the finite solution first order differential equation I have an equation: $xy'+ay=f(x)$, where a is positive constant and $f(x)\rightarrow b<\infty$ as $x \rightarrow 0$. I found the solution:
$y=\frac{C}{|x|^a}+\frac{b}{a}-\frac{1}{|x|^a}\int\limits_0^x f'(t)\frac{t^a}{a}dt$. And I need to show that there is only one finite solution as $x \rightarrow 0$. To do that I have to show that $\frac{1}{|x|^a}\int\limits_0^x f'(t)\frac{t^a}{a}dt \rightarrow 0$ as $x \rightarrow 0$. But don't know how to do it. Can anyone help me?
 A: Using an integration by parts you obtain:
$$\frac{1}{|x|^a}\int\limits_0^x f'(t)\frac{t^a}{a}dt=\frac{1}{|x|^a}\left((f(x)-b)\frac{x^a}{a}-(f(0)-b)\frac{0^a}{a} \right)-\frac{1}{|x|^a}\int\limits_0^x \left(f(t)-f(b) \right)t^{a-1}dt$$
so it is equivalent to show that $\frac{1}{|x|^a}\int\limits_0^x \left(f(t)-f(b) \right)t^{a-1}dt \to 0$. And:
$$\left| \frac{1}{|x|^a}\int\limits_0^x \left(f(t)-b \right)t^{a-1}dt\right| \leq \frac{1}{|x|^a} \sup_{t \in [0,x]} \left|f(t)-b \right| \int\limits_0^x t^{a-1}dt = \sup_{t \in [0,x]} \left|f(t)-b \right|$$
which goes to $0$ as $x \to0$.

Following the comments, here is more details about the integration by parts. 
Let $$u(t)=f(t)-b$$ and $$v(t)=\frac{t^a}{a}$$.
Then $u'(t)=f(t)$ so:
$$\int\limits_0^x f'(t)\frac{t^a}{a}dt=\int_0^x u'(t) v(t) dt = u(x)v(x)-u(0) v(0)-\int_0^x u(t) v'(t)dt$$
and as $u(0)v(0)0$, $u(x)v(x)=(f(x)-b) \frac{x^a}{a}$ and $v'(t)=t^{a-1}$ we obtain:
$$\int\limits_0^x f'(t)\frac{t^a}{a}dt=\left((f(x)-b)\frac{x^a}{a}\right)-0-\int\limits_0^x \left(f(t)-f(b) \right)t^{a-1}dt$$
