Dy/dx = (y^2) /x , form differential equation

In summary: K' is another arbitrary constant.In summary, the given answer (-1/ln x ) + C is likely incorrect and the solution provided by the asker (-1/(ln x - c)) is correct. It can be easily verified by plugging the solution back into the original differential equation. However, it is also important to note that the solution can be further simplified to y = -1/(ln Kx), where K is an arbitrary constant.
  • #1
goldfish9776
310
1

Homework Statement


my ans is lnx = (-1/y) + c
(-1/y) = lnx -c
y = -1/ (lnx -c ) , but the answer given is (-1/ln x )+ C , how to get the answer given ?

Homework Equations

The Attempt at a Solution

 
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  • #2
Your solution seems fine. I believe the given answer may be wrong.
 
  • #3
You can easily verify that your answer is correct and the given answer is wrong by plugging them back into the original DE. By the way, you should post the original DE in the body, not the title of your post.

The only thing I would change is use ##\ln |x|## in the antiderivative.
 
  • #4
goldfish9776 said:

Homework Statement


my ans is lnx = (-1/y) + c
(-1/y) = lnx -c
y = -1/ (lnx -c ) , but the answer given is (-1/ln x )+ C , how to get the answer given ?

Homework Equations

The Attempt at a Solution

When you post a question, please put the problem statement into the first section, not the thread title.

Your answer looks fine to me, except that it should be ln|x|, but you should carry it a step further and solve for y. The given answer should have the constant in the denominator, not just added to the fraction as you show it.
 
  • #5
And more often than not for a result like that you would be writing, since c is an arbitrary constant

y = -1/(ln x + ln K) where K is another arbitrary constant

= -1/(ln Kx) and thence maybe

x = K' e-y
 

Related to Dy/dx = (y^2) /x , form differential equation

What is the meaning of "Dy/dx = (y^2) /x" in the context of a differential equation?

The notation "Dy/dx" represents the derivative of the function y with respect to x. In other words, it represents the rate of change of y with respect to x. The expression "(y^2)/x" is the function that is being differentiated, and it is equal to the derivative of y with respect to x.

How do you determine the general solution of the differential equation "Dy/dx = (y^2) /x"?

To determine the general solution of this differential equation, we need to separate the variables and integrate both sides. This involves bringing all terms with y to one side and all terms with x to the other side. Then, we can use the power rule for integration to solve for y. The result will be the general solution, which includes a constant of integration.

Can you give an example of a specific solution to the differential equation "Dy/dx = (y^2) /x"?

One example of a specific solution to this differential equation is y = 1/x. This solution satisfies the equation because when we substitute it into the equation, we get the statement "1/x = (1/x)^2/x", which is true.

What is the significance of the "D" in the notation "Dy/dx"?

The "D" in "Dy/dx" stands for "derivative." It represents the operation of finding the rate of change of a function with respect to its independent variable. In this context, it is used to denote that the equation is a differential equation, which involves derivatives.

How is the differential equation "Dy/dx = (y^2) /x" used in real-world applications?

This differential equation is commonly used to model exponential growth and decay processes, such as population growth or radioactive decay. It can also be used to describe the flow of fluids and the behavior of electrical circuits. In general, any situation that involves a quantity changing at a rate proportional to its current value can be modeled by this type of differential equation.

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