Existence & Uniqueness Theorem, Ex1.5

[@TheGeneralThings]

Should we also check if δf/δx is continuous around that point? Is there a reason why we wouldn't need to solve for that?

[@georgerothbart4579]

It seems there is a second solution: y=(x^2/4 - 6)^2 +3. We see that this function passes through the initial value (-2,28), and if you take the derivative you will see it satisfies the differential equation. So there are *two* solutions: y=(x^2/4+4)^2 + 3 and y=(x^2/4-6)^2 + 3 What am I doing wrong?

[@badrunna-im]

george rothbart that second solution is for dy/dx=-xsqrt(y-3)

[@giorgiozat]

Yes. If you substitute x=-2 in y=(1/4 x^2 + C)^2+3=28 you obtain 2 solutions: C= 4 and C=-6

[@nandinisharma9742]

When we'll solve √25 we will get 2 answers +5 and - 5 he only solved for +5 how. Come the solution can be unique

[@muntahassan6384]

WHAT ABOUT Y=3 ? BIG FAN FROM SAUDI ARABIA ♥

[@JohnCena12355]

Remember the solution is a function. So what you are saying is y(x)=3, this means that for all x that y = 3... This is incorrect because it does not satisfy the initial condition given as y(-2)=28