can someone elxplai to me how to solve this differential equation.
dy/dx= (x^2+3y^2)/2xy
be careful with parentheses
like use
dy/dx= (x^2+3y^2)/ ( 2xy )
http://www.wolframalpha.com/widgets/view.jsp?id=e602dcdecb1843943960b5197efd3f2a
y = +/- x (cx-1)^.5
Note Mathmate response to the question below titled
Calculus AB
Certainly! To solve this differential equation, we can use the method of separation of variables. Here's how you can do it step by step:
Step 1: Rearrange the equation.
Start by multiplying both sides of the equation by 2xy to get rid of the denominator:
2xy * dy/dx = x^2 + 3y^2
Step 2: Separate the variables.
Now, we will separate the variables by moving all the terms involving y to one side of the equation and all the terms involving x to the other side:
2y * dy = (x^2 + 3y^2) * dx
Step 3: Integrate both sides.
Next, integrate both sides of the equation with respect to their respective variables:
∫(2y * dy) = ∫((x^2 + 3y^2) * dx)
The integral of 2y with respect to y is simply y^2, so the left side becomes:
y^2 = ∫((x^2 + 3y^2) * dx)
The integral on the right side might be a bit tricky, but let's focus on it. To integrate (x^2 + 3y^2) with respect to x, we will treat y^2 as a constant. We can rewrite the right side as follows:
∫(x^2 + 3y^2) * dx = ∫x^2 * dx + ∫3y^2 * dx
The integral of x^2 with respect to x is (1/3)x^3, and since y^2 is a constant with respect to x, we can write:
∫(x^2 + 3y^2) * dx = (1/3)x^3 + 3y^2x + C
(where C is the constant of integration).
Step 4: Solve for y.
Now, we have the equation:
y^2 = (1/3)x^3 + 3y^2x + C
This is a quadratic equation in terms of y^2. To solve for y, we can take the square root of both sides:
y = ±√((1/3)x^3 + 3y^2x + C)
Note that there are two solutions for y due to the ± sign.
And that's it! You have solved the given differential equation and obtained the general solution for y in terms of x.