Summary of the Key Ideas.
Separation of Variables Method
The general solution of a differential equation that is first-order and separable can be found using the separation of variables method.
Section 4.5 Summary & Exercises
Exercises Exercises
Separable.
Determine whether the given differential equation is separable or not. As demonstrated in the examples, if the equation is separable, use parentheses to explicitly show the separable form.
1.
\(\ds \frac{dz}{dt} = \sin(z+t) \)Answer.
not separable
2.
\(\ds s' = t\ln(s^{2t}) + 8t^2 \)Answer.
separable
3.
\(\ds \frac{dy}{dx} = 2y^3 + y + 4 \)Answer.
separable
4.
\(\ds y' - xy = 0 \)Answer.
separable
5.
\(\ds y' = x+y \)Answer.
not separable
6.
\(\ds y'' + y' + y = 0 \)Answer.
not separable
Step-by-Step.
Maybe write 1 or 2 simple scaffolded problems here?
7.
Given the differential equation \(\displaystyle \frac{dy}{dx} = xy^2 \text{,}\) determine if it is separable. If so, rewrite it in the separated form.
Solution.
The equation is separable because we can write it as:
\begin{equation*}
\frac{1}{y^2} dy = x dx
\end{equation*}
Answer.
\(\displaystyle \frac{1}{y^2} dy = x dx \)
Continuing from the previous problem, integrate the left side of the separated equation with respect to \(y\text{.}\)
Solution.
Integrating \(\displaystyle \frac{1}{y^2} dy \) with respect to \(y\text{,}\) we get:
\begin{equation*}
-\frac{1}{y} + C_1
\end{equation*}
Answer.
\(\displaystyle -\frac{1}{y} + C_1 \)
Now, integrate the right side of the separated equation with respect to \(x\text{.}\)
Solution.
Integrating \(x dx \) with respect to \(x\text{,}\) we get:
\begin{equation*}
\frac{1}{2}x^2 + C_2
\end{equation*}
Answer.
\(\displaystyle \frac{1}{2}x^2 + C_2 \)
Combine the results from the previous exercises to form the general solution to the differential equation.
Solution.
Equating the two integrals, we get:
\begin{equation*}
-\frac{1}{y} + C_1 = \frac{1}{2}x^2 + C_2
\end{equation*}
By grouping constants, we can represent them with a single constant:
\begin{equation*}
-\frac{1}{y} = \frac{1}{2}x^2 + C
\end{equation*}
Where \(C = C_2 - C_1\text{.}\)
Answer.
\(\displaystyle -\frac{1}{y} = \frac{1}{2}x^2 + C \)
Warm Ups.
Solve using separation of variables, if possible.
8.
\(\displaystyle \frac{dy}{dx} = y^2 \)
9.
\(\displaystyle \frac{dy}{dx} = x^2y \)
10.
\(\displaystyle \frac{dy}{dx} = e^{x+y} \)
11.
\(\displaystyle \frac{dy}{dx} = \frac{x}{y} \)
12.
\(\displaystyle \frac{dy}{dx} = \cos(x) \sec(y) \)
13.
\(\displaystyle \frac{dy}{dx} = \frac{2x}{1+y^2} \)
14.
\(\displaystyle \frac{dy}{dx} = 2-y \)
15.
\(\displaystyle \frac{dy}{dx} = \frac{y-x}{y+x} \)
16.
\(\displaystyle \frac{dy}{dx} = y^2 \sec^2(x) \)
17.
\(\displaystyle \frac{dy}{dx} = \frac{1}{xy} \)
Further Practice.
Solve using separation of variables, if possible.
18.
\(\displaystyle \frac{dy}{dx} = x y \)Answer.
\(\displaystyle y = Ce^{\frac{1}{2}x^2} \)
19.
\(\displaystyle \frac{dy}{dx} = \frac{x^2 + 1}{y} \)Answer.
\(\displaystyle y^2 = \frac{2}{3}x^3 + 2x + C_2 \)
20.
\(\displaystyle \frac{dy}{dx} = \frac{1 - x^2}{y^2} \)Answer.
\(\displaystyle y = \sqrt[3]{3x - x^3 + C_2} \)
21.
\(\displaystyle y' - 2y = y\sin x \)Answer.
\(\displaystyle y = C_2e^{2x - \cos x} \)
22.
\(\displaystyle x\frac{dv}{dx} = \frac{1-4v^2}{3v} \)Answer.
