From Separation to a Family of Curves
After variables have been separated, the differential equation has one group of terms involving \(y\) and \(dy\), and another involving \(x\) and \(dx\). The next step is to integrate each side. The result is usually not one curve but a family of curves, indexed by an arbitrary constant.
For example, if separation gives \(A(y)\,dy=B(x)\,dx\), integrating both sides produces an equation relating \(y\) and \(x\). Sometimes that equation can be rearranged to write \(y\) explicitly as a function of \(x\). Sometimes keeping it implicit is simpler and fully sufficient.
The General Integration Pattern
Suppose a differential equation is separated on a domain where the division used in separation is valid, giving
Let \(H\) be an antiderivative of \(A\), so \(H'(y)=A(y)\), and let \(G\) be an antiderivative of \(B\), so \(G'(x)=B(x)\). Integrating the two sides gives
The two constants can be combined into one: set \(C=C_2-C_1\). The family can then be written as
There is no requirement to isolate \(y\). The equation \(H(y)=G(x)+C\) is an implicit description of the solution family. If it is convenient to solve for \(y\), that can be done; otherwise, the relation itself can be checked and used.
Why the Integrated Relation Works
The reason integration produces a solution relation is the chain rule. Suppose separation gives \(A(y)\,dy=B(x)\,dx\), or, for a differentiable function \(y=y(x)\),
For antiderivatives \(H'=A\) and \(G'=B\), the chain rule gives
Therefore the derivative of \(H(y(x))-G(x)\) is zero. On any connected interval on which the solution is defined, a differentiable function with zero derivative is constant. Hence
which is the integrated relation. Conversely, differentiating that relation gives \(A(y)y'=B(x)\). This checks the differential equation wherever the relation is differentiable and the original separation is valid.
Worked Examples
Worked Example: An Implicit Cubic Relation
Integrate the separated equation
An antiderivative of \(1+y^2\) with respect to \(y\) is \(y+\frac{y^3}{3}\), since
An antiderivative of \(1+x^2\) with respect to \(x\) is \(x+\frac{x^3}{3}\), since
Including the arbitrary constant gives the implicit family
There is no need to solve this cubic equation for \(y\). To check the result, differentiate both sides with respect to \(x\):
Since \(1+y^2\) is positive for every real \(y\), divide by it to obtain
which is the differential equation corresponding to the separated equation. The implicit relation is therefore a valid way to describe the family.
Worked Example: A Logarithm in an Implicit Solution
Consider the separated equation
The restriction \(y>0\) makes \(1/y\) defined and allows the antiderivative of \(1/y\) to be written as \(\ln y\). Integrate each side:
The left antiderivative is \(y+\ln y\), because its derivative is \(1+1/y\). The right antiderivative is \(x\). Including the arbitrary constant gives
This relation is a useful implicit form; isolating \(y\) is not necessary. Check it by differentiating:
Because \(y>0\), the factor \(1+1/y\) is nonzero. Solving for \(y'\) gives
Thus the implicit relation differentiates to the differential equation \(y'=y/(1+y)\) on the stated domain.
Worked Example: Combining Two Antiderivative Constants
Suppose separation gives
Integrate the left side with respect to \(y\) and the right side with respect to \(x\):
Subtract \(C_1\) from both sides and define \(C=C_2-C_1\). Then
To check the relation, differentiate on an interval where \(x\ne 0\):
Since the separated equation is \(y\,dy=(1/x)\,dx\), this is exactly its derivative form. In particular, the check does not require choosing a sign and writing \(y\) as a square root. The integrated relation is the solution form, on intervals where the original equation and separation are valid.
Constants and Domains
The constant \(C\) is arbitrary: different values generally describe different solution curves. Do not replace it with a particular number unless there is information that determines that number. Finding a value from an initial condition is a separate step.
The domain also matters. Separation may require dividing by a factor involving \(x\) or \(y\). If that factor is zero, the separated equation may not represent the original differential equation there. An antiderivative can introduce further domain restrictions: for example, \(1/x\) is defined only for \(x\ne 0\), and \(\ln|x|\) describes antiderivatives separately on the intervals \(x<0\) and \(x>0\). Keep the domain conditions from the separation step when interpreting the integrated relation.
A useful check is to differentiate the proposed relation and compare the result with the differential equation before separation. If a factor was divided out, also remember that the check applies only where that factor is nonzero. Any cases excluded by the division must be considered against the original equation, not silently added to or removed from the integrated family.
Common Pitfall: Dropping the Constant or Misusing the Logarithm
A frequent error is to integrate correctly but omit the arbitrary constant. For example, from \(y\,dy=(1/x)\,dx\), writing only
leaves out the family of possible vertical shifts in the integrated relation. Differentiating this equation gives \(yy'=1/x\), but it represents only the particular case \(C=0\), not the general family. The general integrated relation is
Another error is to write \(\int (1/x)\,dx=\ln x\) without accounting for negative \(x\). The real-valued antiderivative on either interval that avoids zero is \(\ln|x|\), since
The absolute value is not decoration: it records that the antiderivative is valid on both sides of zero, while the equation still has separate domains on \(x<0\) and \(x>0\).
What an Implicit Solution Tells You
An implicit relation can be a complete and convenient answer. It records how \(x\) and \(y\) are related without requiring a formula with \(y\) alone on one side. To verify such an answer, differentiate it using the chain rule and check that the result matches the original differential equation on the relevant domain.
Implicit form is particularly useful when rearranging the relation would require unnecessary algebra, or when the resulting expression cannot be written using the elementary functions being used. The cubic relation in the first example illustrates the practical point: its verification is direct, even though explicitly solving a cubic for \(y\) would add work without improving the description of the family.
Check Your Understanding
For each question, pay attention to the antiderivatives, the arbitrary constant, and the domain.
- Integrate \(y^2\,dy=2x\,dx\) and write the result with one arbitrary constant.
- Why can the two constants in \(H(y)+C_1=G(x)+C_2\) be replaced by one constant?
- Differentiate the relation \(y+\frac{y^3}{3}=x+C\) and state the resulting equation for \(y'\).
- What domain restriction must be kept when integrating \(\frac{1}{x}\,dx\)?
- Why can an implicit relation be a sufficient description of a solution family even when \(y\) has not been isolated?
