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 \(c\text{.}\).
- \(\quad c = -0.7\)
- 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.
