Differential equations: how to quickly distinguish Separable equations, Linear equations, and differential equations with variables swapped about X

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0:00At the beginning of the video, the topic is narrowed down to "first-order differential equations." The screen displays the title "First-Order Differential Equations" and two common forms: separable equations $\frac{dy}{dx}=f(x)g(y)$ and first-order linear equations $y'+P(x)y=Q(x)$. The instructor explains in parallel that "first-order" means the derivative appears only as $\frac{dy}{dx}$ or $y'$.

0:08Next, a classification order is given: upon encountering a first-order equation, first check whether it is linear; if not linear, treat it as separable. This logic is not a formal proof but rather a problem-solving workflow summarized in the video to aid rapid classification.

0:18The screen then scrolls upward to a summary page, which begins explaining how to determine "linearity." Two core criteria are stated on the page: first, all occurrences of $y,y',y''$ in the equation must be to the first power; second, the equation must contain no composite structures—especially $y$ must not serve as the argument of an outer function. For first-order cases, attention focuses specifically on $y$ and $y'$.

0:30The instructor further clarifies that $y$ or $y'$ may be multiplied by a function of $x$ or by a constant, but not by any quantity involving $y$. That is, expressions like $xy$ and $x y'$ still conform to the idea that coefficients depend only on the independent variable, whereas $yy'$, $y^2$, and $(y')^2$ violate linearity.

0:48To reinforce memory, the screen highlights counterexamples in green: $y^2$, $(y')^2$, $yy'$, $e^y$, $\ln y$, $\cos y$, $\sin y$. These examples share one or more features: either higher powers appear, or products of unknown functions occur, or $y$ is embedded inside a composite function. The video collectively classifies them as "nonlinear."

1:14The screen then returns to the concrete example problem $2y\,dy-(1+\cos x)(1+y^2)\,dx=0$. The instructor first points out that the expression contains $y^2$, so it is not a linear equation. Following the classification workflow introduced earlier in the video, the next step proceeds to handling it as a separable equation.

1:28The key in this step is to rearrange the equation into the form "terms containing $y$ paired with $dy$, terms containing $x$ paired with $dx$." The screen directly shows the transformed result: $\frac{2y}{1+y^2}\,dy=(1+\cos x)\,dx$, labeled beside it as "separable." This corresponds to the general method $\frac{dy}{dx}=f(x)g(y)\Rightarrow g(y)dy=f(x)dx\Rightarrow \int g(y)dy=\int f(x)dx$.

1:44The segment ends without proceeding to integration to find the general solution, stopping instead at the point where separation has been successfully achieved. Overall, the video's emphasis is not on solving the entire problem, but on training recognition: first test for linearity, then eliminate linearity using higher powers, products, or composite structures, and finally convert separable equations into differential forms where each side contains only one variable.

2:00The video begins with a first-order differential equation example: $2ydy - (1+\cos x)(1+y^2)dx = 0$. First, the instructor moves the term containing $dx$ to the right-hand side, obtaining $2ydy = (1+\cos x)(1+y^2)dx$. This step aims to gather variables $x$ and $y$ on opposite sides of the equation.

2:18Next, the instructor points out that the equation is separable and demonstrates the separation process: dividing both sides by $(1+y^2)$ to obtain $\frac{2y}{1+y^2}dy = (1+\cos x)dx$. At this point, the left-hand side contains only $y$ and $dy$, while the right-hand side contains only $x$ and $dx$, fully matching the standard form $g(y)dy = f(x)dx$ of a separable-variable differential equation. The instructor also adds that if direct separation proves difficult, one may first rewrite the equation in the form $\frac{dy}{dx} = f(x)g(y)$ before separating variables.

2:50After completing separation of variables, the instructor explains that the next step is to integrate both sides of the equation indefinitely, i.e., $\int \frac{2y}{1+y^2}dy = \int (1+\cos x)dx$. This transforms solving the differential equation into computing two indefinite integrals.

3:14Next, the video switches to the second example: $\cos x \sin y dy = \cos y \sin x dx$. The instructor observes that the equation contains trigonometric terms such as $\sin y$ and $\cos y$, and thus concludes it is a nonlinear differential equation.

3:35Although it is nonlinear, the instructor points out that it can still be converted into a separable form through simple algebraic manipulation. Since $dy$ and $dx$ already appear on opposite sides of the equation, we need only move the term containing $y$ to the left-hand side and the term containing $x$ to the right-hand side. Specifically, divide both sides by $\cos x \cos y$ to obtain $\frac{\sin y}{\cos y}dy = \frac{\sin x}{\cos x}dx$. At this point, the equation has again been successfully separated, and integration on both sides may proceed.

4:00Let us examine the second example; the equation is $\cos x \sin y dy = \cos y \sin x dx$. To separate the variables, we need to place terms containing $y$ and $dy$ on one side of the equation, and terms containing $x$ and $dx$ on the other side.

