DE-BOOK Implicit General Solutions

Section 5.4 Implicit General Solutions

Notice that Step 3 of the SOV method states: “If possible, explicitly solve for \(y\text{.}\) If not, leave the solution in implicit form.” In the examples we’ve seen, it was always possible to explicitly isolate \(y\) after integrating. However, in some cases, this is not possible (or practical), and we call such solutions implicit solutions.

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Subsection Implicit Solutions

The best way to understand why and how you can get implicit solutions using the separation of variables method is to see it in a concrete example. So, let’s solve the following initial-value problem:

\begin{equation*} Q Q' + e^Q Q' = t, \quad Q(2) = 0. \end{equation*}

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Notice the variables separate naturally once we factor out \(Q'\text{:}\)

\begin{equation*} (Q + e^Q)\frac{dQ}{dt} = t\text{.} \end{equation*}

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Now the SOV method proceeds as usual:

\begin{align*} (Q + e^Q)\ dQ \amp = t\ dt \\ \int(Q + e^Q)\ dQ \amp = \int t\ dt \\ \frac{Q^2}{2} + e^Q \amp = \frac{t^2}{2} + c_1\\ Q^2 + 2e^Q \amp = t^2 + 2c_1. \end{align*}

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The general implicit solution is:

\begin{equation*} Q^2 + 2e^Q = t^2 + c.\qquad (c=2c_1) \end{equation*}

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Although we can’t isolate \(Q\) neatly, we can still find \(c\) using \(Q(2) = 0\text{:}\)

\begin{align*} \text{set}\ t=2, Q=0: \quad (0)^2 + 2e^0 \amp = 2^2 + c\\ 2 \amp = 4 + c \quad\Rightarrow\quad c=-2. \end{align*}

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Thus, the particular implicit solution is:

\begin{equation*} Q^2 + 2e^Q = t^2 - 2. \end{equation*}

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Checkpoint 57. 📖❓ Select the Implicit Solutions.

Let \(y\) be a function of the independent variable, \(t\text{,}\) and let \(c\) be an integration constant. If each expression below is a general solution to some DE, click on the implicit solutions.

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\(\large y = t\tan(t) + c\)		\(\large y^2 + \ln y = t + c\)
\(\)
\(\large y = t + \cos(y)\)		\(\large y = \sqrt{\sin(t^2+25)}\)
\(\)

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Subsection Handling Absolute Values in General Solutions

Checkpoint 58. 📖❓ Review the Absolute Value.

Select the solutions to the equation

\begin{equation*} |x| = 14\text{.} \end{equation*}

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\(\pm 14\)
\(-14\)
\(7\)
\(1\)
\(0\)

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When solving differential equations, it’s common to encounter absolute values in the general solution. For instance, suppose you want to solve for \(y\text{,}\) but arrive at:

\begin{equation} |y| = e^{\sin(x) + c}.\tag{5.2} \end{equation}

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To fully isolate \(y\text{,}\) you need to replace the absolute value with “\(\pm\)”, like so:

\begin{equation} y = \pm e^{\sin(x) + c}.\tag{5.3} \end{equation}

To understand why, consider a simplified problem with the same goal:

\begin{equation*} |y| = 5. \end{equation*}

Clearly, both \(5\) and \(-5\) satisfy this equation, so \(\ |y| = 5\ \Rightarrow\ y=\pm 5\text{.}\)

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🌌 Example 59.

Find the general solution to the differential equation

\begin{equation*} x\frac{dy}{dx} + 10x^2y = 6x^2 \end{equation*}

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Solution.

Start by verifying the equation is separable:

\begin{align*} x\frac{dy}{dx} + 10x^2y \amp= 6x^2\\ \frac{dy}{dx} + 10xy \amp= 6x\\ \frac{dy}{dx} \amp= 6x - 10xy = 2x(3 - 5y). \end{align*}

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Now, we apply the separation of variables method.

