Solve ODE Exact Equations: Initial Value Problem

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In summary, the author found c to be -7 and found y to be 1 + 2 + 4 - 7 = (1)2 - (1)(-2) + (-2)2 - 7.
  • #1
newtomath
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Can you guys point me in the right direction on the problem below?

Solve the given initial value problem and determine at least approx. where the solution is valid:

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

So I have My =-1 and Nx= -1

x^2-xy+ h(y) => -x+h'(y) = 2y-x => h(y)= y^2

=> x^2 -xy+ y^2

where would I go from here to solve the initial value prob?
 
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  • #2
You have the right idea but is written poorly and that is why you don't have an equation in your last step. After you have checked exactness you should arrange your work like this. You have an unknown function f(x,y) satisfying this exact differential and you know

fx(x,y) = 2x - y
(*) f(x,y) = x2 - xy + h(y)
Differentiating this with respect to y and using the equation:
-x + h'(y) = 2y - x
h'(y) = 2y
h(y) = y2 + C

Substitute his in for the h(y) in (*) above which gives

f(x,y) = x2 - xy + y2 + C

This is the function that satisfies df(x,y) = 0 so your solution is

x2 - xy + y2 + C = 0

Now use your initial conditions. Notice at each step of the writeup you have an equation with an = and two sides. No sloppy use of => symbol.
 
  • #3
Thanks. I found c to be 7. But the answer in the text states y as = (x + sqrt(28-3x^2))/2. Do you have any idea how they manipulated into that?
 
  • #4
C is -7, not 7. The text is correct. Use the quadratic formula and solve for y in the equation y2 - (x)y + (x2 - 7) = 0.

Your solution does not define a function; it is a formal solution (according to Spiegel's Applied Differential Equations) only because it satisfies the original differential equation. The way you have it now, (1, 3) and (1, -2) are both points on the curve: (1)2 - (1)(3) + (3)2 - 7 = 1 - 3 + 9 - 7 = 0 = 1 + 2 + 4 - 7 = (1)2 - (1)(-2) + (-2)2 - 7. The initial value problem states y(1) = 3 only, so take the positive branch.
 
Last edited:
  • #5
got it now, thanks
 

Related to Solve ODE Exact Equations: Initial Value Problem

What is an exact equation in the context of solving ODEs?

An exact equation is a type of ordinary differential equation (ODE) in which the coefficients of both the dependent variable and its derivative are functions of the independent variable. This means that the equation can be written in the form of an exact differential, making it easier to solve by hand.

How do you determine if an ODE is exact?

To determine if an ODE is exact, you can use the "exactness test." This involves checking if the partial derivatives of the function with respect to both the dependent and independent variable are equal. If they are equal, then the equation is exact and can be solved using the method of solving exact equations.

What is the process for solving an exact equation and initial value problem?

The process for solving an exact equation and initial value problem involves three steps: 1) determining if the equation is exact, 2) finding the integrating factor, and 3) solving the equation using the integrating factor and initial conditions. This process allows for the exact solution of the ODE and initial value problem.

Can an ODE be solved without using the method of exact equations?

Yes, there are other methods for solving ODEs, such as the method of separation of variables, the method of integrating factors, and the method of variation of parameters. However, for certain ODEs, the method of exact equations may be the most efficient and straightforward method for finding the solution.

What are some real-life applications of solving ODEs using the method of exact equations?

The method of exact equations has many real-life applications, such as in physics, engineering, and economics. For example, it can be used to model the growth of a population, the spread of a disease, or the movement of a particle under the influence of a force. It can also be used to solve problems involving electrical circuits, chemical reactions, and fluid dynamics.

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