4:05Observing the equation, we can divide both sides by $\cos x \cos y$. This cancels $\cos x$ on the left-hand side and $\cos y$ on the right-hand side.

4:12After rearrangement, we obtain $\frac{\sin y}{\cos y} dy = \frac{\sin x}{\cos x} dx$. At this point, the left-hand side contains only a function of $y$ and $dy$, while the right-hand side contains only a function of $x$ and $dx$.

4:18Variable separation has thus been successfully achieved. Here, the instructor emphasizes a quick identification technique: notice that the original equation contains $\cos y$, a nonlinear term in $y$. Within the pedagogical context of this video, such a nonlinear structure usually signals a separable-variable differential equation.

4:30Next, consider the third example; the equation is $y' = e^{2x-y}$. Its exponential part is $2x-y$, a composite expression.

4:38We can apply the inverse operation of the quotient rule for powers with the same base, $\frac{a^m}{a^n} = a^{m-n}$, to split $e^{2x-y}$ into $\frac{e^{2x}}{e^y}$.

4:48Simultaneously, rewrite $y'$ in differential form as $\frac{dy}{dx}$. The equation then becomes $\frac{dy}{dx} = \frac{e^{2x}}{e^y}$.

4:55Examining this new form, we see $e^y$ appearing in the denominator—a nonlinear term in $y$. According to the earlier technique, this is again a separable-variable differential equation.

5:05To completely separate the variables, we multiply both sides of the equation by $e^y dx$.

5:12On the left, $\frac{dy}{dx}$ multiplied by $dx$ leaves $dy$, and then multiplied by $e^y$ becomes $e^y dy$; on the right, $\frac{e^{2x}}{e^y}$ multiplied by $e^y$ cancels the denominator, and then multiplied by $dx$ becomes $e^{2x} dx$.

5:25We finally obtain $e^y dy = e^{2x} dx$. The variables are fully separated, and we can now integrate both sides separately to solve the equation.

6:00In this segment, we first address problem 4 $(y-x\sin x)dx+xdy=0$. The instructor observes that $y$ and $dy$ appear independently in the equation, with no structures such as $y\,dy$, $y^2$, or $\sin y$ that would render the equation nonlinear; thus, the equation is classified as 'first-order linear' rather than 'separable'. The key point here is to perform type identification first, then decide which method to apply next.

6:21The camera then pans up to the summary area at the top, clearly stating that the standard form of a first-order linear differential equation is $y'+P(x)y=Q(x)$, and simultaneously presenting the general solution formula $y=e^{-\int P(x)dx}\left[\int Q(x)e^{\int P(x)dx}dx+C\right]$. A crucial premise is established here: all subsequent algebraic manipulations aim to rearrange the original equation into a form directly comparable to $P(x)$ and $Q(x)$.

6:30Next, concrete simplification begins. First, rearrange the original equation to obtain $xdy=-(y-x\sin x)dx$; then divide both sides by $dx$ to write it as $\frac{dy}{dx}=-\frac{y-x\sin x}{x}$. The instructor emphasizes the order: first position $dx$ and $dy$ correctly, then perform division uniformly—avoiding confusion caused by moving terms and dividing simultaneously within mixed expressions.

7:00Proceed to split the right-hand-side fraction, yielding $\frac{dy}{dx}=-\frac{y}{x}+\sin x$. Then move the term containing $y$ to the left-hand side, resulting in the standard form $y'+\frac{1}{x}y=\sin x$. At this point, the original equation fully aligns with the formula displayed at the top, confirming that it can indeed be handled using the first-order linear method.

7:34Then, identify coefficients by matching against $y'+P(x)y=Q(x)$: $P(x)=\frac{1}{x}$ and $Q(x)=\sin x$. These two terms are circled in green on the board, visually linking the 'standard form' directly to the 'formula parameters', facilitating subsequent substitution into the general solution formula.

7:44Finally, a common pitfall is emphasized: the equation must first be converted into standard form, and $Q(x)$ must appear on the right-hand side of the equation. Mistakenly treating $\sin x$ as $Q(x)$ while leaving it on the left-hand side leads to incorrect application of the formula. The video ends at this preparatory stage: the method is confirmed applicable and ready for formula substitution, but the final solution is not computed.

8:00When classifying a differential equation, we sometimes encounter an equation that exhibits multiple characteristics simultaneously. For instance, it may be both a first-order linear differential equation and a separable-variable differential equation. In such cases, how should we choose the solution method?

8:15The instructor recommends prioritizing the 'first-order linear' method. Although the 'separable-variable' method also works, handling the integration constant $C$ afterward tends to be cumbersome. In contrast, first-order linear differential equations have a fixed general solution formula; substituting into this formula directly yields the final result containing $C$, avoiding tedious intermediate algebraic manipulations.

