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

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.