From a Separated Equation to an Integrated Relation
A separated differential equation has its \(x\)-dependent quantities on one side and its \(y\)-dependent quantities on the other. The next step is to integrate each side. This turns a differential equation into a relation involving \(x\), \(y\), and an arbitrary constant.
The constant matters: integrating both sides does not produce two independent parameters in the final family. It produces one, because only the difference between the two antiderivative constants affects the relation.
Integrating Both Sides
Suppose a differential equation has been separated into the form
Here \(dy\) and \(dx\) indicate which variable is integrated on each side. If \(H\) is an antiderivative of \(1/h\) and \(G\) is an antiderivative of \(g\), integrating gives
where \(C\) is an arbitrary real constant. This is an integrated relation between \(x\) and \(y\). For different values of \(C\), it generally describes different curves.
The notation \(H(y)\) means that the antiderivative \(H\), whose input is the dependent variable, is evaluated at \(y\). Likewise, \(G(x)\) is an antiderivative in the independent variable. Keeping track of which variable is being integrated helps prevent errors in both the antiderivative and the final relation.
Why There Is One Arbitrary Constant
One way to see why the integrated relation has a single constant is to use definite integrals along a solution. Suppose \(y'=g(x)h(y)\), and consider an interval where \(h(y)\ne 0\). Choose a point \(x_0\) on the interval and write \(y_0=y(x_0)\). Separating and integrating from the base values to \(x\) and \(y(x)\) gives
Let \(H\) be an antiderivative of \(1/h\), and let \(G\) be an antiderivative of \(g\). The Fundamental Theorem of Calculus turns this equality into
Rearranging gives
The quantities \(H(y_0)\), \(G(x_0)\), and the chosen base point are fixed for this particular solution. Their net contribution is one constant. Thus each solution has a relation of the form \(H(y)=G(x)+C\).
The same point appears if indefinite integrals are written with constants on both sides. Suppose one writes \(H(y)+C_1=G(x)+C_2\). Subtracting \(C_1\) from both sides gives \(H(y)=G(x)+(C_2-C_1)\). Since the difference \(C_2-C_1\) can be any real number, it is represented by one arbitrary constant \(C\), not two independent parameters.
Worked Examples
Worked Example: A Product of \(x\) and \(y\)
Integrate the separated equation obtained from \(y'=xy\), considering first solutions for which \(y\ne 0\).
Divide by \(y\), which is allowed on an interval where \(y\ne 0\):
Write the separated form and integrate each side:
The antiderivatives are \(\ln|y|\) and \(x^2/2\). Therefore
Check the relation by differentiating with respect to \(x\). On an interval where \(y\ne 0\), the chain rule gives
This recovers the original differential equation. The relation describes nonzero solution branches. The constant function \(y=0\) also satisfies \(y'=xy\), but it was excluded when we divided by \(y\).
Worked Example: A Polynomial in the Dependent Variable
Integrate the separated equation
Integrate term by term on each side:
so
To check, differentiate both sides with respect to \(x\):
Rearranging yields \(y'=(1+x)/(1+y^2)\), which is the differential equation represented by the separated form. The integrated relation gives a family of curves indexed by \(C\); this step does not require solving the relation explicitly for \(y\).
Worked Example: Logarithms on a Restricted Interval
Integrate the separated equation from \(y'=y/x\). Consider an interval on which \(x\ne 0\) and \(y\ne 0\), so division by both is valid:
Integrating gives
The logarithms are defined here because \(x\) and \(y\) are nonzero. Check the relation by differentiating:
Multiplication by \(y\) gives \(y'=y/x\), as required. The relation applies on intervals that do not include \(x=0\), and this derivation covers only branches with \(y\ne 0\). In the original equation, \(y=0\) is also a solution on any interval where \(x\ne 0\); it is not represented by a derivation that divided by \(y\).
Antiderivative Constants: A Common Pitfall
A common mistake is to treat the two constants that appear when integrating separately as two freely varying parameters in the final relation. For example, one might write
and conclude that the family depends independently on both \(C_1\) and \(C_2\). That overcounts the constants: rearrangement shows that only their difference appears.
Another error is to drop a constant altogether by treating two indefinite integrals as automatically equal. From \(\int f(y)\,dy=\int g(x)\,dx\), the correct general relation includes an arbitrary constant after the two antiderivative constants are combined. Omitting it usually removes most of the solution family.
Reading the Result as a Family of Curves
The constant \(C\) labels the different relations produced by integration. For instance, \(H(y)=G(x)+2\) and \(H(y)=G(x)-1\) generally describe different curves. Each curve represents a possible solution wherever it forms a differentiable function \(y\) of \(x\) in the domain of the original equation.
The integrated relation is not automatically a claim that every choice of \(C\) gives a solution on every interval. The original equation may be undefined at some \(x\)- or \(y\)-values, and the separation step may have required \(h(y)\ne 0\). The interval and any excluded values must be consistent with the equation and with the steps used to obtain the relation.
It is also useful to distinguish integrating from choosing one particular solution. Integration gives a family with an arbitrary constant. A condition specifying a particular point on a solution curve can select a value of that constant, but no such selection is made during integration alone.
Identify which factors depend on \(x\) and which depend on \(y\), and note any values that cannot be divided by.
Use \(dy\) on the side involving \(y\) and \(dx\) on the side involving \(x\).
Move the antiderivatives into one relation and represent the net constant by a single \(C\).
Differentiate the relation where appropriate to recover the equation, and check values excluded by division in the original equation.
Why This Step Matters
Integration converts a differential equation, which describes how a solution changes, into a relation that connects the variables along a solution curve. That relation can make the family of possible curves easier to describe and analyze. The constant records which member of the family is under consideration.
Care with constants and restrictions is part of preserving the meaning of the original equation. A relation with no arbitrary constant may describe too little; division by a zero factor may omit a solution; and a relation can only represent solutions on intervals where its assumptions hold. These checks keep the integrated result tied to the differential equation that produced it.
Check Your Understanding
For each question, pay attention to both the integration and the conditions under which the steps are valid.
- If \(\frac{1}{h(y)}\,dy=g(x)\,dx\), what form does the integrated relation take?
- Why do two antiderivative constants reduce to one arbitrary constant?
- In integrating \(y'=xy\) by dividing by \(y\), what condition is needed, and what solution must be checked separately?
- Differentiate \(y+y^3/3=x+x^2/2+C\) to recover the corresponding differential equation.
- Why does an integrated relation with an arbitrary constant describe a family rather than one selected solution?
