr 16 at 0:03