the solution of this problem says that since $y^2$ is an even function and so is $cosx-1$ therefore $y'$ is an even function which implies $y$ is odd function and from there on the solution builds up
I can not understand how is $y(x)^2$ even??
here is the full solution https://www.youtube.com/watch?v=WI3UBYJ21iw&t=98s
I am getting more confused now here is one more example where $y^2$ is taken to be even
this is the solution https://www.youtube.com/watch?v=vnqOCpqUebQ
what's going on???
On
As best I can tell, the channels are conflating two tests.
If a differential equation of the form $\frac{dy}{dx} = F(x)$ has a unique solution on $(-c,c)$ such that $y(0) = a$, then
- $F(-x) = -F(x)$ if and only if $y(x)$ is an even function
- $F(-x) = F(x)$ if and only if $y(x) - a$ is an odd function
If a differential equation of the form $\frac{dy}{dx} = F(x,y)$ has a unique solution on $(-c,c)$ such that $y(0) = a$, then
- If $F(-x,y) = -F(x,y)$, then $y(x)$ is an even function.
- If $F(-x,2a-y) = F(x,y)$, then $y(x)-a$ is an odd function.
The two tests are related, but definitely distinct, and while it sounds like they are using the former test based on their explanations, only the latter applies.
We are given $$y'=y^2-1+\cos x $$ $$y(0)=0$$ Putting $x=0$ gives $$y'(0)=0$$ Differentiating $$y''=2yy'-\sin x $$ Putting $x=0$ gives $$y''(0)=0$$ Differentiating $$y'''=2yy''+2(y')^2-\cos x $$ Putting $x=0$ gives $$y'''(0)=-1$$ So by Taylor's theorem, $$y(x)=y(0)+xy'(0)+\frac12x^2y''(0)+\frac16x^3y'''(0)+O(x^4) =-\frac16x^3+O(x^4) $$ and we see that $y(x)$ is strictly decreasing close to $x=0$.