Differential equation

Question

Solve the differential equation: $$x \frac{dy}{dx} = y(3-y)$$ where x=2 when y=2, giving y as a function of x.

Can someone solve this and then explain what the second line about $x=2$, $y=2$ means?

Answer 1

So you want to "solve" the initial value problem:

$\begin{cases} x y^\prime (x) = y(x)\ (3-y(x)) \\ y(2)=2 \end{cases}$.

Your ODE has a singular point in $x=0$ (for the coefficient of the $y^\prime (x)$ term vanishes), hence if the IVP has a solution it will have not to be defined in $x=0$.

Put the ODE in normal form, i.e. rewrite:

$\displaystyle y^\prime (x) =\frac{y(x)\ (3-y(x))}{x}$;

the function $f(x,y):= \frac{y (3-y)}{x}$ is of class $C^\infty$ in $(x,y)\in \Big(]-\infty ,0[\cup ]0,+\infty[\Big)\times \mathbb{R}$, hence the existence and uniqueness theorem applies and your IVP has unique local solution $y(x)$ whose graph passes through the point $(2,2)$.

The solution $y(x)$ is continuous in a neighbourhood $I_1$ of $x=2$ (because it has to be differentiable to satisfy the ODE), hence the composite function $f(x,y(x))$ is continuous in $I_1$; as $y^\prime (x)=f(x,y(x))$, then $y^\prime (x)$ is continuous in $I_1$, therefore $y(x)$ is a $C^1$ function in $I_1$. But then $y^\prime (x)$ is of class $C^1$ in $I_1$, for the composite function $f(x,y(x))$ is of class $C^1$ (apply the chain rule); therefore $y(x)$ is of class $C^2$... Bootstrapping, you see that $y(x)$ is of class $C^\infty$ in the neighbourhood $I_1$ of the initial point $2$.

Moreover, the solution $y(x)$ is also strictly increasing in a neighbourhood of $2$: in fact, $y(2)=2>0$ hence by continuity you can find a neighbourhood $I_2\subseteq I_1$ of $2$ in which $0<y(x)<3$, so:

$\displaystyle y^\prime (x)=\frac{y(x)\ (3-y(x))}{x} >0$,

thus $y(x)$ increases strictly.

Now you have all the ingredients to properly solve your problem: in fact, in $I_2$ you can divide both sides of the ODE by $y(x)\ (3-y(x))$ and rewrite:

$\displaystyle \frac{y^\prime (x)}{y(x)\ (3-y(x))} =\frac{1}{x}$;

now fix a point $x \in I_2$ and integrate both sides from $2$ to $x$:

$\displaystyle \int_2^x \frac{y^\prime (t)}{y(t)\ (3-y(t))}\ \text{d} t =\int_2^x \frac{1}{t}\ \text{d} t$

(I've introduced a dummy variable in the integrals); now the RHside gives you $\ln x -\ln 2$, hence you have to work on the LHside. Keeping in mind that $y(t)$ is strictly monotone hence invertible in $I_2$, we can make the change of variable $\theta =y(t)$: as $y(2)=2$ and $\text{d} \theta = y^\prime (t)\ \text{d} t$, you get:

$\displaystyle \begin{split}\int_2^x \frac{y^\prime (t)}{y(t)\ (3-y(t))}\ \text{d} t &= \int_2^{y(x)} \frac{1}{\theta (3-\theta)}\ \text{d} \theta \\ &=\frac{1}{3} \ln \theta - \frac{1}{3} \ln (3-\theta) \Big|_2^{y(x)} \\ &=\frac{1}{3} \left(\ln y(x) -\ln (3-y(x)) -\ln 2\right)\end{split}$.

Therefore the solution to your problem is implicitly determined by the equation:

$\displaystyle \ln \left( \frac{y(x)}{3-y(x)}\right) = \ln \frac{x^3}{4}$,

i.e.:

$\displaystyle \frac{4y(x)}{3-y(x)}=x^3$.

The latter equation is a rational algebraic equation w.r.t. $y(x)$ and can be solved with the usual tools, which yield:

$\displaystyle y(x)=\frac{3x^3}{4+x^3}$.

There is more that can be said, e.g. how the local solution $y(x)$ can be extended to a maximal solution... But that's another story.

Answer 2

This is just an Initial Value Problem. You use the techniques you know to solve for the general solution. Here is a good resource: http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx. More specifically, you should look at "Separable Equations"

The general solution in this case will have one arbitrary constant. You use the I.V.P to solve for this constant.

Answer 3

This is easy because it's separable variables. So solve $dy/(y(3-y))=dx/x$.