# How to show that a given differential equation has no solution?

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

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

• Should your ODE be $y'' + P(x)y'(x) + Q(x)y(x) = 0$? – Robert Lewis Aug 13 '19 at 22:23
• 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? – Joe Aug 14 '19 at 2:19
• @RobertLewis yes, I have edited it – user483672 Aug 14 '19 at 14:39
• @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 – user483672 Aug 14 '19 at 14:40

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.