First-order equations · Definition
Separable equation
Before you start you need to know
This is the first family of first-order equations that can be solved completely just by integrating, and it serves as a model for the other methods.
Definition
A separable equation is a first-order differential equation whose derivative is the product of a function \(g\) of \(x\) alone and a function \(h\) of \(y\) alone.
The method moves everything that depends on \(y\) to one side and everything that depends on \(x\) to the other: write \(dy/h(y)=g(x)\,dx\) and integrate both sides. The constant of integration is written only once. For example, \(y'=2xy\) gives \(\ln|y|=x^2+c\), and hence \(y=Ce^{x^2}\).
The result is often implicit, like \(y^2=x^2+C\) for \(y'=x/y\), and solving for \(y\) is not always convenient. The constant is chosen with the initial condition, and the branch of the solved form (the sign of a root) is dictated by the initial value.
Dividing by \(h(y)\) assumes \(h(y)\ne0\). Every root \(y_0\) of \(h\) gives a constant solution \(y=y_0\) that the procedure does not produce and that must be added by hand. Nor should we assume the solution exists for all \(x\): \(y'=y^2\) with \(y(0)=1\) gives \(y=1/(1-x)\), which stops existing at \(x=1\).
Not every equation is separable: \(y'=x+y\) does not factor as a product. Sometimes a rewrite is enough, as in \(y'=e^{x+y}=e^{x}e^{y}\). When it is not possible, other methods are needed, such as the one for linear equations that comes next.
Think about it
Is \(y'=x(1+y)\) separable? If so, what is its general solution?
Yes: it is \(g(x)=x\) times \(h(y)=1+y\). Then \(\ln|1+y|=x^2/2+c\), that is \(y=Ce^{x^2/2}-1\), which includes the constant solution \(y=-1\) when \(C=0\).
Exercises 10
- Which of these equations is separable?
- When separating variables in \(y'=2xy\), which equality of integrals results?
- Find the general solution of \(y'=3y\).
- Separating variables in \(y'=x/y\), which implicit relation do the solutions satisfy?
- Solve the problem \(y'=y\), \(y(0)=2\).
- Solve \(y'=-2xy\), \(y(0)=3\).
- Solve \(y'=y^2\), \(y(0)=1\).
- When solving \(y'=y(y-2)\) we divide by \(y(y-2)\). Which constant solutions might be lost?
- Solve \(y'=-x/y\), \(y(0)=2\).
- Solve \(y'=e^{x+y}\), \(y(0)=0\).
References
- William E. Boyce, Richard C. DiPrima, Douglas B. Meade, Elementary Differential Equations and Boundary Value Problems, 12th edition, Wiley, 2021. §2.2
- Dennis G. Zill, A First Course in Differential Equations with Modeling Applications, 11th edition, Cengage, 2018. §2.2