Differential Equation Roadblock

In summary, the differential equation y*y'(1+x^2)-x(1+y^2)=0 has a solution of y = Sqrt(4x^2 + 3) when y(0) = Sqrt(3). The approach used was to separate the variables and integrate, and the error in the initial attempt was due to confusion between y(0) and y(x).
  • #1
jinksys
123
0

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.
 
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  • #2
jinksys said:

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.
 
  • #3
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 :rolleyes:
 

1. What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives. It describes the relationship between a function and its rate of change.

2. What is the purpose of solving differential equations?

Solving differential equations allows us to model real-world phenomena and predict future behavior. It is used in many fields such as physics, engineering, economics, and biology.

3. What are the different types of differential equations?

There are several types of differential equations, including ordinary differential equations, partial differential equations, and stochastic differential equations. They differ based on the number of variables and the types of functions involved.

4. What are some common methods for solving differential equations?

Some common methods for solving differential equations include separation of variables, substitution, and using specific formulas such as the Euler method or the Runge-Kutta method. These methods vary in complexity and are chosen based on the specific equation and its properties.

5. What are some applications of differential equations in real life?

Differential equations are used in a wide range of applications, such as predicting population growth, modeling the spread of diseases, analyzing chemical reactions, and designing circuits. They are also used in finance to model stock prices and in physics to describe the motion of objects.

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