Trouble with first-order exact equation

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In summary, the conversation discusses a problem with differentiating a given equation and the use of the chain rule. The original equation involves a factor of [dy/dx] which comes from the chain rule when differentiating the product of xy^2. The individuals in the conversation offer helpful tips and suggest more practice with identifying when to use the chain rule.
  • #1
lukka
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Hi everyone, any help with this would be greatly appreciated..

I have been practicing simple differential equations for a couple months now and kinda just taking it easy and enjoying building my intuition. i have encountered a chapter in my text by Backhouse (pure mathematics 2) involving first order exact equations as a prelude to using integrating factors. It shows by example an inseparable differential equation..

2xy [dy/dx] +y^2 = e^(2x)

whose LHS is said to be equal to the derivative of the product (xy^2). The trouble I'm having here is that when i check and differentiate (xy^2) by product rule, i wind up with just 2xy + y^2. My question is where does the factor of [dy/dx] in the original equation come from? i suspect that i might be doing something wrong here?
 
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  • #2
You must use chain rule.
 
  • #3
lukka said:
Hi everyone, any help with this would be greatly appreciated..

I have been practicing simple differential equations for a couple months now and kinda just taking it easy and enjoying building my intuition. i have encountered a chapter in my text by Backhouse (pure mathematics 2) involving first order exact equations as a prelude to using integrating factors. It shows by example an inseparable differential equation..

2xy [dy/dx] +y^2 = e^(2x)

whose LHS is said to be equal to the derivative of the product (xy^2). The trouble I'm having here is that when i check and differentiate (xy^2) by product rule, i wind up with just 2xy + y^2. My question is where does the factor of [dy/dx] in the original equation come from? i suspect that i might be doing something wrong here?
Never say "differentiate" without specifying "differentiate with respect to which variable?"

Here, you are differentiating with respect to x. The derivative of y^2 with respect to x is NOT 2y. That is the derivative of y^2 with respect to y. The derivative of y^2 with respect to x is (by the chain rule that szynkasz mentions) 2y dy/dx.
 
  • #4
I see where I'm going wrong, thanks to you both for pointing this out for me.. clearly need more practice with identifying when to use the chain rule! Thanks again!
 
  • #5


First-order exact equations can be tricky, so don't worry if you're having trouble with them. It sounds like you are on the right track by using the product rule to differentiate (xy^2). However, you may want to double check your work to make sure you are not missing any terms or making any mistakes in your calculations.

Another helpful tip for solving first-order exact equations is to always check if the equation is actually exact by verifying that the partial derivatives of the coefficients with respect to x and y are equal. If they are not equal, then the equation is not exact and you may need to use a different method to solve it.

Additionally, the factor of [dy/dx] in the original equation comes from the fact that we are dealing with a differential equation, where the derivative of the dependent variable (y) with respect to the independent variable (x) is involved. This is why we use the notation [dy/dx] to represent the derivative.

I hope this helps and keep practicing! Solving differential equations takes time and practice to master, so don't get discouraged. Good luck!
 

FAQ: Trouble with first-order exact equation

What is a first-order exact equation?

A first-order exact equation is a type of differential equation that can be solved by finding a function whose derivative is equal to the given equation. It contains both a dependent and an independent variable, and the solution involves finding a function that satisfies both the equation and its partial derivatives.

How do you know if an equation is first-order exact?

An equation is first-order exact if the coefficients of the dependent and independent variables are equal and the equation satisfies the partial derivatives test. This means that the partial derivatives of the equation with respect to both variables are equal, indicating that the equation can be solved using the method of exact equations.

What is the method for solving first-order exact equations?

The method for solving first-order exact equations involves finding a function that satisfies both the equation and its partial derivatives. This can be done by integrating the equation with respect to the dependent variable and then checking if the resulting function satisfies the partial derivatives. If not, the process is repeated by integrating the equation with respect to the independent variable until a solution is found.

What are some applications of first-order exact equations?

First-order exact equations have many applications in science and engineering, particularly in fields such as physics, chemistry, and biology. They can be used to model various physical phenomena, such as heat transfer, fluid flow, and chemical reactions. They are also commonly used in population dynamics, economics, and other fields that involve rate equations.

Are there any limitations to using first-order exact equations?

While first-order exact equations are powerful tools for solving certain types of differential equations, they do have some limitations. They can only be used for equations with two variables, and the coefficients of these variables must be equal. Additionally, the method can become quite complex and time-consuming for more complicated equations, so alternative methods may be more efficient in those cases.

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