Recover an explicit family when possible
From a logarithm to a formula
The slope field shows \(\frac{dy}{dx} = -2xy\). Integrating the separated equation gave \(\ln|y| = -x^2 + C\). A formula that starts with \(y =\) is an explicit solution, and this skill finds it.
Raise \(e\) to the power of both sides. This removes the logarithm and leaves the absolute value, so \(|y| = e^{C}e^{-x^2}\).
The absolute value allows two signs: \(y = e^{C}e^{-x^2}\) above the x-axis, or \(y = -e^{C}e^{-x^2}\) below it. One symbol, \(A = \pm e^{C}\), covers both.
Move the slider for \(A\). A positive \(A\) gives a curve above the x-axis, and a negative \(A\) gives its mirror image below it. At \(x = 0\) the curve has height \(A\), because \(e^{0} = 1\).
Undo the logarithm, keep the sign
An integrated equation relates \(y\) to \(x\). To solve for \(y\), you undo what was done to \(y\), in the reverse order. If the equation has \(\ln|y|\), you undo it with \(e\) to the power of both sides, which gives \(|y|\) and not \(y\).
Here the equation is \(\frac{dy}{dx} = g(x)\,y\) for a different function \(g\). Drag the orange point. The curve through it does not cross the x-axis, so the sign of \(y\) stays the same along it.
Press the button. The mirror image of the chalk curve in the x-axis appears, faint. At every \(x\) the two heights have the same size and opposite signs, so one constant \(A\), positive or negative, describes both.
Notation
Integrating \(\frac{dy}{y} = g(x)\,dx\) gives an equation with a logarithm of \(y\), where \(G\) is the antiderivative of \(g\) with \(G(0) = 0\).
\[\ln|y| = G(x) + C\]Raise \(e\) to the power of both sides. The logarithm disappears, because \(e^{\ln|y|} = |y|\), and \(e^{a + b} = e^a e^b\) splits the right side.
\[|y| = e^{C}\,e^{G(x)}\]The absolute value allows two signs, so write both.
\[y = \pm e^{C}\,e^{G(x)}\]Give the constant one name, \(\cy{A} = \pm e^{C}\). Because \(e^{C}\) is positive, \(A\) can be any number except 0. At \(x = 0\) the curve has height \(A\), because \(e^{G(0)} = 1\).
\[y = \cy{A}\,e^{G(x)}\]The start point fixes \(A\), and the sign of \(A\) matches the sign of \(y\) there.
When it fails
Some integrated equations cannot be solved for \(y\). Press the button. The field changes to \(\frac{dy}{dx} = \frac{1}{1 + e^{y}}\), which separates as \((1 + e^{y})\,dy = dx\) and integrates to \(y + e^{y} = x + C\). No sequence of familiar steps isolates \(y\).
The equation still gives a solution curve for each \(C\), such as the chalk curve. Leave the answer as the equation itself, an implicit solution.
Example 1: a shifted exponential
Integrating a separated equation gave \(\ln|y - 2| = 3x + C\). The curve passes through \((0, 5)\). Solve for \(y\).
Raise \(e\) to the power of both sides, so the logarithm disappears. The two faint curves show the two signs of \(y - 2\), one above the dashed line \(y = 2\) and one below it.
\[|y - 2| = e^{C}\,e^{3x}\]Write \(\pm e^{C}\) as one constant \(A\). Then add 2 to both sides.
\[y - 2 = A\,e^{3x}, \qquad y = 2 + A\,e^{3x}\]The curve passes through \((0, 5)\), so substitute \(x = 0\) and \(y = 5\) to find \(A\). The point lies above the dashed line \(y = 2\), so \(A\) is positive.
\[5 = 2 + A\,e^{0} = 2 + A, \qquad A = 3\]The explicit solution is the formula with \(A = 3\).
\[y = 2 + 3e^{3x}\]
Example 2: choosing the branch
Integrating \(\dfrac{dy}{dx} = -\dfrac{x}{4y}\) gave \(4y^2 + x^2 = K\). The curve passes through \((0, -1)\). Solve for \(y\).
The curve passes through \((0, -1)\), so substitute to find \(K\). Then solve for \(y^2\). The faint curves are the two halves of the ellipse \(4y^2 + x^2 = 4\), called branches. Each half is a separate solution, because a solution is a function of \(x\).
\[\begin{aligned} K &= 4(-1)^2 + 0^2 = 4 \\ y^2 &= \dfrac{4 - x^2}{4} \end{aligned}\]A tempting wrong move: Take the square root of both sides.
\[y = \dfrac{\sqrt{4 - x^2}}{2}\]The square root sign gives only the positive value. At \(x = 0\) this formula gives \(y = 1\), but the curve passes through \((0, -1)\). The positive root is the upper half of the ellipse, shown in red.
The equation \(y^2 = \frac{4 - x^2}{4}\) has two solutions, with opposite signs. The start point lies below the x-axis, so choose the negative one.
\[y = -\dfrac{\sqrt{4 - x^2}}{2}\]The root needs \(4 - x^2 \ge 0\), so \(-2 \le x \le 2\). A later skill treats the ends of this interval.
Practice
Integrating \(\dfrac{dy}{dx} = \dfrac{2y}{x + 1}\) gave the equation below. The curve passes through \((0, 5)\). Solve for \(y\). \[\begin{aligned} \ln|y| &= 2\ln|x + 1| \\ &\quad + C \end{aligned}\]
Write \(2\ln|x + 1|\) as one logarithm. Move the slider for \(A\) until the curve passes through \((0, 5)\).
Solution
By the power rule for logarithms, \(2\ln|x + 1| = \ln(x + 1)^2\). Raise \(e\) to the power of both sides.
\[|y| = e^{C}\,(x + 1)^2\]Write \(\pm e^{C}\) as one constant \(A\), so that \(y = A(x + 1)^2\). The curve passes through \((0, 5)\), so substitute to find \(A\).
\[5 = A(0 + 1)^2, \qquad A = 5\]With \(A = 5\) the chalk curve passes through the orange point, so this is the explicit solution.
\[y = 5(x + 1)^2\]
Questions
The graph shows the slope field of \(\dfrac{dy}{dx} = -\dfrac{y}{2}\) and the curves \(y = A\,e^{-x/2}\) for several values of \(A\). The orange curve passes through \((0, -2)\).
Which formula describes the orange curve through \((0, -2)\)?
- \(y = -2e^{-x/2}\)
- \(y = 2e^{-x/2}\)
- \(y = -2e^{x/2}\)
Answer
\(y = -2e^{-x/2}\)
One formula, \(y = A\,e^{-x/2}\), describes the curves through \((0, 3)\) and \((0, -3)\). How do the two curves relate?
- They are mirror images across the x-axis, with \(A = 3\) and \(A = -3\).
- They are the same curve.
- One is the other moved down by 6.
Answer
They are mirror images across the x-axis, with \(A = 3\) and \(A = -3\).
Integrating a separated equation gave \(\ln|y + 1| = 2x + C\). The curve passes through \((0, -4)\). Which formula is correct?
- \(y = -1 - 3e^{2x}\)
- \(y = -1 + 3e^{2x}\)
- \(y = -4e^{2x}\)
- \(y = -1 - 4e^{2x}\)
Answer
\(y = -1 - 3e^{2x}\)
Summary
You raise e to the power of both sides, write both signs, and give the constant one name.
This works because the absolute value inside a logarithm allows two signs, and the start point shows which one applies.
You use it when an integrated equation has a logarithm of y or a square of y and you want a formula for y.
This is one of five skills in topic 7.6 of Which & Why: AP Calculus.