Are there multiple valid solutions for the given initial value problem?

[footballxpaul]

(2x-y)dx+(2y-x)dy=0 y(1)=3

solve the given initial value problem and determine at least approx where the solution is valid?


this should be simple right? I got 2y+2x=C and then 2y+2x=8. It doesn't match the answer in the back of the book at all, and I do see how they could have gotten the answer in the back. Is my answer right? The book says y=[x+sqrt(28-3x^2)]/2, abs(x)<sqrt(28/3)? or is the book right, and if so how do you get to their answer?

 

[g_edgar]

How did you get your answer, and what's that y(1) on the end?

 

[footballxpaul]

its y(1)=3, I just solve the ivp by using the exact method. It seemed simple but it doesn't match the books answer.

*my bad posted in wrong forum, if a mod wants to move this go ahead

 

[djeitnstine]

This should be in the homework forum. How about showing us your work, this is an extremely easy problem since it is already exact.

Hint you can directly integrate.

 

[HallsofIvy]

(2x-y)dx+(2y-x)dy=0 y(1)=3

solve the given initial value problem and determine at least approx where the solution is valid?


this should be simple right? I got 2y+2x=C and then 2y+2x=8. It doesn't match the answer in the back of the book at all, and I do see how they could have gotten the answer in the back. Is my answer right? The book says y=[x+sqrt(28-3x^2)]/2, abs(x)<sqrt(28/3)? or is the book right, and if so how do you get to their answer?

If 2y+ 2x= 8, then y= 4- x dy= -dx so (2x-y)dx+ (2y-x)dy= (2x- 4+ x)dx+ (8- 2x- x)(-dx)= (3x- 4)dx- (8- 3x)dx= -12, not 0. Your solution is clearly wrong.

Now, how did you get that?

As djeitnstine said, this is an "exact" differential equation that can be integrated relatively easily.

 

[djeitnstine]

Indeed, not even the solution is incorrect. I worked it out this morning...however even without working it out its clear the solution seems extraneous.

Edit: both the OP's solution and the one you claim to have found in the book.