Finding General Solutions Using Separation of Variables

AP Course

AP Calculus AB/BC

Study the complete College Board sequence for AP Calculus AB and BC, from limits through infinite series.

Choose an official unit and topic to open its lecture, concept check, or focused practice.

AP Calculus AB/BC · Unit 7 · Topic 7.6

Finding General Solutions Using Separation of Variables

Separate x- and y-factors, integrate both sides, and include a constant.

1. Topic Focus

Model rates with differential equations, read slope fields, approximate solutions, solve separable equations, and interpret exponential or logistic models.

This topic: Separate x- and y-factors, integrate both sides, and include a constant.

2. Key Relationship

\(\frac{1}{g(y)}dy=f(x)dx\)

Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.

3. Visual Connection

Separation workflowPreserve equilibrium solutions, integrate both variables, and let the initial value determine the constant and branch.

4. Worked Example

For y′=xy, integrate dy/y=x dx to obtain y=Ce^(x²/2).

Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.

5. Concept Development

A first-order differential equation is separable when algebra can express it as a product of a function of \(x\) and a function of \(y\):

\(\frac{dy}{dx}=f(x)g(y).\)

On a region where \(g(y)\ne0\), move the \(y\)-expression with \(dy\) and the \(x\)-expression with \(dx\), then antidifferentiate:

\(\frac{1}{g(y)}\,dy=f(x)\,dx\quad\Longrightarrow\quad \int\frac{1}{g(y)}\,dy=\int f(x)\,dx.\)

The resulting one-parameter family is a general solution. Topic 7.6 stops before an initial condition selects one particular member; that step is developed in Topic 7.7.

Recognizing a separable equation

Equation	Decision	Useful rewrite
\(y'=x(y+1)\)	separable	\(dy/(y+1)=x\,dx\)
\(y'=6x^2+4x\)	separable	take \(g(y)=1\)
\(y'=xy+3x-2y-6\)	separable after factoring	\(y'=(x-2)(y+3)\)
\(y'=x+y\)	not separable by ordinary factoring	the sum cannot become \(f(x)g(y)\)

Always simplify and factor before deciding. An autonomous equation \(y'=g(y)\) is separable with \(f(x)=1\), and an equation \(y'=f(x)\) is separable with \(g(y)=1\).

The complete separation workflow

Factor and classify. Put the equation in the form \(y'=f(x)g(y)\).
Preserve equilibrium solutions. Solve \(g(y)=0\) before dividing by \(g(y)\).
Separate variables. Write all \(y\)-factors with \(dy\) and all \(x\)-factors with \(dx\).
Integrate both sides. Use one arbitrary constant, usually on the right.
Simplify the constant. Absorb nonzero multipliers and signs into a renamed arbitrary constant.
Solve for \(y\) when practical. An implicit relation is acceptable when isolation is difficult or creates unnecessary branches.
Restore omitted solutions and state restrictions. Include equilibria lost during division and note singularities or branch intervals.
Verify. Differentiate the final family and substitute it into the original equation.

Why equilibrium solutions can disappear

Dividing by \(g(y)\) assumes \(g(y)\ne0\). If \(g(c)=0\), then the constant function \(y=c\) may solve the original equation but is excluded from the separated algebra. Record these values first. Sometimes the explicit family later includes an equilibrium through \(C=0\); sometimes it does not, so each equilibrium must be listed separately.

Managing logarithms and constants

Keep absolute values when integrating \(1/u\):

\(\int\frac{du}{u}=\ln|u|+C.\)

If \(\ln|u|=H(x)+C\), exponentiation gives \(|u|=e^Ce^{H(x)}\). The positive factor \(e^C\) and the possible sign combine into one nonzero constant \(K\), so \(u=Ke^{H(x)}\). If \(K=0\) also produces a valid solution of the original equation, it can be admitted afterward.

