MHB Ewaz's question at Yahoo Answers regarding a Bernoulli Equation

[MarkFL]

Here is the question:

Solve this Differential Equations problem? (Bernoulli's)?

y' + xy = y2

answer should be

y = 1/[ -e^(x2/2) ∫ e^(-x2/2) dx]

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Ewaz,

We are given to solve:

$$\frac{dy}{dx}+xy=y^2$$

Dividing through by $$y^2$$ (observing we are losing the trivial solution $y\equiv0$) we obtain:

$$y^{-2}\frac{dy}{dx}+xy^{-1}=1$$

We want to use the substitution:

$$v=y^{-1}$$

Differentiating with respect to $x$, we then obtain:

$$\frac{dv}{dx}=-y^{-2}\frac{dy}{dx}$$

And so our ODE becomes:

$$\frac{dv}{dx}-xv=-1$$

This is a linear ODE, and thus computing our integrating factor, we obtain:

$$\mu(x)=e^{-\int x\,dx}=e^{-\frac{x^2}{2}}$$

Multiplying the ODE by this factor, we obtain:

$$e^{-\frac{x^2}{2}}\frac{dv}{dx}-xe^{-\frac{x^2}{2}}v=-e^{-\frac{x^2}{2}}$$

Observing that the left side is not the differentiation of a product, we obtain:

$$\frac{d}{dx}\left(e^{-\frac{x^2}{2}}v \right)=-e^{-\frac{x^2}{2}}$$

Integrating with respect to $x$, there results:

$$\int\,d\left(e^{-\frac{x^2}{2}}v \right)=-\int e^{-\frac{x^2}{2}}\,dx$$

$$e^{-\frac{x^2}{2}}v=-\int e^{-\frac{x^2}{2}}\,dx$$

Multiplying through by $$e^{\frac{x^2}{2}}$$, we obtain:

$$v=-e^{\frac{x^2}{2}}\int e^{-\frac{x^2}{2}}\,dx$$

Back-substituting for $v$, we have:

$$\frac{1}{y}=-e^{\frac{x^2}{2}}\int e^{-\frac{x^2}{2}}\,dx$$

Inverting both sides, we the find:

$$y(x)=\frac{1}{-e^{\frac{x^2}{2}}\int e^{-\frac{x^2}{2}}\,dx}$$