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