Integrate both sides
Two regions of equal area
The equation \(\frac{dy}{dx} = \frac{3x^2}{2y}\) separates into \(2y\,dy = 3x^2\,dx\). Below the field is the graph of \(3x^2\). To the left, turned on its side, is the graph of \(2y\).
The solution starts at \((0, 1)\). Drag the orange point along it. The lower teal region lies under \(3x^2\), from 0 to the point's x-value. The left teal region lies under \(2y\), from 1 to the point's y-value.
At \((2, 3)\), both regions have area 8. The integral of \(3x^2\) from 0 to 2 is 8, and the left trapezoid has parallel sides 2 and 6 and width 2.
Along the solution, the two areas stay equal. Integrating both sides of the separated equation turns this equality into a formula.
Equal areas, one constant
The separated equation is now \(\frac{1}{h(y)}\,dy = g(x)\,dx\), so the lower graph is \(g(x)\) and the left graph is \(\frac{1}{h(y)}\). The area under each graph, from the start to the point, is the integral of that side.
Show the thin strip. Then drag the point along the solution. Each small move adds the strip to each region, and the two strips have equal area, so the two areas stay equal.
Move the start. Each start gives its own solution, so the formula needs one constant, \(C\), to tell them apart.
Notation
Integrate the left side with respect to \(y\). Integrate the right side with respect to \(x\). The teal regions, from the start to the point, show these two integrals.
\[\int \frac{1}{h(y)}\,dy = \int g(x)\,dx\]Let \(\cb{H}\) be an antiderivative of \(\frac{1}{h(y)}\) and \(\cb{G}\) an antiderivative of \(g(x)\). The two constants of integration combine into one constant, \(C\).
\[\cb{H(y)} = \cb{G(x)} + \cq{C}\]This is an implicit solution, an equation in \(x\) and \(y\) that defines the solutions without solving for \(y\). In this scene the left graph is \(2y\) and the lower graph is \(3x^2\), so the equation is \(y^2 = x^3 + C\). Each value of \(C\) gives one curve, and each curve comes from a different start.
An integral that gives a logarithm needs an absolute value, as in \(\int \frac{1}{y}\,dy = \ln|y|\).
When it fails
Without \(C\), the equation is \(y^2 = x^3\), the one red curve. It passes through \((0, 0)\) and misses the start \((0, 1)\).
The faint curves are gone too: without a constant there is one curve, not a family that reaches every start.
Example 1: powers on both sides
A curve satisfies \(\dfrac{dy}{dx} = \dfrac{4x^3}{3y^2}\) and passes through \((1, 2)\). Find an implicit solution.
To separate, you multiply both sides by \(3y^2\) and by \(dx\).
\[3y^2\,dy = 4x^3\,dx\]Integrate each side with respect to its own variable: \(3y^2\) gives \(y^3\) and \(4x^3\) gives \(x^4\). Write one constant. The two teal regions, from \(x = 1\) and from \(y = 2\), have equal area.
\[y^3 = x^4 + C\]The curve passes through \((1, 2)\), so substitute \(x = 1\) and \(y = 2\) to find \(C\).
\[2^3 = 1^4 + C, \qquad C = 7\]With \(C = 7\), this is the implicit solution, and its curve passes through \((1, 2)\).
\[y^3 = x^4 + 7\]
Example 2: a logarithm
A curve satisfies \(\dfrac{dy}{dx} = 3x^2 y\) and passes through \((0, -2)\). Find an implicit solution.
To separate, you divide both sides by \(y\). You cannot divide by 0, so this step needs \(y \ne 0\). Then you multiply both sides by \(dx\).
\[\frac{1}{y}\,dy = 3x^2\,dx\]A tempting wrong move: Integrate the left side and write \(\ln y\).
\[\ln y = x^3 + C\]The solution lies below the x-axis, and the teal region on the left runs from \(y = -2\) downward. The logarithm \(\ln y\) is not defined for negative \(y\), so this equation fails everywhere on the solution.
The integral of \(\frac{1}{y}\) is \(\ln|y|\), which is defined for negative \(y\) too.
\[\ln|y| = x^3 + C\]The curve passes through \((0, -2)\), so substitute \(x = 0\) and \(y = -2\) to find \(C\). The implicit solution is \(\ln|y| = x^3 + \ln 2\).
\[\ln|-2| = 0 + C, \qquad C = \ln 2\]
Practice
A curve satisfies \(\dfrac{dy}{dx} = \dfrac{e^x}{2y}\) and passes through \((0, 3)\). Find an implicit solution.
Multiply both sides by \(2y\) and by \(dx\), so each side has one variable. The start \((0, 3)\) fixes the constant.
Solution
To separate, you multiply both sides by \(2y\) and by \(dx\).
\[2y\,dy = e^x\,dx\]Integrate each side with respect to its own variable: \(2y\) gives \(y^2\) and \(e^x\) gives \(e^x\). Write one constant.
\[y^2 = e^x + C\]The curve passes through \((0, 3)\), so substitute \(x = 0\) and \(y = 3\). Because \(e^0 = 1\), \(C = 8\).
\[9 = 1 + C, \qquad y^2 = e^x + 8\]
Questions
The graph shows the slope field of \(\dfrac{dy}{dx} = \dfrac{x}{y}\), which separates as \(y\,dy = x\,dx\). The solution through \((0, 3)\) is marked. Below the field is the graph of \(x\), and to the left, turned on its side, the graph of \(y\).
The teal region under the graph of \(y\), from \(y = 3\) to \(y = 5\), is a trapezoid. Its parallel sides are 3 and 5, and its width is 2. What is its area?
- 8
- 16
- 10
Answer
8
Along the solution the two regions have equal area. The teal region under the graph of \(x\) is a triangle with base \(X\) and height \(X\). For which \(X\) is its area 8?
- 4
- 8
- 2
Answer
4
A separable equation is written as \((2y + 1)\,dy = x\,dx\). Which equation comes from integrating both sides?
- \(y^2 + y = \dfrac{x^2}{2} + C\)
- \(y^2 + y = \dfrac{x^2}{2}\)
- \(y^2 + y = x^2 + C\)
- \(2y^2 + y = \dfrac{x^2}{2} + C\)
Answer
\(y^2 + y = \dfrac{x^2}{2} + C\)
Summary
After you separate the variables, you integrate each side with respect to its own variable and add one constant.
This works because the two teal areas stay equal along a solution.
The start fixes that constant, and here the solution through \((0, 1)\) is \(y^2 = x^3 + 1\).
This is one of five skills in topic 7.6 of Which & Why: AP Calculus.