- Thread starter
- Admin
- #1

I have posted a link there to this thread so the OP can view my work.Solve this Differential Equations problem? (Bernoulli's)?

y' + xy = y2

answer should be

y = 1/[ -e^(x2/2) ∫ e^(-x2/2) dx]

- Thread starter MarkFL
- Start date

- Thread starter
- Admin
- #1

I have posted a link there to this thread so the OP can view my work.Solve this Differential Equations problem? (Bernoulli's)?

y' + xy = y2

answer should be

y = 1/[ -e^(x2/2) ∫ e^(-x2/2) dx]

- Thread starter
- Admin
- #2

We are given to solve:

\(\displaystyle \frac{dy}{dx}+xy=y^2\)

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

\(\displaystyle y^{-2}\frac{dy}{dx}+xy^{-1}=1\)

We want to use the substitution:

\(\displaystyle v=y^{-1}\)

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

\(\displaystyle \frac{dv}{dx}=-y^{-2}\frac{dy}{dx}\)

And so our ODE becomes:

\(\displaystyle \frac{dv}{dx}-xv=-1\)

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

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

Multiplying the ODE by this factor, we obtain:

\(\displaystyle e^{-\frac{x^2}{2}}\frac{dv}{dx}-xe^{-\frac{x^2}{2}}v=-e^{-\frac{x^2}{2}}\)

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

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

Integrating with respect to $x$, there results:

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

\(\displaystyle e^{-\frac{x^2}{2}}v=-\int e^{-\frac{x^2}{2}}\,dx\)

Multiplying through by \(\displaystyle e^{\frac{x^2}{2}}\), we obtain:

\(\displaystyle v=-e^{\frac{x^2}{2}}\int e^{-\frac{x^2}{2}}\,dx\)

Back-substituting for $v$, we have:

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

Inverting both sides, we the find:

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