8:48To verify this, let us consider the concrete example $(1+e^x)dy = ye^x dx$ and see how to convert it into the standard form of a first-order linear equation. First, divide both sides of the equation by $dx$ and $(1+e^x)$, obtaining the derivative expression: $\frac{dy}{dx} = \frac{ye^x}{1+e^x}$.

9:10Next, rearrange terms by moving all terms involving $y$ to the left-hand side so that the equation matches the structure $y' + p(x)y = q(x)$. After rearrangement, we obtain: $\frac{dy}{dx} - \frac{e^x}{1+e^x}y = 0$.

9:25We can now clearly identify each component in the standard form. Note that the negative sign must be included; thus, $p(x) = -\frac{e^x}{1+e^x}$, while the nonhomogeneous term on the right-hand side is $q(x) = 0$. This successfully transforms the original equation into a first-order linear differential equation, after which we simply apply the general solution formula to solve it.

10:00This segment first handles Problem 5 $(1+e^x)dy=ye^xdx$. The instructor aims to identify which class of first-order differential equation it belongs to and demonstrates how to rewrite it into a suitable form for solving.

10:10Following the idea of separation of variables, factors containing $y$ should be grouped on the $dy$ side, and factors containing $x$ on the $dx$ side. Thus, divide both sides by $1+e^x$ to obtain $dy=\frac{ye^x}{1+e^x}dx$.

10:27Next, divide both sides by $y$ to obtain $\frac{dy}{y}=\frac{e^x}{1+e^x}dx$. This step completes separation of variables, indicating that the equation is separable.

10:40The same problem can also be viewed from the perspective of "first-order linear." Rewrite the original expression as $\frac{dy}{dx}=\frac{ye^x}{1+e^x}$, then rearrange terms to get $\frac{dy}{dx}-\frac{e^x}{1+e^x}y=0$, which matches the standard form $\frac{dy}{dx}+p(x)y=q(x)$, where $p(x)=-\frac{e^x}{1+e^x}$ and $q(x)=0$.

10:55Therefore, the video emphasizes that Problem 5 is "both linear and separable," meaning one need not choose only one classification; in practice, one may select whichever method is more convenient.

11:02Next comes Problem 6 $\frac{dy}{dx}=\frac{y}{x+y^2}$. Here, first observe the structure of the original expression with respect to $y$; the denominator contains $y^2$, so it is not a first-order linear differential equation in $y$.

11:23The instructor further explains that if an equation is neither easily recognizable as separable nor convertible into the first-order linear standard form in $y$, one may consider the "extended form": swapping the roles of independent and dependent variables.

11:39For Problem 6, this means treating $x$ as a function of $y$. The rewritten form shown on screen is $\frac{dx}{dy}-\frac{1}{y}x=y$, which is precisely the first-order linear standard form in $x$.

11:52Thus, the key point of this segment is: when classifying a differential equation, do not rely solely on its superficial form in $\frac{dy}{dx}$; when it is nonlinear in $y$, try rewriting it as a first-order linear equation in $x$.

12:00This segment begins with Problem 6: the whiteboard displays $\frac{dy}{dx}=\frac{y}{x+y^2}$. Instead of solving immediately, the instructor first asks whether it can be treated as a 'separable variables' equation.

12:11Next, complete the criterion for separable variables: the left-hand side of the equation must be $\frac{dy}{dx}$, and the right-hand side must be expressible as $f(x)\cdot g(y)$, that is, a factor depending only on $x$ multiplied by a factor depending only on $y$. The board work circles $g(y)$ in green to emphasize "multiplication is required".

12:30Returning to problem 6, the denominator is $x+y^2$, where $x$ and $y$ are coupled by addition and cannot be split into $f(x)g(y)$. Therefore, the video explicitly concludes: this problem is not a separable variables equation.

13:03Since the approach with respect to $y$ leads nowhere, the instructor adopts the method of "taking reciprocals on both sides", treating $x$ as a function of $y$, yielding $\frac{dx}{dy}=\frac{x+y^2}{y}$. The key step here is swapping the roles of independent and dependent variables.

13:10Then, split the right-hand side: $\frac{x+y^2}{y}=\frac{1}{y}x+y$; next, rearrange terms to obtain $\frac{dx}{dy}-\frac{1}{y}x=y$, which is $x'-\frac{1}{y}x=y$. At this point, the equation has become a first-order linear differential equation in $x$.

13:27Comparing with the standard form $x'+p(y)x=q(y)$, the video directly identifies $p(y)=-\frac{1}{y}$ and $q(y)=y$. The instructor specifically cautions that the unknown function is now $x$, and one must not confuse it with the earlier "linear equation in $y$".

