A first order differential equation is separable if it can be written in one of the following forms:
\[\begin{align} \frac{\mathrm{d} y}{\mathrm{d} x} &= f(x,y) = \frac{g(x)}{h(y)}, \\ \frac{\mathrm{d} y}{\mathrm{d} x} &= f(x,y) = \frac{h(y)}{g(x)}. \end{align}\]
A separable equation is solved by separating the variables, that is, rearranging the equation so that everything involving $y$ appears on one side of the equation, and everything involving $x$ appears on the other. The equation can then be integrated directly.
For an equation in the form:
\[\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{g(x)}{h(y)},\]
multiplying both sides by $h(y)\mathrm{d} x$ gives:
\[h(y)\mathrm{d} y = g(x) \mathrm{d} x,\]
which can be integrated directly:
\[\int h(y) \mathrm{d} y = \int g(x) \mathrm{d} x.\]
This will yield a solution for $y(x)$.
Similarly, an equation in the form:
\[\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{h(y)}{g(x)}\]
can be multiplied by $\dfrac{\mathrm{d} x}{h(y)}$ and then integrated:
\[\int \frac{\mathrm{d} y}{h(y)} = \int \frac{\mathrm{d} x}{g(x)},\]
thus yielding a solution for $y(x)$.
Note: The solution obtained for $y$ by computing these integrals may be implicitly defined. Rearranging the solution may be necessary to obtain an explicit solution for $y$, although in some cases it may not be possible to express $y$ explicitly.
Solve the differential equation
\[\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3x^2}{y}\]
subject to the condition $y(0)=2$.
The equation can be rearranged so that everything involving $y$ appears on the left-hand side of the equation, and everything involving $x$ appears on the right.
First multiply both sides by $y$:
\[y\frac{\mathrm{d} y}{\mathrm{d} x} = 3x^2,\]
then multiply both sides by $\mathrm{d} x$ to obtain
\[y \; \mathrm{d} y = 3x^2 \; \mathrm{d} x.\]
This can be integrated directly using standard integrals:
\begin{align} \int y \; \mathrm{d} y &= \int 3x^2 \; \mathrm{d} x, \\ \frac{y^2}{2} &= x^3 + C. \end{align}
Note: A constant of integration arises for both the integral on the left-hand side and the integral on the right-hand side. Since these constants are both arbitrary, then can be absorbed into one constant, denoted here by $C$, which is conventionally included on the side involving the independent variable $x$.
Rearranging the above equation gives an explicit solution for $y$:
\[\begin{align} \frac{y^2}{2} &= x^3 + C \\ y^2 &= 2x^3 + C' & (C' &= 2C)\\ y &= \pm\sqrt{2x^3 + C'}. \end{align}\]
Note: Since multiplying an arbitrary constant by a number gives an arbitrary constant, the quantity $2C$ can be relabelled $C'$, another arbitrary constant.
To find the solution that satisfies $y(0)=2$, substitute $x=0$ and $y=2$ into the solution and solve for $C'$:
\[\begin{align} 2 &= \pm\sqrt{0+C'} \\ 2 &= \pm\sqrt{C'} \end{align}.\]
Clearly this can only be satisfied by taking the positive square root. Then:
\[2 = \sqrt{C'} \Rightarrow C'=4,\]
and the solution satisfying the condition $y(0)=2$ is:
\[y=\sqrt{2x^3+4}.\]
Note: Since the condition $y(0)=2$ is only satisfied by taking the positive square root, the solution is only valid when the positive square root is taken, so the $\pm$ sign is no longer included.
Solve the differential equation
\[\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{x+3}\]
subject to the condition $y(0)=1$.
Multiplying both sides of the equation by $\mathrm{d} x$ gives
\[\frac{\mathrm{d} y}{y} = \frac{\mathrm{d} x}{x+3}.\]
Every term involving $x$ now appears on the right-hand side, and every term involving $y$ now appears on the left-hand side. Each side of the equation can therefore be integrated directly:
\[\begin{align} \int \frac{\mathrm{d} y}{y} &= \int\frac{\mathrm{d} x}{x+3}, \\ \ln\lvert y\rvert &= \ln\lvert x+3 \rvert + C. \end{align}\]
To find a solution which gives $y$ explicitly, exponentiate both sides to obtain:
\[y=A(x+3),\]
where $A$ is an arbitrary constant.
Note: The constant $A$ comes from applying the laws of logarithms and powers:
\[e^{\large{\ln\mid x+3\mid+C} } = e^C e^{\large{\ln\mid x+3\mid} } = e^C(x+3) = A(x+3),\]
where $A= e^C$. Since $ e$ is a number, $ e$ raised to a constant power is also a constant, and it is permissible to denote that constant by a single letter $A$.
Hence, the general solution to the differential equation is
\[y=A(x+3).\]
To find the solution that satisfies $y(0)=1$, substitute $x=0$ and $y=1$ into the solution and solve for $A$:
\[1=A(0+3) \Rightarrow 3A=1 \Rightarrow A=\frac{1}{3}.\]
Hence the solution to the given differential equation that satisfies the condition $y(0)=1$ is
\[y=\frac{1}{3}(x+3).\]
Prof. Robin Johnson solves $\dfrac{\mathrm{d}y}{\mathrm{d}x}=-2y^3$, subject to the condition $y(0)=1$.
Prof. Robin Johnson solves $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1+y^2}{2+3x}$, subject to the condition $y\left(-\dfrac{1}{3}\right)=0$.
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.