Question

2y^3y'=x^2+4 дайте решение уравнения пожалуйста

Answer 1

Решение смотрите в приложении

Answer 2

Solving the Equation 2y^3y' = x^2 + 4

To solve the equation 2y^3y' = x^2 + 4, we can use separation of variables. The first step is to rewrite the equation in a form that allows us to separate the variables.

The equation can be rewritten as:

y^3y' = (x^2 + 4)/2

Now, we can separate the variables by multiplying both sides of the equation by dx:

y^3dy = (x^2 + 4)/2 dx

Next, we integrate both sides of the equation:

∫y^3dy = ∫(x^2 + 4)/2 dx

Integrating the left side gives us:

∫y^3dy = (1/4)y^4 + C1

Integrating the right side gives us:

∫(x^2 + 4)/2 dx = (1/2)∫x^2 dx + ∫2 dx = (1/2)(1/3)x^3 + 2x + C2

Combining the results, we have:

(1/4)y^4 + C1 = (1/6)x^3 + 2x + C2

To find the particular solution, we need to determine the values of the constants C1 and C2. This requires additional information, such as initial conditions or boundary conditions.

Without additional information, we cannot determine the specific solution to the equation. However, we have obtained the general solution in terms of the constants C1 and C2:

(1/4)y^4 + C1 = (1/6)x^3 + 2x + C2

Please provide any additional information or specific conditions if you would like a more precise solution.