Skip to main content

Section 3.5 Visualizing Solutions

An effective way to deepen your understanding of solutions is through visualization. Although a family of solutions includes infinitely many curves, plotting only a few helps reveal how the general solution, particular solutions, and initial conditions are related.
Think of a family of solutions like a map full of side-by-side paths. The map shows all possible routes a traveler could take, each one representing a particular solution. The general solution defines the layout of all these paths, and choosing an initial condition is like dropping a pin on the map: the curve that passes through that point is the specific path (or solution) you follow.

Subsection 📈 Interactive: Visualizing Solutions

Consider the differential equation:
\begin{equation*} \frac{dy}{dx} = 2xy - 6x, \end{equation*}
which has the general solution:
\begin{equation*} y = c e^{x^2} + 3. \end{equation*}
Since the constant \(c\) can take any value, there are many possible solutions (green curves in Figure 39). Adjust the initial condition (i.e., point on the \(y\)-axis) to see the specific solutions it corresponds to (blue curve).
A phase-plane display of the solution family y equals 3 plus c times e to the x squared, drawn as green curves; the reader drags an initial-condition point along the y-axis to highlight the matching particular solution, shown in blue, of dy dx equals 2xy minus 6x.
Figure 39. Particular solutions with different initial conditions.

Checkpoint 40. 🤔💭 Interactive Follow-up Questions.

Adjust the initial condition in the interactive to help you answer the following:

(a) 📖❓ Find the particular solution.

Find the particular solution that satisfies \(y(0)=5\text{.}\)
  • \(\quad y = -2e^{x^2}+3\)
  • Incorrect. The value of \(c\) must make the solution pass through \((0, 5)\text{.}\) Hover over the curve in the figure that passes through \((0,5)\text{.}\)
  • \(\quad y = 0.5e^{x^2}+3\)
  • Incorrect. Remember that at \(x = 0\text{,}\) the exponential term \(e^{x^2}\) equals 1, so \(y(0) = c + 3\text{.}\) What value of \(c\) gives \(y(0) = 5\text{?}\) Hover over the curve in the figure that passes through \((0,5)\text{.}\)
  • \(\quad y = 2e^{x^2}+3\)
  • Correct! The value \(c = 2\) ensures that \(y(0) = 2 + 3 = 5\text{,}\) so this solution passes through \((0, 5)\text{.}\)
  • \(\quad y = 5e^{x^2}+3\)
  • Incorrect. The general solution would pass through \((0, 4)\) if \(c = 1\text{.}\) Hover over the curve in the figure that passes through \((0,5)\text{.}\)

(b) 📖❓ Find the initial condition.

What is the initial condition for the particular solution
\begin{equation*} y = 1.3e^{x^2} + 3\text{?} \end{equation*}
  • \(\quad y(0)=4.3\)
  • Correct! Moving \(y(0)\) to this point gives the particular solution above.
  • \(\quad y(0)=5\)
  • Incorrect. Hint: move \(y(0)\) around until you see the particular solution above.
  • \(\quad y(0)=1.3\)
  • Incorrect. Hint: move \(y(0)\) around until you see the particular solution above.
  • \(\quad y(1)=-1.7\)
  • Incorrect. Hint: move \(y(0)\) around until you see the particular solution above.

(c) 📖❓ Find the constant \(c\text{.}\).

Find the \(c\)-value for the particular solution that approximately satisfies \(y(1)=1\text{.}\)
  • \(\quad c = -0.7\)
  • Correct! Moving \(y(0)\) to \(2.3\) gives a blue curve that nearly passes through the point \((1,1)\text{.}\)
  • \(\quad c = -2\)
  • Incorrect. Hint: identify the point \((1,1)\) and move \(y(0)\) until the blue curve intersects with this point.
  • \(\quad c = 0.5\)
  • Incorrect. Hint: identify the point \((1,1)\) and move \(y(0)\) until the blue curve intersects with this point.
  • \(\quad c = -1\)
  • Incorrect. Hint: identify the point \((1,1)\) and move \(y(0)\) until the blue curve intersects with this point.

(d) 📖❓ Role of Initial Conditions.

What role do initial conditions play in solving differential equations?
  • They determine the general form of the solution.
  • Incorrect. Initial conditions are not used to find the general solution.
  • They are used to determine the constants in the general solution.
  • Correct! Initial conditions are used to find specific values for constants in the general solution.
  • They are used to find the particular solution.
  • Correct! Initial conditions are used to find the specific solution that applies to a particular scenario.
  • They are not needed if the general solution is already known.
  • Incorrect. If provided, initial conditions are always needed to get the particular solution from the general solution.
You have attempted of activities on this page.
PrevTopNext