Differential Equations Quiz: Identifying Separable Des

What this quiz covers

This quiz focuses on Identifying Separable Des, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\frac{dy}{dx} = \sin(x) - \cos(y)$
2. $\frac{dy}{dx} = \sin(x)\cos(y) - \cos(x)\sin(y)$
3. $\frac{dy}{dx} = y\sin(x) - \sin(x)$ (correct answer)
4. $\frac{dy}{dx} = \sin(xy)$

Explanation: A differential equation is separable if $\frac{dy}{dx}$ can be expressed as a product of a function of $x$ and a function of $y$. Choice C is $\frac{dy}{dx} = y\sin(x) - \sin(x) = \sin(x)(y-1)$. This is of the form $g(x)h(y)$, so it is separable. Choice A is a sum of a function of $x$ and a function of $y$, which is not separable. Choice B is equivalent to $\frac{dy}{dx} = \sin(x-y)$ by a trigonometric identity, which is not separable. Choice D has $x$ and $y$ coupled inside the sine function and cannot be separated.

Question 2

1. $\frac{dy}{dx} = e^{x-y}$
2. $e^y \frac{dy}{dx} = x+1$
3. $\frac{dy}{dx} = x^2e^y + e^y$
4. $\frac{dy}{dx} = e^{xy}$ (correct answer)

Explanation: A separable differential equation can be written as $\frac{dy}{dx} = g(x)h(y)$. We must check each option. A: $\frac{dy}{dx} = e^{x-y} = e^x e^{-y}$, which is separable. B: $e^y \frac{dy}{dx} = x+1$ can be written as $\frac{dy}{dx} = (x+1)e^{-y}$, which is separable. C: $\frac{dy}{dx} = x^2e^y + e^y = e^y(x^2+1)$, which is separable. D: $\frac{dy}{dx} = e^{xy}$. The term $xy$ in the exponent prevents the expression from being factored into a function of $x$ times a function of $y$. Therefore, this equation is not separable.

Question 3

1. $F(x,y) = \frac{y^2}{x^2} + \frac{2y}{x}$
2. $F(x,y) = \sec(y/x)$
3. $F(x,y) = \frac{x+y}{x-1}$ (correct answer)
4. $F(x,y) = \frac{x^2+y^2}{xy}$

Explanation: A function $F(x,y)$ is homogeneous of degree zero if $F(tx, ty) = F(x,y)$ for any $t \neq 0$. This is equivalent to saying $F$ can be written as a function of $y/x$. Let's test each option. A: $F(x,y) = (y/x)^2 + 2(y/x)$, depends on $y/x$. B: $F(x,y) = \sec(y/x)$, depends on $y/x$. C: $F(tx,ty) = \frac{tx+ty}{tx-1} = \frac{t(x+y)}{tx-1} \neq F(x,y)$. This is not homogeneous. D: $F(x,y) = \frac{x^2+y^2}{xy} = \frac{x}{y} + \frac{y}{x}$, which depends on $y/x$. Since the function in C is not homogeneous, the substitution $u=y/x$ will not make the DE separable.

Question 4

1. The equation is separable because it can be rewritten in the form $g(x)dx = h(y)dy$. (correct answer)
2. The equation is not separable because the $dx$ term contains a function of $y$.
3. The equation is linear in $y$.
4. The equation is homogeneous.

Explanation: The given equation is $(xy+x)dx = (x^2y^2+x^2)dy$. We can factor both sides: $x(y+1)dx = x^2(y^2+1)dy$. Assuming $x \neq 0$, we can divide by $x$: $(y+1)dx = x(y^2+1)dy$. Now, we can separate the variables by grouping all $x$ terms with $dx$ and all $y$ terms with $dy$. This gives $\frac{1}{x}dx = \frac{y^2+1}{y+1}dy$. Since the variables can be separated, the equation is separable. Distractor B is a common misconception before algebraic manipulation. The equation is not linear due to the $y^2$ term. It is not homogeneous because the degrees of the terms are not consistent.

