Need help to solve y'=(x^2+y^2)^(3/2)

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In summary, the conversation is about a person seeking help for solving an equation involving the derivative of a function. They have attempted to use a coordinate transformation and the chain rule, but are struggling with the new equation that arises. The expert suggests developing a new equation and finding a solvable equation for y before taking the cube root to solve for y.
  • #1
pociners
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i need some help to solve this equation..

[tex]\frac{dy}{dx}=(x^{2}+y^{2})^{\frac{3}{2}}[/tex]

thanks.
 
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  • #2
Welcome to PF

This should be posted in the "Calculus & Beyond" section of the Homework Help forum, and it should be posted using the template for homework help threads which includes:

1. An exact statement of the problem

2. Any equations that might be relevant to solving the problem

3. YOUR ATTEMPT AT A SOLUTION SO FAR

What have you done so far for this problem? Hint: can you think of a coordinate transformation that might help here?
 
  • #3
I'm sorry if I was in the wrong section. I don't know about that.

I've tried using the coordinate transformation.
I suppose [tex]x^{2}+y^{2}=r^{2}[/tex]
where [tex]x=r\cos{\theta}[/tex] and [tex]y=r\sin{\theta}[/tex]

but I do not know what is the [tex]\frac{dy}{dx}[/tex]?

can you help me with this?
 
  • #4
double post...
 
  • #5
I was thinking you could use the chain rule ie

(dy/dx) = (dy/dr)(dr/dx)

But it's working out to be a bit messy for me. Anyway, see if that helps.
 
  • #6
cepheid said:
I was thinking you could use the chain rule ie

(dy/dx) = (dy/dr)(dr/dx)

But it's working out to be a bit messy for me. Anyway, see if that helps.
i've tried it..but it just added a new problem..
[tex]r^{3}=\sin(\theta)\cos(\theta)[/tex]

then what should i do with this??
 
  • #7
just develop a new equation of a circle of r^3, then write a new equation that's solvable
 

1. What is the given equation asking me to do?

The given equation, y'=(x^2+y^2)^(3/2), is asking you to find the derivative of y with respect to x.

2. How do I solve this type of equation?

This is a separable differential equation, which means you can separate the variables and integrate both sides to solve for y.

3. Can you explain the steps to solve this equation?

First, rewrite the equation as y'=(x^2+y^2)^(3/2) as y'=((x^2+y^2)^2)^(1/2). Then, separate the variables by dividing both sides by (x^2+y^2)^(3/2) to get y'/(x^2+y^2)^(3/2)=1. Next, integrate both sides with respect to x to get the general solution: y=1/5(x^2+y^2)^(5/2)+C. Finally, you can use initial conditions to solve for the constant C and get the specific solution.

4. Are there any specific tips or tricks to solve this equation?

One tip is to use substitution to simplify the equation. For example, letting u=x^2+y^2 can make the equation easier to integrate. Also, pay attention to your initial conditions as they are crucial in finding the specific solution.

5. Can you provide an example of solving this equation?

For example, if the initial condition is y(0)=1, then the specific solution would be y=1/5(x^2+y^2)^(5/2)-1/5. To check, you can take the derivative of this solution and substitute in x=0 and y=1 to get y'(0)=1, which satisfies the initial condition.

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