13:50Finally, the screen shifts upward to reveal the general solution formula $x=e^{-\int p(y)dy}\left[\int q(y)e^{\int p(y)dy}dy+C\right]$. This segment stops at the point "ready to substitute into the formula", without computing the final solution.

14:00The clip begins paused at Problem 6. The board first writes $\frac{dy}{dx}=\frac{y}{x+y^2}$ and marks in red "nonlinear in y"; it is then rewritten as $\frac{dx}{dy}-\frac{1}{y}x=y$ and labeled "first-order linear differential equation in y". The mathematical maneuver here is swapping the independent and dependent variables: instead of treating $x$ as the independent variable and $y$ as the unknown function, we treat $x$ as a function of $y$, making $\frac{dx}{dy}$ the new object of differentiation.

14:03Next, the screen remains on the lower half of problem 6; the board writes $X'-\frac{1}{y}X=y$ and labels $P(y)=-\frac{1}{y}$ and $Q(y)=y$ beside it. Below, the corresponding general solution formula $x=e^{-\int P(y)dy}\left[\int Q(y)e^{\int P(y)dy}dy+C\right]$ is given. This shows that once successfully converted into a linear equation in $y$, we can directly apply the linear formula—only with the independent variable changed from $x$ to $y$, and the unknown function changed from $y$ to $x$.

14:06The screen scrolls upward to reveal Problem 5and Problem 4Problem 5starts from $(1+e^x)dy=ye^x dx$ and is first rewritten as $\frac{dy}{y}=\frac{e^x}{1+e^x}dx$ so it is a separable equation;$\frac{dy}{dx}=\frac{ye^x}{1+e^x}$ then it is rewritten as $\frac{dy}{dx}-\frac{e^x}{1+e^x}y=0$ and rearranged as $y'+P(x)y=Q(x)$ which also matches the standard linear first order form, where $P(x)=\frac{e^x}{1+e^x}$、$Q(x)=0$The board clearly states 'both linear and separable,' emphasizing that the same equation can be viewed from multiple perspectives.

14:10After scrolling further upward, the top summary area fully appears, titled "First-Order Differential Equations", followed by "First, determine the type." The first category is "Separable-Variable", corresponding to the form $g(y)dy=f(x)dx$, with an alternative written as "or $\frac{dy}{dx}=f(x)g(y)$", and explained as "integrate both sides indefinitely: $\int g(y)dy=\int f(x)dx$." The second category is "First-Order Linear Differential Equation", corresponding to the form $y'+P(x)y=Q(x)$, whose general solution formula is $y=e^{-\int P(x)dx}\left[\int Q(x)e^{\int P(x)dx}dx+C\right]$. A note at the top adds "Extension: $X'+P(y)X=Q(y)$", precisely echoing the variable-swapping approach used in Problem 6.

14:14Below the top summary, Examples 1 through 3 collectively demonstrate separable-variable equations. Problem 1 transforms $2ydy-(1+\cos x)(1+y^2)dx=0$ into $\frac{2y}{1+y^2}dy=(1+\cos x)dx$; Problem 2 transforms $\cos x\sin y\,dy=\cos y\sin x\,dx$ into $\frac{\sin y}{\cos y}dy=\frac{\sin x}{\cos x}dx$; Problem 3 transforms $y'=e^{x+y}$ into $e^y dy=e^x dx$. These three problems share the feature that the right-hand side—or the differential form—can be split into a factor depending solely on $x$ and another depending solely on $y$, allowing direct separation and subsequent integration.

14:18Problem 4 demonstrates first-order linear identification: $(y-x\sin x)dx+x dy=0$ is rewritten as $\frac{dy}{dx}+\frac{1}{x}y=\sin x$, labeled "first-order linear." The key step here is rearranging the equation into the standard structure $y'$ plus a linear term equals a right-hand-side function, rather than remaining in a mixed differential form. The audio commentary synchronously emphasizes: "How to distinguish... at a glance... you can quickly identify its type," fully aligning with the classification framework on the board.

14:23In the latter part of the clip, the instructor condenses the method into a workflow: first determine the type, then apply the formula. The audio states: "Once you've identified the type, just plug it into this formula—then it's much easier. And after plugging into the formula, it becomes simply an integration process." This echoes the order of the two formulas displayed at the top: separable-variable equations correspond to integrating both sides, while first-order linear equations correspond to substituting into the general solution formula and then computing an integral.

14:27Finally, the instructor offers study advice: "So, regarding integration—be sure to practice it thoroughly!" Mathematically, this is not a new theorem but a concluding reminder about the solution chain: type identification determines "which formula to use," while integration proficiency determines "whether you can compute the result." Thus, the entire segment forms a clear central thread: identify the type of first-order differential equation, swap variables if necessary, apply the corresponding general solution formula, and ultimately carry out integration.

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