Question 5

1. For $n=0$ only. (correct answer)
2. For $n=1$ only.
3. For all integer values of $n$.
4. For no integer values of $n$.

Explanation: First, rewrite the equation in the form $\frac{dy}{dx} = F(x,y)$: $\frac{dy}{dx} = \frac{ny}{x} + x^n e^x$. For this equation to be separable, $F(x,y)$ must be factorable into a product $g(x)h(y)$. The expression is a sum of two terms. It can only be factored if the terms share a common factor related to $y$ or $x$. If $n=0$, the equation becomes $\frac{dy}{dx} = \frac{0 \cdot y}{x} + x^0 e^x$, which simplifies to $\frac{dy}{dx} = e^x$. This can be written as $\frac{dy}{dx} = g(x)h(y)$ where $g(x)=e^x$ and $h(y)=1$. Thus, it is separable for $n=0$. For any other integer $n \neq 0$, the right side $\frac{ny}{x} + x^n e^x$ cannot be factored into a product of a function of $x$ and a function of $y$. Therefore, the equation is separable only for $n=0$.

Question 6

1. $\frac{dy}{dx} = \ln(x+y^2)$
2. $\frac{dy}{dx} = \ln(x) + \ln(y^2)$
3. $\frac{dy}{dx} = (\ln(xy))^2$
4. $\frac{dy}{dx} = x\ln(y^2) + \ln(y^2)$ (correct answer)

Explanation: We test each equation for separability. A: The sum $x+y^2$ inside the logarithm prevents separation. B: Using logarithm properties, $\frac{dy}{dx} = \ln(x) + 2\ln(y)$. This is a sum of a function of $x$ and a function of $y$, not a product, so it is not separable. C: Using logarithm properties, $\frac{dy}{dx} = (\ln(x) + \ln(y))^2 = (\ln x)^2 + 2(\ln x)(\ln y) + (\ln y)^2$. This expression is not factorable into $g(x)h(y)$. D: We can factor the right side: $\frac{dy}{dx} = x\ln(y^2) + \ln(y^2) = (x+1)\ln(y^2) = (x+1)(2\ln y)$. This is in the form $g(x)h(y)$, so it is separable.

Question 7

1. $f(x,y) = xy^2$ (correct answer)
2. $f(x,y) = xy$
3. $f(x,y) = x+y$
4. $f(x,y) = y/x$

Explanation: A singular solution is a solution that cannot be obtained from the general solution by choosing a value for the integration constant. First, we identify which equations are separable. A, B, and D are separable. C is not. Now we analyze the solutions. A: $\frac{dy}{dx} = xy^2 \implies \frac{dy}{y^2} = x dx \implies -\frac{1}{y} = \frac{x^2}{2} + C \implies y = \frac{-2}{x^2+2C}$. The constant solution $y=0$ (which makes $h(y)=y^2=0$) cannot be obtained from this general solution for any finite value of $C$. Thus, $y=0$ is a singular solution. B: $\frac{dy}{dx} = xy \implies \frac{dy}{y} = x dx \implies \ln|y| = \frac{x^2}{2} + C \implies y=Ae^{x^2/2}$. Here, $y=0$ is obtained when $A=0$, so it's a particular solution. D: $\frac{dy}{dx}=y/x \implies \frac{dy}{y}=\frac{dx}{x} \implies \ln|y| = \ln|x|+C \implies y=Ax$. Here, $y=0$ is obtained when $A=0$, so it's a particular solution.

Question 8

1. $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 0$
2. $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$
3. $f \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}$ (correct answer)
4. $f$ is a homogeneous function of degree one.

Explanation: If $f(x,y) = g(x)h(y)$, then $\frac{\partial f}{\partial x} = g'(x)h(y)$, $\frac{\partial f}{\partial y} = g(x)h'(y)$, and $\frac{\partial^2 f}{\partial x \partial y} = g'(x)h'(y)$. Let's check the condition in C: $f \frac{\partial^2 f}{\partial x \partial y} = (g(x)h(y))(g'(x)h'(y))$. And $\frac{\partial f}{\partial x} \frac{\partial f}{\partial y} = (g'(x)h(y))(g(x)h'(y))$. These two expressions are equal. This condition is a known test for separability. Option A is related to Euler's theorem for homogeneous functions of degree 0. Option B is Laplace's equation. Option D describes a class of homogeneous equations, which are not necessarily separable without a substitution.

