$y''=y $ implies an exponential solution. Try $ke^x $ where $k $ is an arbitrary constant. You get $y''=ke^x $. Then you plug in $y(0)=1$ and find $k=1$. So, I get $y=e^x $ is a solution.
$y''=y $ implies an exponential solution. Try $ke^x $ where $k $ is an arbitrary constant. You get $y''=ke^x $. Then you plug in $y(0)=1$ and find $k=1$. So, I get $y=e^x $ is a solution.
sorry I am not really sure of you method.
this is what i am used to .
question 1 is the worked answer to the question. i am not sure where they get v=1 as it is not given in the question.