There is no need to write \(C_1\) on one side and \(C_2\) on the other: their difference is another arbitrary constant.

Explicit and implicit general solutions

An explicit family has the form \(y=G(x,C)\). An implicit family such as \(y^2-x^2=C\) is also a valid general solution when differentiation recovers the differential equation. Solving an implicit relation may create separate branches, and each branch must be restricted to intervals where it is differentiable and the original equation is defined.

Domain and interval checks

Separation can introduce logarithms, roots, inverse trigonometric functions, or denominators. Final solutions are valid only on intervals that avoid forbidden inputs and singularities. For example, a family containing \(1/(C+\cos x)\) must be restricted to intervals where its denominator is nonzero.

AP writing habit: Show the separated differential form and the antiderivatives before presenting the general solution. List lost equilibria explicitly rather than relying on the reader to infer them.

6. Detailed Worked Example and Error Check

Example 1: Exponential-type family

Solve \(y'=2xy\). First, \(y=0\) is an equilibrium. For \(y\ne0\),

\(\frac{dy}{y}=2x\,dx\quad\Longrightarrow\quad\ln|y|=x^2+C.\)

Exponentiating and absorbing the sign gives

\(y=Ce^{x^2}.\)

Allowing \(C=0\) includes the equilibrium, so this one family gives all solutions on real intervals.

Example 2: Factor before separating

Consider \(y'=xy+3x-2y-6\). Factor by grouping:

\(xy+3x-2y-6=(x-2)(y+3).\)

The equilibrium is \(y=-3\). For other solutions,

\(\frac{dy}{y+3}=(x-2)\,dx\Rightarrow\ln|y+3|=\frac{x^2}{2}-2x+C.\)

Thus

\(y=-3+Ce^{x^2/2-2x}.\)

The choice \(C=0\) restores the equilibrium.

Example 3: An implicit family with branches

For \(y'=x/y\), the differential equation is undefined when \(y=0\). Separate without dividing by a possible solution:

\(y\,dy=x\,dx\Rightarrow\frac{y^2}{2}=\frac{x^2}{2}+C\Rightarrow y^2-x^2=C.\)

This implicit family may be written \(y=\pm\sqrt{x^2+C}\) on intervals where the chosen branch is real, nonzero, and differentiable. The two signs are different branches, not one function switching signs freely.

Example 4: Inverse tangent appears

Solve \(y'=x(1+y^2)\). Since \(1+y^2\) never vanishes, there are no equilibrium solutions.

\(\frac{dy}{1+y^2}=x\,dx\Rightarrow\arctan y=\frac{x^2}{2}+C.\)

Therefore

\(y=\tan\left(\frac{x^2}{2}+C\right),\)

on intervals that avoid the vertical asymptotes of the tangent expression.

Example 5: Exponential expressions on both variables

For \(y'=e^{x-y}=e^xe^{-y}\), multiply by \(e^y\):

\(e^y\,dy=e^x\,dx\Rightarrow e^y=e^x+C.\)

An explicit form is

\(y=\ln(e^x+C),\)

valid on any interval where \(e^x+C>0\).

Example 6: A logistic-type equation

For \(y'=y(1-y)\), the equilibria \(y=0\) and \(y=1\) must be recorded first. For \(y\ne0,1\), partial fractions give

\(\int\frac{dy}{y(1-y)}=\int dx\Rightarrow\ln|y|-\ln|1-y|=x+C.\)

After exponentiating and solving,

\(y=\frac{Ce^x}{1+Ce^x}.\)

The choice \(C=0\) includes \(y=0\), but no finite \(C\) produces \(y=1\), so \(y=1\) must be listed separately. Also exclude inputs where \(1+Ce^x=0\).

Example 7: Leaving the answer implicit

For \(\sin y\,y'=x\), write

\(\sin y\,dy=x\,dx\Rightarrow-\cos y=\frac{x^2}{2}+C.\)

This implicit relation is a complete general-solution form. Solving with an inverse cosine would require multiple branches and additional interval restrictions without improving the answer.