Question 9

1. $\frac{dy}{dx} = \frac{y}{x}$
2. $\frac{dy}{dx} = x^2 y$
3. $\frac{dy}{dx} = y\cos(x) + \sin(x)$
4. $\frac{dy}{dx} = \frac{\cos(x)}{y}$ (correct answer)

Explanation: We must check both conditions (separable, not linear) for each option. A: $\frac{dy}{dx} = \frac{y}{x}$ is separable ($\frac{dy}{y} = \frac{dx}{x}$) and linear ($\frac{dy}{dx} - \frac{1}{x}y = 0$). B: $\frac{dy}{dx} = x^2 y$ is separable ($\frac{dy}{y} = x^2 dx$) and linear ($\frac{dy}{dx} - x^2 y = 0$). C: $\frac{dy}{dx} = y\cos(x) + \sin(x)$ is linear, but it is not separable due to the term $\sin(x)$. D: $\frac{dy}{dx} = \frac{\cos(x)}{y}$ is separable ($y dy = \cos(x) dx$). It is not linear because the dependent variable $y$ appears as $1/y$, which is not a linear function of $y$. Therefore, this is the correct answer.

Question 10

1. $g(x,y) = x+y$, leading to $\frac{du}{dx} = u^2$
2. $g(x,y) = x+y$, leading to $\frac{du}{dx} = \frac{1+u^2}{u^2}$ (correct answer)
3. $g(x,y) = (x+y)^2$, leading to $\frac{du}{dx} = \frac{1}{u}$
4. The equation cannot be transformed into a separable one with a linear substitution.

Explanation: When you encounter a differential equation that isn't separable, look for substitutions that can simplify the structure. The presence of $(x+y)$ terms suggests trying $u = x + y$ as a substitution. Let's work through this systematically. If $u = x + y$, then taking the derivative with respect to $x$: $\frac{du}{dx} = 1 + \frac{dy}{dx}$ This means $\frac{dy}{dx} = \frac{du}{dx} - 1$. Substituting both $u = x + y$ and this expression for $\frac{dy}{dx}$ into the original equation: $\frac{du}{dx} - 1 = \frac{1}{u^2}$ Rearranging: $\frac{du}{dx} = 1 + \frac{1}{u^2} = \frac{u^2 + 1}{u^2}$ This is now separable since we can write it as $\frac{u^2}{u^2 + 1} du = dx$. Answer A uses the correct substitution but gets the wrong differential equation—it's missing the "$1$" term that comes from differentiating $u = x + y$. Answer C attempts $u = (x+y)^2$, but this leads to a more complex derivative involving the chain rule and doesn't simplify the equation effectively. Answer D is incorrect because we've just demonstrated that a linear substitution does work. The key insight is recognizing when the form of a differential equation suggests a particular substitution, then carefully applying the chain rule to transform the equation. Always double-check your derivative calculations when making substitutions—the extra terms that appear (like the "$1$" here) are crucial for getting the correct transformed equation.

Question 11

1. Let $u = x + y$; then $\frac{du}{dx} = 1 + \frac{dy}{dx}$, leading to a separable equation in $u$ and $x$ (correct answer)
2. Let $v = \frac{y}{x}$; then $\frac{dv}{dx} = \frac{1}{x}\left(\frac{dy}{dx} - v\right)$, leading to a homogeneous equation that becomes separable
3. The equation is already separable without substitution since it can be written as $\frac{dy}{dx} = \tan\left(\frac{\pi}{4} + x + y\right)$
4. The equation cannot be made separable by any elementary substitution because the argument $x + y$ creates an inseparable coupling

