Question

x'+xy=y^3 \ \to \ \frac{C'(y)e^{\frac{y^2}{2} }-yC(y)e^{\frac{y^2}{2} }}{ e^{y^2} } }+\frac{yC(y)}{ e^{\frac{y^2}{2} } }=y^3

Answer 1

Ответ:

Объяснение:

Извнени что так долго вот все во вложении не благодари

Answer 2

It seems like you are trying to solve a first-order ordinary differential equation (ODE) of the form:

x' + xy = y^3.

To solve this ODE, we can use the method of integrating factors. The integrating factor for this equation is given by e^(y^2/2). Multiplying both sides of the equation by this integrating factor, we get:

e^(y^2/2) * x' + e^(y^2/2) * xy = e^(y^2/2) * y^3.

Now, notice that the left-hand side of the equation can be simplified using the product rule for differentiation:

d/dy(e^(y^2/2) * x) = e^(y^2/2) * x' + (1/2) * y * e^(y^2/2) * x.

So, the equation becomes:

d/dy(e^(y^2/2) * x) = e^(y^2/2) * y^3.

Now, we can integrate both sides with respect to y:

∫ d/dy(e^(y^2/2) * x) dy = ∫ e^(y^2/2) * y^3 dy.

Integrating the left-hand side gives us:

e^(y^2/2) * x = ∫ e^(y^2/2) * y^3 dy.

Now, let's find the integral of the right-hand side. This requires some integration by parts. Let u = y^2 and dv = y * e^(y^2/2) dy.

Then, du = 2y dy and v = e^(y^2/2). The integration by parts formula is given by:

∫ u dv = uv - ∫ v du.

So, we have:

∫ e^(y^2/2) * y^3 dy = ∫ y^2 * e^(y^2/2) dy = y^2 * e^(y^2/2) - ∫ 2y * e^(y^2/2) dy = y^2 * e^(y^2/2) - 2 ∫ y * e^(y^2/2) dy.

Now, we notice that the integral on the right-hand side is just equal to -d/dy(e^(y^2/2)). So, we can write:

∫ e^(y^2/2) * y^3 dy = y^2 * e^(y^2/2) + 2 * e^(y^2/2) + C(y),

where C(y) is an arbitrary constant of integration that may depend on y.

Going back to our original equation, we have:

e^(y^2/2) * x = y^2 * e^(y^2/2) + 2 * e^(y^2/2) + C(y).

Now, we can solve for x:

x = y^2 + 2 + C(y) * e^(-y^2/2).

And there you have the general solution for the given differential equation. C(y) is an arbitrary function of y that depends on the initial conditions of the problem.