Solving the Equation y^2- xy + (x+1)y' = 0

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In summary, the given equation is a Bernoulli equation and can be solved by putting it in standard form and making a substitution. Examples can be found in any introductory differential equations textbook.
  • #1
psholtz
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Homework Statement


Solve the following equation:

[tex]y^2- xy + (x+1)y' = 0[/tex]

The Attempt at a Solution


The equation isn't exact, and it isn't homogeneous.

I've tried a range of different substitutions, including v = y - x, v = y^2, v = y^2 - xy, none of which seem to lead down a fruitful path.

I've tried differentiating this expression, to obtain a second-order ODE, and then eliminate either the nonlinear term, or the y' term, etc, between the two expressions, but that doesn't seem to lead anywhere fruitful either..

Any hints? :-)
 
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  • #2
psholtz said:

Homework Statement


Solve the following equation:

[tex]y^2- xy + (x+1)y' = 0[/tex]

The Attempt at a Solution


The equation isn't exact, and it isn't homogeneous.

I've tried a range of different substitutions, including v = y - x, v = y^2, v = y^2 - xy, none of which seem to lead down a fruitful path.

I've tried differentiating this expression, to obtain a second-order ODE, and then eliminate either the nonlinear term, or the y' term, etc, between the two expressions, but that doesn't seem to lead anywhere fruitful either..

Any hints? :-)

That' a Bernoulli equation. Put it in standard form and make the usual substitution. Check out any intro DE textbook for an example.
 

1. How do you solve the equation y^2-xy+(x+1)y' = 0?

To solve this equation, you can use the quadratic formula or factor the equation. If you use the quadratic formula, let a = 1, b = -x, and c = (x+1)y' and solve for y. If you choose to factor, you can rewrite the equation as (y-x)(y+(x+1)y') = 0 and solve for y.

2. What are the different types of solutions for this equation?

The solutions for this equation can be real or complex numbers. If the discriminant (b^2-4ac) is positive, there will be two real solutions. If the discriminant is zero, there will be one real solution. If the discriminant is negative, there will be two complex solutions.

3. Can this equation be graphed?

Yes, this equation can be graphed by setting it equal to y and using a graphing calculator or software. The resulting graph will be a parabola.

4. How can this equation be applied in real life?

This equation can be used to model various situations in physics and engineering, such as projectile motion or electrical circuits. It can also be used in economics to model relationships between variables.

5. Is there a specific method to solve this equation?

There are multiple methods to solve this equation, such as using the quadratic formula, factoring, or using differential equations techniques. The best method to use will depend on the specific equation and the desired outcome.

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