Explanation: With $u = x + y$, we have $\frac{du}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{du}{dx} - 1$. The equation becomes $\frac{du}{dx} - 1 = \frac{\sin u + \cos u}{\cos u - \sin u}$, or $\frac{du}{dx} = 1 + \frac{\sin u + \cos u}{\cos u - \sin u} = \frac{\cos u - \sin u + \sin u + \cos u}{\cos u - \sin u} = \frac{2\cos u}{\cos u - \sin u}$. This separates as $\frac{\cos u - \sin u}{\cos u}\,du = 2\,dx$. Choice B uses the wrong substitution for this type. Choice C is incorrect trigonometry. Choice D is wrong because the substitution does work.

Question 12

1. The equation is separable after dividing both sides by $(x^2 + 1)$ and recognizing that $\frac{x + \sqrt{x^2 + 1}}{x^2 + 1} = \frac{d}{dx}[\operatorname{sinh}^{-1}(x)]$
2. The equation is not separable because the term $y\sqrt{x^2 + 1}$ creates a mixed radical expression that cannot be separated into functions of $x$ and $y$ independently
3. The equation becomes separable only after the substitution $u = x^2 + 1$, which eliminates the radical and allows factorization
4. The equation is separable after factoring the right side as $y(x + \sqrt{x^2 + 1})$ and rearranging to $\frac{dy}{y} = \frac{x + \sqrt{x^2 + 1}}{x^2 + 1}\,dx$ (correct answer)

Explanation: When you encounter a differential equation and need to determine separability, always look for ways to factor and rearrange terms so that all functions of x appear on one side and all functions of y on the other. Starting with $(x^2 + 1)\frac{dy}{dx} = xy + y\sqrt{x^2 + 1}$, the key insight is to factor the right side. Notice that both terms contain y: $xy + y\sqrt{x^2 + 1} = y(x + \sqrt{x^2 + 1})$. This factorization allows you to rewrite the equation as $(x^2 + 1)\frac{dy}{dx} = y(x + \sqrt{x^2 + 1})$. Now you can separate variables by dividing both sides by $y(x^2 + 1)$: $\frac{dy}{y} = \frac{x + \sqrt{x^2 + 1}}{x^2 + 1}\,dx$. This confirms that choice D is correct. Choice A incorrectly focuses on recognizing a specific derivative formula rather than the fundamental separability process. While the derivative relationship mentioned is mathematically correct, it's not necessary for determining separability. Choice B incorrectly concludes the equation isn't separable, missing the crucial factorization step that makes separation possible. Choice C suggests an unnecessary substitution—the equation is already separable without any substitution once you factor properly. The key strategy here is to always look for common factors when you have multiple terms. In differential equations, factoring often reveals hidden separability that isn't immediately obvious. Don't get distracted by complex-looking expressions; focus on the algebraic structure first.

Question 13

1. When $|n - m| = 1$, because this ensures that after appropriate algebraic manipulation, the variables can be separated on opposite sides
2. When $n = m + 1$, because this creates the proper degree relationship for direct separation without substitution
3. When $n$ and $m$ are both even, because even-degree homogeneous functions have special separation properties
4. When $n = m$, because the equation becomes homogeneous and can be made separable using the substitution $v = \frac{y}{x}$ (correct answer)

Explanation: When you encounter a differential equation where both the numerator and denominator are homogeneous functions (meaning they satisfy the scaling property given), you're dealing with a homogeneous differential equation. The key insight is recognizing what degree of homogeneity makes the equation separable. Answer D is correct because when $n = m$, both $P(x,y)$ and $Q(x,y)$ have the same degree of homogeneity. This makes $f(x,y) = \frac{P(x,y)}{Q(x,y)}$ a homogeneous function of degree zero, since $f(tx,ty) = \frac{t^n P(x,y)}{t^n Q(x,y)} = f(x,y)$. Such equations can always be made separable using the substitution $v = \frac{y}{x}$ (so $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$), which transforms the equation into one where variables separate. Answer A is incorrect because $|n-m| = 1$ doesn't guarantee separability through standard algebraic manipulation alone. Answer B is wrong because $n = m + 1$ creates a homogeneous function of degree 1, not degree 0, which doesn't lead to direct separation. Answer C is incorrect because the parity (even/odd nature) of $n$ and $m$ is irrelevant to separability; it's their equality that matters. Remember this pattern: homogeneous differential equations (where the right side is a function of $\frac{y}{x}$ only) are the ones that become separable with the $v = \frac{y}{x}$ substitution. Look for equal degrees of homogeneity as your signal.

