Question

2xydx+(x^2+3y^2)dy=0

Answer 1

Ответ:

либо d=0, x и y-любое число

либо y=0, x и b- любое число

Объяснение:

оба уравнения умножается на d и y, так что они должны быть равны 0

али 2x^2*dy=-(x^2+3y^2)dy dy- сокращается

2x^2=-x^2-3y^2 переносим -x^2

3x^2=-3y^2

x^2=-y^2

|x|=-|y|- это невозможно, поэтому пишем только первые 2 ответа...

Answer 2

The given differential equation is:

2xy dx + (x^2 + 3y^2) dy = 0

To solve this equation, we can use the method of separation of variables. First, let's rewrite the equation as follows:

2xy dx + (x^2 + 3y^2) dy = 0

Divide both sides of the equation by (x^2 + 3y^2) to isolate the variables:

(2xy / (x^2 + 3y^2)) dx + dy = 0

Now, we can separate the variables x and y:

(2xy / (x^2 + 3y^2)) dx = -dy

Next, let's integrate both sides:

∫ (2xy / (x^2 + 3y^2)) dx = ∫ -dy

The left-hand side requires a substitution. Let u = x^2 + 3y^2, then du = (2x dx + 6y dy). Solving for dx in terms of du and dy:

2x dx = du - 6y dy

Now we can substitute this back into the integral on the left-hand side:

∫ (1/u) du = ∫ -dy

Now, integrate both sides:

ln|u| = -y + C₁

Where C₁ is the constant of integration. Now, substitute back for u:

ln|x^2 + 3y^2| = -y + C₁

To solve for y, we'll first get rid of the natural logarithm by exponentiating both sides:

|x^2 + 3y^2| = e^(C₁) * e^(-y)

Since e^(C₁) is just a constant, we can write it as another constant, let's call it K:

|x^2 + 3y^2| = K * e^(-y)

Now, we can consider the absolute value. This equation can have two forms depending on the sign of K:

1. If K is positive, then we have:

x^2 + 3y^2 = K * e^(-y)

2. If K is negative, then we have:

x^2 + 3y^2 = -K * e^(-y)

These are the general solutions to the given differential equation. Depending on the initial conditions or specific context of the problem, you may need to further simplify or work with one of these forms.