How to Solve the Differential Equation dy/dx= x^2/y?

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In summary, the user is asking for help solving a differential equation and is reminded to post their question in the appropriate forum with their attempted solution using the homework template.
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bilb27
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Anyone help me with this please.

Solve the differential equation

dy/dx= x^2/y
 
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Hi bilb27, welcome to PF. If you click on the Rules tab above, you will find, among other things, this:

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  • #3
http://www.wolframalpha.com/input/?i=dy%2Fdx%3D+x^2%2Fy
 

1. What is the meaning of dy/dx in the equation "dy/dx = x^2/y"?

The notation dy/dx represents the derivative of the function y with respect to x. It measures the instantaneous rate of change of y with respect to x at a specific point.

2. How do you find the derivative of "x^2/y"?

To find the derivative of x^2/y, we use the quotient rule, which states that the derivative of a quotient is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. In this case, the derivative of x^2 is 2x and the derivative of y is 1, so the final answer is (2xy - x^2)/y^2.

3. What is the significance of the equation "dy/dx = x^2/y"?

The equation represents a relationship between the rate of change of y with respect to x and the variables x and y themselves. It can be used to calculate the slope of a tangent line to the graph of the function y = x^2/y at a given point.

4. How can "dy/dx = x^2/y" be applied in real-life situations?

The equation can be applied in various fields of science and engineering, such as physics, economics, and biology. For example, it can be used to determine the rate of change of a physical quantity, such as velocity or acceleration, in a given system.

5. Can "dy/dx = x^2/y" be solved for y?

Yes, it is possible to solve for y by rearranging the equation to y = x^2/(dy/dx). This gives an implicit expression for y in terms of x and the derivative dy/dx. However, in most cases, it is more useful to leave the equation in its original form and use it to calculate the derivative at specific points.

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