Question 14

1. Use the substitution $u = x + y$ and $v = x - y$ to create a system of first-order equations that can be solved independently
2. Translate the coordinate system by setting $X = x + h$ and $Y = y + k$ where $h$ and $k$ are chosen to eliminate the constant terms (correct answer)
3. Factor the numerator and denominator to find common terms that can be cancelled, revealing the underlying separable structure
4. Apply the integrating factor method directly since the equation is linear in the transformed variables after appropriate substitution

Explanation: The equation has the form $\frac{dy}{dx} = \frac{ax + by + c}{dx + ey + f}$ with $a=1, b=1, c=1, d=1, e=-1, f=-1$. To eliminate constants, we solve $x + y + 1 = 0$ and $x - y - 1 = 0$ simultaneously, giving $x = 0, y = -1$. Setting $X = x - 0 = x$ and $Y = y - (-1) = y + 1$, the equation becomes $\frac{dY}{dX} = \frac{X + (Y-1) + 1}{X - (Y-1) - 1} = \frac{X + Y}{X - Y}$, which is homogeneous and can be made separable with $v = \frac{Y}{X}$. Choice A creates an unnecessarily complex system. Choice C won't work since there are no common factors. Choice D misidentifies the equation type.

Question 15

1. The equation is not separable because the trigonometric functions $\sin(x)$ and $\cos(x)$ appear with different powers and cannot be factored appropriately
2. The equation is separable by first dividing by $\cos(x)$: $\tan(x)\frac{dy}{dx} = y + y^2$, then separating as $\frac{dy}{y(1+y)} = \frac{dx}{\tan(x)}$
3. The equation is separable after factoring: $\sin(x)\frac{dy}{dx} = y\cos(x)(1 + y)$, which gives $\frac{dy}{y(1+y)} = \frac{\cos(x)}{\sin(x)}\,dx = \cot(x)\,dx$ (correct answer)
4. The equation requires the substitution $u = \tan(x)$ to become separable, since $\frac{du}{dx} = \sec^2(x) = 1 + \tan^2(x)$

Explanation: When you encounter a differential equation, the first step is determining whether it's separable—meaning you can rewrite it so all terms involving one variable (and its differential) are on one side, and all terms involving the other variable are on the opposite side. Looking at $\sin(x)\frac{dy}{dx} = y\cos(x) + y^2\cos(x)$, the key insight is recognizing that the right side can be factored. Notice that both terms contain $\cos(x)$ and involve $y$: $y\cos(x) + y^2\cos(x) = y\cos(x)(1 + y)$. This gives us $\sin(x)\frac{dy}{dx} = y\cos(x)(1 + y)$. Now you can separate variables by dividing both sides appropriately: $\frac{dy}{y(1+y)} = \frac{\cos(x)}{\sin(x)}dx = \cot(x)dx$. This confirms the equation is separable, making choice C correct. Choice A incorrectly assumes that different trigonometric functions automatically prevent separability—this misses the crucial factoring step. Choice B makes an error by dividing by $\cos(x)$ first, which creates $\tan(x)\frac{dy}{dx} = y + y^2$. While this factors the right side, you cannot separate $\frac{dy}{y(1+y)} = \frac{dx}{\tan(x)}$ cleanly since $\frac{1}{\tan(x)} = \cot(x)$ is integrable, but the approach shown is unnecessarily complicated. Choice D suggests a substitution that's completely unnecessary since the equation is already separable. Strategy tip: Always look for factoring opportunities on both sides of a differential equation before concluding it's not separable. Common factors often make separation possible.