Example 8: A rational power

Solve \(y'=y^2\sin x\). The equilibrium \(y=0\) is set aside. For \(y\ne0\),

\(y^{-2}\,dy=\sin x\,dx\Rightarrow-\frac1y=-\cos x+C.\)

Renaming the constant yields

\(y=\frac{1}{C+\cos x},\)

together with \(y=0\). Each nonzero solution is restricted to intervals where \(C+\cos x\ne0\).

Common separation errors

Declaring an equation separable before simplifying or factoring it.
Dividing by a \(y\)-factor without recording its zeros as possible equilibria.
Writing \(dy/g(y)=f(x)\) and omitting \(dx\).
Forgetting the absolute value in a logarithmic antiderivative.
Keeping two independent integration constants instead of combining them.
Forcing an implicit relation into an invalid single explicit branch.
Ignoring denominator, radical, logarithm, or tangent restrictions.

7. AP Reasoning Routine

Translate the context into a rate equation, verify candidate solutions by substitution, carry constants through integration, and apply initial conditions last.

Identify the representation and requested quantity.
State the rule or theorem and verify its conditions.
Keep exact values until the final requested approximation.
Interpret sign, units, interval, and context.

AP Checkpoint

Find each general solution. Include equilibrium solutions and interval restrictions when needed.
(a) Decide whether \(y'=x(y-1)\) and \(y'=x+y\) are separable.
(b) \(y'=3xy\).
(c) \(y'=(x+1)(y-2)\).
(d) \(y'=x/y\).
(e) \(y'=x(1+y^2)\).
(f) \(y'=e^{x-y}\).
(g) \(y'=2x/(y+1)\).
(h) \(y'=y/x\).
(i) \(y'=y^2\sin x\).
(j) Factor and solve \(y'=xy+2x-y-2\).

Check the solution

(a) \(y'=x(y-1)\) is separable: \(dy/(y-1)=x\,dx\). The equation \(y'=x+y\) is not separable by writing its right side as \(f(x)g(y)\).
(b) Record \(y=0\). For \(y\ne0\), \(dy/y=3x\,dx\), so \(y=Ce^{3x^2/2}\). Allowing \(C=0\) includes the equilibrium.
(c) The equilibrium is \(y=2\). Integration gives \(\ln|y-2|=x^2/2+x+C\), hence \(y=2+Ce^{x^2/2+x}\); \(C=0\) includes the equilibrium.
(d) \(y\,dy=x\,dx\), so \(y^2-x^2=C\). Explicit branches are \(y=\pm\sqrt{x^2+C}\) on intervals where they are real, nonzero, and differentiable.
(e) \(dy/(1+y^2)=x\,dx\), giving \(\arctan y=x^2/2+C\), so \(y=\tan(x^2/2+C)\) between its vertical asymptotes.
(f) \(e^y dy=e^x dx\), so \(e^y=e^x+C\) and \(y=\ln(e^x+C)\), where \(e^x+C>0\).
(g) \((y+1)dy=2x\,dx\), so \((y+1)^2=2x^2+C\). Branches must avoid \(y=-1\), where the original equation is undefined.
(h) On an interval not containing \(x=0\), \(dy/y=dx/x\), giving \(\ln|y|=\ln|x|+C\), hence \(y=Cx\). The value \(C=0\) includes the equilibrium \(y=0\).
(i) The equilibrium is \(y=0\). For nonzero solutions, \(y^{-2}dy=\sin x\,dx\), so \(y=1/(C+\cos x)\) on intervals where the denominator is nonzero.
(j) \(xy+2x-y-2=(x-1)(y+2)\). Thus \(dy/(y+2)=(x-1)dx\), so \(y=-2+Ce^{x^2/2-x}\). The choice \(C=0\) includes \(y=-2\).