Mistake Master

Finding General Solutions Using Separation of Variables AB & BC

An equation separates when its right side factors as a function of $x$ times a function of $y$: $xy$, $y\cos x$ and $\frac{x + 1}{y^{2}}$ all do, while $x + y$ and $\sin(x + y)$ do not, and $xy + x$ does because it factors as $x(y + 1)$. Separating means every $y$ on the side with $dy$ and every $x$ on the side with $dx$; a stray $y$ left on the $dx$ side means integrating is not yet legal.

Integrating both sides introduces exactly ONE constant, and where it ends up is the whole difficulty. From $\frac{dy}{dx} = xy$ comes $\ln|y| = \frac{x^{2}}{2} + C$, and exponentiating turns that into $|y| = e^{C}e^{x^{2}/2}$, so the constant becomes a FACTOR: $y = Ce^{x^{2}/2}$, not $e^{x^{2}/2} + C$. Those two families are stretches and shifts of the same curve and are not interchangeable. Other separable forms leave the constant elsewhere, as in $y^{2} = 2x^{2} + C$ or $y = -\frac{1}{x + C}$, and an implicit answer is a complete general solution.

SEPARATING dy/dx = xy, ONE STEP AT A TIME 1. dy/y = x dx EVERY y ON THE LEFT 2. ∫ dy/y = ∫ x dx INTEGRATE BOTH SIDES 3. ln|y| = x²/2 + C ONE CONSTANT, HERE 4. |y| = e^C · e^(x²/2) e^C BECOMES A FACTOR 5. y = Ce^(x²/2) THE PLUS OR MINUS IS ABSORBED THE CONSTANT ENTERS AT STEP 3 AND NOWHERE ELSE. ADDING IT AFTER STEP 4 GIVES y = e^(x²/2) + C, A DIFFERENT FAMILY. ONE C IS ENOUGH: TWO WOULD COMBINE INTO ONE ANYWAY. CHECK BY DIFFERENTIATING: y′ = Cx e^(x²/2) = xy.
Five lines, and the only one that can go wrong twice is the third. Put the constant in earlier and it has nothing to be the constant of; put it in after step 4 and it shifts the curve instead of stretching it.
THE TEST: DOES THE RIGHT SIDE FACTOR INTO x STUFF TIMES y STUFF? dy/dx = xy SEPARABLE ALREADY A PRODUCT dy/dx = y cos x SEPARABLE A PRODUCT TOO dy/dx = (x + 1)/y² SEPARABLE DIVIDING IS A PRODUCT dy/dx = xy + x SEPARABLE IT FACTORS: x(y + 1) dy/dx = x + y NOT SEPARABLE A SUM THAT RESISTS dy/dx = sin(x + y) NOT SEPARABLE THE SUM IS INSIDE TRY THE FACTORING BEFORE DECIDING A SUM IS HOPELESS. ROW FOUR IS A SUM. THE TWO THAT FAIL STILL HAVE SOLUTIONS, FOUND BY OTHER MEANS. THEY APPEAR ON EXAMS AS SLOPE FIELD AND EULER QUESTIONS INSTEAD.
Six equations and one question asked of each. The fourth row is the one that catches people: it is a sum, and it factors anyway, so the test is factoring rather than the shape of the expression.

The work

3 ways in · any order
Lesson
Finding General Solutions Using Separation of Variables

Tests whether an equation separates by trying to factor its right side, then runs one example from separation through integration to the solved form with the constant tracked at every line, showing why it becomes a multiplicative factor after exponentiating and covering the power, reciprocal and logarithmic endings.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items on separation of variables: recognising which equations separate at all, completing the separation before integrating, and placing the single constant of integration where the algebra puts it rather than adding it to the finished answer.

Not yet available · 10 items
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.

Take the diagnostic to identify your misconceptions