does anyone know why y'=2x, y(0)=0, y(1)=100 has no solution?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

does anyone know why y'=2x, y(0)=0, y(1)=100 has no solution?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

If dy/dx = 2x, what is the general solution of that equation? It's y(x) = x^2 + C. Now is there ANY choice of the number C such that y(0) = 0 and y(1) = 100?
Ya general solution y=x^2 + c1 no problem. but the book says "Show that the problem has no solution. Is this an Initial value problem?" I don't how this statement can be correct when c can take on any value since it is arbitrary.
right you have a contradiction so there is no y with those conditions that satisfy the differential equation

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Given that y(x) = x^2 + c, you now what to find ONE value of c such that it both true that y(0) = 0 AND y(1) = 100 Does there exist one value of c such that both of those conditions are satisfied?
nope
Therefore the initial value problem y' = 2x, y(0) = 0, y(1) = 100 has no solution.
why does it have to be so opaque, but yet so simple.
it's not opaque now. That's why we do exercises! :-)

Not the answer you are looking for?

Search for more explanations.

Ask your own question