# I with differential equation(3x^2-2y^2)(dy/dx)=2xy

• I
• edgarpokemon
In summary, the conversation discusses a problem that the speaker has been stuck on for two days. They are trying to solve for a solution using a specific method, but are struggling to understand why their answer does not match the given solution. After providing their attempted steps and integrating, the speaker realizes their mistake and includes the constant of integration to get the correct answer. The conversation ends with a thank you from the speaker.
edgarpokemon

edgarpokemon said:
What about the constant of integration?

Simon Bridge
haruspex said:
What about the constant of integration?
if v^2-2=u, then du=2v, which cancels the 2v that is above of the denominator right? help! how would you do it using the method i used?

edgarpokemon said:
du=2v
2dv.
But you did not answer my question. Go through the integration step and make sure to include a constant of integration. If you still do not see your error, please post all your steps in detail.

haruspex said:
2dv.
But you did not answer my question. Go through the integration step and make sure to include a constant of integration. If you still do not see your error, please post all your steps in detail.
I don't see my error! i will post my steps: (3x^2-2y^2)(dy/dx)=2xy. (3x^2-2y^2)/(2xy)=dx/dy. (3x^2/2xy)-(2y^2/2xy)=dx/dy. 3(x/y)/2-(y/x)=dx/dy. x=yv, v=x/y.
(3/2)v-(1/v)=(dv/dy)y+v. (3/2)v-(1/v)-v=(dv/dy)y. find lcd for left side of equation. lcd is 2v, so (3v^2-2-2v^2)/(2v)=(dv/dy)y. make dv on one side and dy on the other side, so (2v/(v^2-2))dv=dy/y. integral of (2v/(v^2-2))dv: u=v^2-2, du=2vdv. dv=du/2v, the 2v cancels out, leading to a resulting integral of ln|v^2-2|=ln|y|+c. multiply both sides by e leads to, y=(x^2/y^2)-2+c. the values are y=-1 and x=0. so c=y-(x^2/y^2)+2. c=-1+2=1. y=(x^2/y^2)-2+1. y=(x^2/y^2)-1. y=(x^2-y^2)/(y^2).
y^3=x^2-y^2. y^3+y^2=x^2. y^2(y+1)=x^2. but the answer is 2y^2(y+1)=x^2. my question would be the factor of two on the right side. help!

edgarpokemon said:
ln|v^2-2|=ln|y|+c. multiply both sides by e
No, you do not multiply by e to remove the ln()s. What do you do?
edgarpokemon said:
That is wrong.

haruspex said:
No, you do not multiply by e to remove the ln()s. What do you do?

That is wrong.
I thought that it was possible to cancel the natural log by e^x since it is the opposite of the natural log. so will the constant c also be e^c? and I solve for c? help me already! jaja

oh I got it! so ln|y|=ln|c((x^2/y^2)-2). x=0. y=-1. I use e^x on both side of the equation. so c=1/2. then y=1/2((x^2/y^2)-2). then y=(x^2-2y^2)/(2y^2). finally 2y^3+2y^2=x^2. correct? I have never worked a problem were I have to found what c is, so maybe that was my problem?

edgarpokemon said:
ln|y|=ln|c((x^2/y^2)-2)
Where ln(|c|) is the constant of integration, yes.

haruspex said:
Where ln(|c|) is the constant of integration, yes.
thank you very much :P

## What is a differential equation?

A differential equation is an equation that relates an unknown function to its derivatives. It is commonly used in mathematical models to describe the behavior of systems that change over time.

## What is the order of this differential equation?

The order of a differential equation is determined by the highest derivative of the unknown function that appears in the equation. In this case, the highest derivative is dy/dx, so the order is 1.

## How do you solve this differential equation?

There are various methods for solving differential equations, depending on the type of equation. In this case, the equation is separable, meaning that the variables can be separated on opposite sides of the equation. This allows us to solve for y in terms of x.

## What is the general solution to this differential equation?

The general solution to a differential equation is an expression that includes all possible solutions to the equation. In this case, the general solution is y = Cx^2, where C is a constant.

## What is the particular solution to this differential equation?

The particular solution to a differential equation is a specific solution that satisfies any initial conditions given. In this case, if we are given the initial condition y(0) = 0, then the particular solution would be y = 0. If no initial conditions are given, then the general solution is also the particular solution.

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