Related questions with answers
Question
Does the initial value problem (x-2)y'=y, y(2)=1 have a solution? Does your result contradict our present theorems?
Solution
VerifiedAnswered 2 years ago
Answered 2 years ago
Step 1
1 of 2Since the initial condition is given at , we obtain
So, the cofficient of is at . That indicates that something will likely go wrong.
Let's solve the given ODE.
Now, the variables are separated. Integrate the left side in relation to , and the right side in relation to .
By taking exponents, we obtain the general solution
where .
We can see that
However, this doesn't contradict our present theorems, because before applying them you need to write the equation in the standard form
Now, we can see that is not defined for .
Create a free account to view solutions
By signing up, you accept Quizlet's Terms of Service and Privacy Policy
Create a free account to view solutions
By signing up, you accept Quizlet's Terms of Service and Privacy Policy
Recommended textbook solutions

Advanced Engineering Mathematics
10th Edition•ISBN: 9780470458365 (5 more)Erwin Kreyszig4,134 solutions

Advanced Engineering Mathematics
9th Edition•ISBN: 9780471488859 (3 more)Erwin Kreyszig4,201 solutions

Advanced Engineering Mathematics
6th Edition•ISBN: 9781284105902 (2 more)Dennis G. Zill5,294 solutions

More related questions
1/4
1/7