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The function comes out to be $$ c_1\cos(\ln x)+c_2\sin(\ln x)+\frac{x^2}{5}+1$$

Someone pointed out that by putting $x=0$ in the given equation, we get $f(0)=1$ but it is given that $f(0)=0$ hence there are no solutions.

Is this a valid argument?

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    $\begingroup$ It must be $$x>0$$ see the condition above! $\endgroup$ Jun 2 '19 at 10:00
  • $\begingroup$ Is your $y$ another $f(x)$? Then yes, this argument shows that you can not give initial conditions in a singular point without restrictions. $\endgroup$ Jun 2 '19 at 15:39
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The given equation has to be only satisfied for $x>0$. You can take $\lim_{x\rightarrow 0}$ to obtain $$ \lim_{x\rightarrow 0} \big(x^2 f''(x) + xf'(x) + f(x) \big) = 1$$ $$ \lim_{x\rightarrow 0} \big(x^2 f''(x)\big) + 0\cdot f'(0) + f(0) = 1$$ $$ \lim_{x\rightarrow 0} \big(x^2 f''(x)\big) + f(0) = 1$$ It can be satisfied even with $f(0)=0$ if $\lim_{x\rightarrow 0} f''(x) = \infty $. The argument that inserting $x=0$ to the eqaution gives you $f(0)=1$ is invalid, because it assumes that $\lim_{x\rightarrow 0} f''(x)$ is finite, which (as you can check from the solution) is not true.

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Put $x=e^{t}$ in the given d.e. and you will get $$y''(t)+y(t)=e^{2t} +1$$ Solve this linear differential equation with constant coefficients and you will have $y(t)=acos(t)+bsin(t)+\dfrac{e^{2t}}{5}+1$ then use the initial conditions to find $a$ and $b$ and then substitute $t=ln x$.

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  • $\begingroup$ The initial conditions translate to conditions on values at $t=-\infty$, which is slightly impractical for the oscillating part. $\endgroup$ Jun 2 '19 at 15:38

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