Solving Equations: Choosing the Right Method | Homogeneous Equation Example

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In summary, the conversation is about finding the correct method to solve the equation (t^3)(y^2)dt + (t&4)(y^-6)dy = 0 and determining if the equation dy/dx - y/x = (x^2)sin2x is homogeneous. Dao Tuat asks for help with the first equation and someone suggests separating the variables. Dao Tuat then shares their solution and asks if it looks correct. They also ask for help with the second equation and someone else shares their solution. The conversation ends with Dao Tuat confirming if their solutions and approach are correct.
  • #1
Dao Tuat
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Could somone help me out and tell me what method to use to solve:

(t^3)(y^2)dt + (t&4)(y^-6)dy = 0

Also, the equation dy/dx - y/x = (x^2)sin2x is homogeneous, right?

Thanks,
Dao Tuat
 
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  • #2
for the top one try seperating the variables
 
  • #3
Ok, so I separated them and integrated and ended up with:
y=[7^(6/7)]/(7(ln(t)-c))

Does this look right?

Thanks,
Dao Tuat
 
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  • #4
Can anyone help me out with the second equation? I'm completely lost.
 
  • #5
that equation for post #3 is not what i got. What did your separation of variables look like?

EDIT: I'm assuming that equation you have is either [tex] y=\frac {7^\frac {6}{7}}{7(ln(t)-c)}[/tex] or [tex] y=7^\frac{6/7}{7(ln(t)-c))}[/tex]
 
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  • #6
Yes, it looked like the first one, which is the same as
y=7*sqrt[-1/(7[ln(t)+c])], correct?

What I did was end up with -[(t^3)/(t^4)]dt=[(y^-6)/(y^2)]dy after seperating

I then integrated and got -ln(t)+c=-1/(7y^7), and then solved for y.

For the other equation, dy/dx - y/x = (x^2)sin2x, I got:
y=[(xsin2x)/4]-[([x^2]cos2x)/2]
 
  • #7
Do these answeres and approach look right?
 

FAQ: Solving Equations: Choosing the Right Method | Homogeneous Equation Example

What is a homogeneous equation?

A homogeneous equation is an equation where all the terms have the same degree, meaning they have the same power of the variables. In other words, there are no constants in a homogeneous equation.

What is the difference between a homogeneous equation and a non-homogeneous equation?

The main difference between a homogeneous equation and a non-homogeneous equation is that a homogeneous equation has all terms with the same degree, while a non-homogeneous equation has at least one term with a different degree.

What is the method for solving homogeneous equations?

The method for solving homogeneous equations is to first identify the degree of the equation and then substitute the variables with a new variable, such as y = vx, to reduce the equation to a simpler form. This new equation can then be solved using methods such as separation of variables or integrating factors.

What are some common applications of homogeneous equations in science?

Homogeneous equations have many applications in science, including in physics, chemistry, and engineering. They are often used to model systems that are in a state of equilibrium, such as chemical reactions or physical forces acting on an object.

Why is it important to choose the right method when solving equations?

Choosing the right method when solving equations is important because different methods may be more efficient or effective for different types of equations. Using the wrong method can lead to errors or make the problem more complex than it needs to be. It is important to understand the characteristics of the equation and choose the appropriate method to solve it accurately and efficiently.

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