Simple IVP problem: (1 + x)y' = x−y, y(1) = 1/4

[m2017]

[FONT=wf_segoe-ui_normal, Segoe UI, Segoe WP, Tahoma, Arial, sans-serif, serif, EmojiFont] [/FONT]The solution curve of the IVP (1 + x)y' = x−y, y(1) = 1/4 passes through which of the following points?

The answer is (-2,-2) but I am not seeing how they got this.

If you substitute (-2,-2) in the IVP you get y'=0

If you sub in y(1)=1/4 you get y'=3/8 so I am not really understanding what is going on.

 

[tkhunny]

The solution curve of the IVP (1 + x)y' = x−y, y(1) = 1/4 passes through which of the following points?

The answer is (-2,-2) but I am not seeing how they got this.

If you substitute (-2,-2) in the IVP you get y'=0

If you sub in y(1)=1/4 you get y'=3/8 so I am not really understanding what is going on.

Let's see...

y'(x = -2) = (-2+2)/(1-2) = 0

y'(x = 1) = (1-1/4)/(1+1) = (3/4)/2 = 3/8

Why is it difficult to understand that the value of the derivative at x = -2 is y' = 0 and the value of the derivative at x = 1 is y' = 1/4? Do you somehow believe thay y' is constant? Are you supposed to find the solution, y?

 

[m2017]

Let's see...

y'(x = -2) = (-2+2)/(1-2) = 0

y'(x = 1) = (1-1/4)/(1+1) = (3/4)/2 = 3/8

Why is it difficult to understand that the value of the derivative at x = -2 is y' = 0 and the value of the derivative at x = 1 and y= 1/4? Do you somehow believe thay y' is constant? Are you supposed to find the solution, y?

Sorry I should have made my question more clear. I know that the gradient between (1,1/4) and (-2,-2) is 0.75. The derivative at (-2,-2) is 0, and the derivative at (1,1/4) is 3/8, the part I struggle with is understanding how all of that proves that is the IVP goes through the point (-2,-2)?

 

[Steven G]

Sorry I should have made my question more clear. I know that the gradient between (1,1/4) and (-2,-2) is 0.75. The derivative at (-2,-2) is 0, and the derivative at (1,1/4) is 3/8, the part I struggle with is understanding how all of that proves that is the IVP goes through the point (-2,-2)?

Were you asked to simply verify that the IVP goes through (-2,-2). Isn't that shown in the previous post?

 

[m2017]

Were you asked to simply verify that the IVP goes through (-2,-2). Isn't that shown in the previous post?

The question is asking which of the following points the IVP goes through? It gives multiple options but (-2,-2) is the correct one and I don't understand how they arrived at this answer. I know the derivative at (-2,-2) is 0 but how would that prove the solution curve goes through (-2,-2)?

 

[Steven G]

The question is asking which of the following points the IVP goes through? It gives multiple options but (-2,-2) is the correct one and I don't understand how they arrived at this answer. I know the derivative at (-2,-2) is 0 but how would that prove the solution curve goes through (-2,-2)?

What does it mean to you for a curve to go through a point????

Try plugging in -2 for x, -2 for y and 0 for y' into (1 + x)y' = x−y. Show us what you get.

 

[m2017]

What does it mean to you for a curve to go through a point????

Try plugging in -2 for x, -2 for y and 0 for y' into (1 + x)y' = x−y. Show us what you get.

Oh that makes perfect sense. I was trying to make some connection between the other pieces of information that I completely overlooked that. In this question the fact that y(1)=1/4 has nothing to do with figuring out that (-2,-2) goes through the curve and I was trying to connect those facts.

Thanks for your help.

 

[Steven G]

Oh that makes perfect sense. I was trying to make some connection between the other pieces of information that I completely overlooked that. Thanks for that.

No problem. Good luck

 

[Dr.Peterson]

The question is asking which of the following points the IVP goes through? It gives multiple options but (-2,-2) is the correct one and I don't understand how they arrived at this answer. I know the derivative at (-2,-2) is 0 but how would that prove the solution curve goes through (-2,-2)?

There will be a family of solutions to the equation, one of which has f(1) = 1/4. I think the only way to tell whether (-2,-2) is in that particular curve is to actually solve the equation and find the value of the constant for f(1)= 1/4. You can't just plug something into the equation; for any point, there will be some curve in this family, but not generally the one you want.

It may help if you tell us the entire question exactly as given, including the list of choices. Sometimes that can give a clue.

 

[m2017]

There will be a family of solutions to the equation, one of which has f(1) = 1/4. I think the only way to tell whether (-2,-2) is in that particular curve is to actually solve the equation and find the value of the constant for f(1)= 1/4. You can't just plug something into the equation; for any point, there will be some curve in this family, but not generally the one you want.

It may help if you tell us the entire question exactly as given, including the list of choices. Sometimes that can give a clue.

Upon further reflection I think you are right. The fact that plugging in (-2,-2) into the equation works doesn't prove anything. I could arbitrarily plug any points I wanted and get a y' value but that would not prove anything. The other options were (1, 1/3), (2, 3/2) ,(1/2, 2/3). So I solved the IVP using integrating factor method and yes (-2,-2) exists on the curve but not for the reasons mentioned above.