Separation, initial conditions, and lost equilibria

Understand the idea

For a separable equation move the factors involving $\displaystyle y$ to the $\displaystyle dy$ side and the factors involving $\displaystyle x$ to the $\displaystyle dx$ side, then integrate. Keep an integration constant until applying the initial condition. Dividing by a function of $\displaystyle y$ can discard equilibrium solutions, so inspect those separately. State the interval containing the initial point on which the expression solves the equation.

\frac{dy}{dx}=g(x)h(y)\quad\Rightarrow\quad\int\frac{dy}{h(y)}=\int g(x)\,dx\quad\text{when }h(y)\ne0

Worked example

Solve $\displaystyle y'=2xy$ with $\displaystyle y(0)=3$ .

For

\displaystyle y\ne0

, separate to get

\displaystyle dy/y=2x\,dx

. Integration gives

\displaystyle \ln|y|=x^2+C

, hence

\displaystyle y=Ae^{x^2}

. Applying the initial condition gives

\displaystyle A=3

. Thus

\displaystyle y=3e^{x^2}

for all real

\displaystyle x

. The discarded equilibrium

\displaystyle y=0

does not meet this initial condition.

Try it yourself

Solve $\displaystyle y'=y^2$ with $\displaystyle y(0)=1$ and state its maximal interval.

Show solution and reasoning

$\displaystyle dy/y^2=dx$ gives $\displaystyle -1/y=x+C$ . With $\displaystyle y(0)=1$ , $\displaystyle C=-1$ , so $\displaystyle y=1/(1-x)$ on $\displaystyle (-\infty,1)$ .