Solution to the differential equation $xy''+y'+xy=0$ 
Show that the differential equation
  $$xy''+y'+xy=0$$
  admits a solution of the form 
  $$\varphi(x)=\int_0^1f(t)\cos(xt)dt$$
  for some function $f(t)$.

Since
$$\varphi'(x)=\frac{d}{dx}\int_0^1f(t)\cos(xt)dt
=\int_0^1\frac{\partial}{\partial x}f(t)\cos(xt)dt
=-\int_0^1t f(t)\sin(xt)dt$$
and 
$$\varphi''(x)=-\frac{d}{dx}\int_0^1t f(t)\sin(xt)dt
=-\int_0^1\frac{\partial}{\partial x}t f(t)\sin(xt)dt
=-\int_0^1t^2 f(t)\cos(xt)dt$$
it must be satisfied 
$$\int_0^1f(t)\left[-xt^2\cos(xt)-t\sin(xt)+x\cos(xt)\right]dt=0.$$
I don't know how to proceed from this point. I've tried to use integration by parts to simplify the expressions for $\varphi'$ and $\varphi''$ but became even worse. 
 A: Note that multiplying by $x$ yields Bessel ODE:
$$
x^2y''+xy'+x^2y=0
$$ 
for $n=0$. A solution for this ODE is known as Bessel function $J_0(x)$. One characterization of $n$-th Bessel function is 
$$
J_n(x)=\frac{1}{\pi}\int_{0}^{\pi}\cos(nt-x\sin t)\,dt
$$
In particular 
$$
J_0(x)=\frac{1}{\pi}\int_{0}^{\pi}\cos(x\sin t)\,dt=
\overbrace{\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\cos(x\sin t)\,dt}^{(1)}+\overbrace{\frac{1}{\pi}\int_{\frac{\pi}{2}}^{\pi}\cos(x\sin t)\,dt}^{(2)}\\
$$
Note that 
$$
(1)\overset{u=t-\frac{\pi}{2}}{=}
\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\cos(x\cos u)\,du
\overset{t=\cos u}{=}\frac{1}{\pi}\int_{1}^{0}\cos(xt)\cdot-\frac{1}{\sqrt{1-t^2}}\,dt
$$
so
$$
(1)=\int_{0}^{1}\frac{1}{\pi\sqrt{1-t^2}}\cos(xt)\,dt
$$
Similarly, 
$$
(2)\overset{\sin t=u}{=}
\frac{1}{\pi}\int_{0}^{1}\cos(xu)\cdot\frac{1}{\sqrt{1-u^2}}du
$$
so
$$
(2)=\int_{0}^{1}\frac{1}{\pi\sqrt{1-t^2}}\cos(xt)\,dt
$$
Thus
$$
J_0(x)=\int_{0}^{1}\frac{2}{\pi\sqrt{1-t^2}}\cos(xt)\,dt
$$
and your desired $f(t)$ is $f(t)=\frac{2}{\pi\sqrt{1-t^2}}$
A: I think you are meant to use some ingenuity in finding a suitable $f(t)$ which might make your last equation hold. Try $$f(t)=\frac{1}{\sqrt{1-t^2}}$$ The integrand becomes $$x\sqrt{1-t^2}\cos xt-\frac{t}{\sqrt{1-t^2}}\sin xt$$ We observe that can be integrated to $\sqrt{1-t^2}\sin xt$ which is 0 at $t=0$ and $t=1$.
