Differential Equation Roadblock

[jinksys]

Homework Statement

Find the solution of the differential equation yy'(1+x²)-x(1+y²)=0 if y(0)=Sqrt(3)

The Attempt at a Solution

I let y=Sqrt(3) and x=0 and solve for y'.
y'=0
Integrate both sides.
y=x +c
Plug my points into find C
Sqrt(3)=0 +c
C=Sqrt(3)
My solution is y=x+Sqrt(3).

Check:
Plug x+Sqrt(3) in for y, and 1 for y' into the DE.
End up with Sqrt(3)=0, so somewhere I went wrong.
Any help appreciated, Thanks.

 

[Mark44]

Homework Statement

Find the solution of the differential equation yy'(1+x²)-x(1+y²)=0 if y(0)=Sqrt(3)

The Attempt at a Solution

I let y=Sqrt(3) and x=0 and solve for y'.
y'=0
Integrate both sides.
y=x +c
Plug my points into find C
Sqrt(3)=0 +c
C=Sqrt(3)
My solution is y=x+Sqrt(3).

Check:
Plug x+Sqrt(3) in for y, and 1 for y' into the DE.
End up with Sqrt(3)=0, so somewhere I went wrong.
Any help appreciated, Thanks.

You are confusing y(0) with y(x). y(0) represents the y-value when x = 0. y(x) represents the y-value for an arbitrary x-value. If you were to plot the graph of y as a function of x, all you know is that the graph goes through (0, sqrt(3)).

In your attempt at a solution, you said:

I let y=Sqrt(3) and x=0 and solve for y'.
y'=0
Integrate both sides.
y=x +c

How did you get y' = 0? This would have been correct if y(x) = sqrt(3). IOW if the graph of y were a horizontal line sqrt(3) units above the x-axis.

If it were true that y' = 0 (it actually isn't true, though), how did you get y = x + c? If you integrate 0 with respect to x, you don't get x + C.

The DE you are given is a separable DE. That means it is possible to get all of the things that involve y (including dy) on one side, and all of the things that involve x (and dx) on the other side. That's the approach you should take with this problem.

 

[jinksys]

Thanks for your help.

I got it now, y=Sqrt( 4x²+3 )

I don't know why I thought the integral of 0 with respect to X was x