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Convert the initial value problem $y''+\lambda x^2y=f(x)$ with $y'(0)=y(0)=0$ into a Volterra integral equation.

Here if we consider the two linearly independent solution of the above equation as $u(x)$ and $v(x)$, then we know that the reduced equation will be $$y(x)=\int_0^x \frac{u(t)v(x)-u(x)v(t)}{A}f(t)dt$$ where $A$ is given by $u(x)v'(x)-u'(x)v(x)=A$. But I could not find the solution of the above equation in terms of $u$ and $v$ explicitly. The method of inspection also did not help me in this regard. Can someone supply a method such that this type of equation can be solved easily? I know that this problem can be handled by Leibniz rule, but I want a help for this technique only. Thanks in advance.

Edit :

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am_11235...
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  • Are you to find a solution formula or an equivalent (in some sense) integral equation? For the latter you can take the Green function of the second derivative operator. – Lutz Lehmann Sep 02 '19 at 10:30
  • Will it suffice? Since the Green's function method is only applicable to BVP. But this is an IVP. – am_11235... Sep 02 '19 at 10:48
  • Actually what I described above, it also appeared to me that it is a solution of the problem rather than a conversion to integral equation. But in the literature it is described as the method of conversion to integral equation. That's where I'm confused. – am_11235... Sep 02 '19 at 10:53
  • Anyways, I have uploaded two attachment regarding the method above. Please explain how this could be a method for conversion to integral equation as the integrand already contains the known function $f(x)$ in eqn. $(16)$ of second attachment. – am_11235... Sep 02 '19 at 11:26
  • In the simplified case you would use the Taylor formula $y(x)=y(0)+y'(0)x+\int_0^x(x-s)y''(s)ds$ to get $$y(x)=\int_0^x(x-s)(f(s)-λs^2y(s))ds.$$ It depends on your further purposes if this is sufficient or not. – Lutz Lehmann Sep 02 '19 at 11:31
  • OK. But what about the attachments above? – am_11235... Sep 02 '19 at 11:34
  • That would require the solution of the homogeneous equations. It might be possible to express them in Bessel functions, but as I said, if that is useful depends on the use you want to make of the integral equation. – Lutz Lehmann Sep 02 '19 at 11:43

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