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Let $h(x)=1+x$ and $z(x)=e^x$ be two solutions of $$y''(x)+P(x)y'+Q(x)y(x)=0$$ Then the set of conditions for which the DE has no solution is

  1. $y(0)=2$, $y'(0)=1$
  2. $y(1) =0$ , $y'(1)= 1$

The answer is option 1

I have found the value of $P(x)$ and the Wronskian but I am not able to go any further. I don't know how to use these conditions to find the solution

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    $\begingroup$ Should your ODE be $y'' + P(x)y'(x) + Q(x)y(x) = 0$? $\endgroup$ – Robert Lewis Aug 13 '19 at 22:23
  • $\begingroup$ I'm confused by the notation. If $h(x) = 1+x$ IS a solution, then how can $y(1)=0$ be a condition, when $h(1)=2$? By 'solution', do you not mean functions which, when substituted for $y$, satisfy both the DE and the conditions? $\endgroup$ – Joe Aug 14 '19 at 2:19
  • $\begingroup$ @RobertLewis yes, I have edited it $\endgroup$ – user483672 Aug 14 '19 at 14:39
  • $\begingroup$ @Joe I have no idea, I posted the ques as it was given in the book. The same thing confused me. I don't understand how to use these initial condition $\endgroup$ – user483672 Aug 14 '19 at 14:40
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The given solutions are largely independent, thus span the solution space. Any other solution is a linear combination of these two.

1) Note that $z(x)-h(x)=e^x-1-x=O(x^2)$, so that $x=0$ can not be a regular point of this ODE. Indeed, the vector $(y(0),y'(0))=(2,1)$ should be a linear combination of the corresponding vectors for $z$ and $h$, but both of them are $(1,1)$.

2) The phase space vectors are $(0,1)$ for $y$, which should be a linear combination of $(2,1)$ for $h$ and $(e,e)$ for $z$, which is solvable.

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