\(\displaystyle v = \pm \sqrt{C_3x^{-8/3} + \frac{1}{4}} \)
23.
\(\displaystyle \frac{dy}{dx} = \frac{4-2x}{3y^2-5} \)Answer.
\(y^3-5y=4x-x^2+C \)
24.
\(\displaystyle \frac{dy}{dx} = 6x(y-1)^{2/3}\)Answer.
\(\displaystyle y(x)=\amp\ \displaystyle 1+(x^2+C)^3\)
Initial Value Problems.
Solve each of the following initial value problems using separation of variables, if possible.
25.
\(\displaystyle \frac{dy}{dx} = -6xy, \hspace{0.5cm} y(0)=7.\)Answer.
\begin{equation*}
y(x) = 7e^{-3x^2}
\end{equation*}
26.
\(\displaystyle \frac{dz}{dt} = 2tz^2, \hspace{0.5cm} z(1) = 2 \)Answer.
\(z(t) = -\frac{1}{t^2 - \frac{3}{2}} \)
27.
\(\displaystyle \frac{dy}{d\theta} = y\sin\theta, \hspace{0.5cm} \displaystyle y(\pi) = -3 \)Answer.
\(\displaystyle y = -3e^{-1}e^{-\cos\theta} \text{,}\) or \(\displaystyle y = -3e^{-\cos\theta-1} \)
28.
\(\displaystyle \frac{dy}{dx} - 8x^3e^{-2y} = 0, \hspace{0.5cm} \displaystyle y(1) = 0 \)Answer.
\(\displaystyle y = \frac{1}{2}\ln(4x^4 - 3) \) , or \(\displaystyle y = \ln\Big[\sqrt{4x^4 - 3}\Big] \)
Applications.
29.
A 4-lb roast, initially at \(50^{\circ}F\text{,}\) is placed in a \(375^{\circ} F\) oven at 5:00 PM After 75 minutes it is found that the temperature \(T(t)\) of the roast is \(125^{\circ}\) F. When will the roast be \(150^{\circ}F\) (medium rate)?Answer.
\(t = -[\ln(225/325)]/[0.0035] \approx 105(min)\)
Conceptual Questions.
30.
Explain why separating variables works as a method for solving ordinary differential equations.31.
Why is it important for the equation to be "separable" in order to use the method of separation of variables? What does it mean for a differential equation to be separable?32.
Provide an example of a differential equation that is not separable and explain why the method of separation of variables cannot be applied to it.33.
In solving a differential equation using separation of variables, why might the constant of integration appear on both sides of the resulting equation?34.
Some differential equations solved using separation of variables yield implicit solutions, while others yield explicit solutions. Explain the difference between implicit and explicit solutions, and provide examples of each.35.
True/False: Every first-order ordinary differential equation can be solved using the method of separation of variables.36.
Which of the following differential equations is not separable?
- \(\displaystyle \frac{dy}{dx} = y \cos(x) \)
- \(\displaystyle \frac{dy}{dx} = y + e^x \)
- \(\displaystyle \frac{dy}{dx} = \sin(x) \sin(y) \)
- \(\displaystyle \frac{dy}{dx} = x^2y \)
37.
True/False: If a differential equation is separable, its solution will always be an explicit function of \(x \text{.}\)
38.
The process of writing a differential equation in the form \(M(y)dy = N(x)dx \) is called _________ of variables.
39.
Which of the following cannot be directly solved by separation of variables?
- \(\displaystyle \frac{dy}{dx} = x + y \)
- \(\displaystyle \frac{dy}{dx} = xy \)
- \(\displaystyle \frac{dy}{dx} = \sin(x)y \)
- \(\displaystyle \frac{dy}{dx} = \frac{x}{y} \)
40.
True/False: The method of separation of variables can be directly applied to higher-order differential equations.
41.
True/False: If you can write a differential equation in the form \(h(y) \frac{dy}{dx} = g(x) \text{,}\) then it is separable.
42.
A differential equation of the form \(\frac{dy}{dx} = g(x) \) without a \(y \) term:
- Is always separable
- Is never separable
- Can be separable depending on \(g(x) \)
- Is an implicit equation
43. A structured exercise.
Here is where we give the student the background information required to start accomplishing tasks.
(a)
Solve a separable DE using separation of variables.
\(y(x) =\)
Answer.
\(0.142857e^{7x}-2x^{2}+C\)
(b)
The second step to do. We’ll be lazy and just include an answer.
Answer.
Just the answer.
A little wrap up.
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