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Separate variables

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\begin{equation*} \frac{1}{3 - 5y}dy = 2x\ dx \end{equation*}

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u-sub \(\quad\left. \begin{array}{c} u = 3-5y\\ du = -5dy \end{array} \right| \rightarrow\)

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\begin{equation*} \int \frac{1}{3 - 5y}dy = \int 2x\ dx \end{equation*}

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Isolate up to \(|3 - 5y|\)

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\begin{align*} \ln|3 - 5y| \amp = x^2 + c_1 \\ |3 - 5y| \amp = e^{x^2 + c_1} \end{align*}

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At this point, we replace the absolute value with \(\pm\text{:}\)

\begin{equation*} 3 - 5y = \pm e^{x^2 + c_1}. \end{equation*}

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With \(y\) free, we can fully isolate it and combine constants:

\begin{align*} -5y \amp = \pm e^{c_1} e^{x^2} - 3\\ y \amp = \frac{\pm e^{c_1}}{-5} e^{x^2} + \frac{3}{5}. \end{align*}

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Setting the combined constant, \(c=\dfrac{\pm e^{c_1}}{5}\text{,}\) we get the general solution:

\begin{equation*} y = ce^{x^2} + \frac{3}{5}\text{.} \end{equation*}

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Subsection IVPs with “\(\pm\)” in the General Solution

For some initial-value problems, the particular solution can be expressed explicitly even when the general solution is implicit. To see this, consider the following example:

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🌌 Example 60.

Solve the initial-value problem

\begin{equation*} \frac{dy}{dx} = \frac{\cos x}{y}, \quad y(0) = -5\text{.} \end{equation*}

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Solution.

Applying the SOV method

\begin{align*} \frac{dy}{dx} \amp = (\cos x)\cdot\frac{1}{y} \\ y\ dy \amp = \cos(x)\ dx \\ \int y\ dy \amp = \int \cos(x)\ dx \\ \frac{y^2}{2} \amp = \sin(x) + c_1 \end{align*}

we arrive at the implicit general solution:

\begin{equation*} y^2 = 2\sin(x) + c \end{equation*}

where \(c=2c_1\text{.}\)

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Explicitly solving for \(y\text{,}\) shows our general solution has two branches:

\begin{equation*} y(x) = \pm\sqrt{2\sin(x) + c} = \left\{ \begin{array}{c} -\sqrt{2\sin(x) + c} \\ \sqrt{2\sin(x) + c} \end{array} \right. \end{equation*}

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Since our solution must satisfy \(y(0)=-5\text{,}\) which has a negative \(y\)-value, our particular solution must come from the negative branch:

\begin{equation*} y(x) = -\sqrt{2\sin x + c}. \end{equation*}

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Finally, we find \(c\) with the initial condition \(y(0) = -5\text{:}\)

\begin{equation*} -5 = -\sqrt{2\sin(0) + c} \ \Rightarrow\ c = 25. \end{equation*}

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Thus, the explicit particular solution is:

\begin{equation*} y = -\sqrt{2\sin x + 25}. \end{equation*}

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The separation of variables method is a flexible and powerful tool for solving a wide range of differential equations, including nonlinear ones. As you’ve seen, the method is systematic: separate, integrate, and isolate. While the mechanics of the method are straightforward, the complexity often lies in the algebraic manipulation and integration.

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Checkpoint 61. 📖❓ Reorder the Solution.

Reorder the scrambled steps used to solve

\begin{equation*} \frac{dq}{dt} - 3qt^2 = 2q \end{equation*}

using the Separation of Variables method.

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Equation is 1st-order ✅
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\(\dfrac{dq}{dt} = q(2 + 3t^2)\)
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\(\dfrac{1}{q}\ dq = (2 + 3t^2)\ dt\)
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\(\ds\int \frac{1}{q} dq = \int (2 + 3t^2) dt\)
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\(\ln|q| = 2t^2 + t^3 + c\)
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\(|q| = e\vphantom{|}^{\large 2t^2 + t^3 + c}\)
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\(q = \pm e^c e\vphantom{|}^{\large 2t^2 + t^3}\)
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\(q = c e\vphantom{|}^{\large 2t^2 + t^3}\)

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