Question 16

1. The equation can be made separable by factoring the numerator as $(x+y)(x^2+y^2)$ and cancelling common terms with the denominator
2. The equation is not homogeneous because the numerator has degree 3 while the denominator has degree 2, so standard homogeneous techniques don't apply
3. The equation is homogeneous of degree 1 and becomes separable after the substitution $v = \frac{y}{x}$ and algebraic simplification (correct answer)
4. The equation is separable in its current form after factoring the numerator as $x^3(1 + \frac{y}{x} + (\frac{y}{x})^2 + (\frac{y}{x})^3)$ and the denominator as $x^2(1 - (\frac{y}{x})^2)$

Explanation: When you encounter a differential equation like this, you should immediately check if it's homogeneous by examining whether both the numerator and denominator can be expressed as functions of the same degree in x and y. Let's verify homogeneity by checking if $f(tx,ty) = t^k f(x,y)$ for some power k. For the numerator: $(tx)^3 + (tx)^2(ty) + (tx)(ty)^2 + (ty)^3 = t^3(x^3 + x^2y + xy^2 + y^3)$. For the denominator: $(tx)^2 - (ty)^2 = t^2(x^2 - y^2)$. So our function becomes $\frac{t^3(x^3 + x^2y + xy^2 + y^3)}{t^2(x^2 - y^2)} = t^1 \cdot \frac{x^3 + x^2y + xy^2 + y^3}{x^2 - y^2}$. This confirms the equation is homogeneous of degree 1. For homogeneous equations, the standard technique is substituting $v = \frac{y}{x}$, which gives $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$. After substitution and simplification, this becomes separable. Option A is wrong because factoring $(x+y)(x^2+y^2)$ doesn't match the numerator $x^3 + x^2y + xy^2 + y^3$, and you can't cancel with the denominator. Option B incorrectly concludes the equation isn't homogeneous—while the degrees differ, the ratio gives degree 1 homogeneity. Option D describes factoring that doesn't lead to immediate separability without the homogeneous substitution. Study tip: Always test for homogeneity by checking if $f(tx,ty) = t^k f(x,y)$. If confirmed, use the substitution $v = \frac{y}{x}$ to make it separable.

Question 17

1. The equation is separable because it can be written as $\frac{dy}{dx} = \frac{x(x+y)}{y(y+x)}$, which simplifies to $\frac{dy}{dx} = \frac{x}{y}$ (correct answer)
2. The equation is not separable because the factored form $\frac{x(x+y)}{y(y+x)}$ contains mixed terms that cannot be separated into functions of $x$ alone and $y$ alone
3. The equation is separable only when $x + y \neq 0$, but becomes non-separable along the line $y = -x$
4. The equation is separable because after factoring, it has the form $\frac{dy}{dx} = f(xy)$ where $f$ is a function of the product $xy$ only

Explanation: Factoring gives $\frac{dy}{dx} = \frac{x(x+y)}{y(y+x)} = \frac{x(x+y)}{y(x+y)}$. Since $x+y$ appears in both numerator and denominator, we can cancel (assuming $x+y \neq 0$) to get $\frac{dy}{dx} = \frac{x}{y}$, which is clearly separable as $y\,dy = x\,dx$. Choice B is wrong because it fails to recognize the cancellation. Choice C incorrectly suggests the separability depends on the constraint $x+y \neq 0$, but this is just a domain restriction. Choice D is wrong because this is not a function of $xy$ alone after simplification.

Question 18

1. Factor out $e^x$ from both numerator and denominator to obtain $\frac{dy}{dx} = \frac{e^y - e^{-y}}{e^y + e^{-y}} = \tanh(y)$, which is separable (correct answer)
2. Use the substitution $u = x + y$ and $v = x - y$ to transform the equation into a system that can be analyzed for separability
3. Recognize that $e^{x+y} - e^{x-y} = e^x(e^y - e^{-y})$ and $e^{x+y} + e^{x-y} = e^x(e^y + e^{-y})$, so the equation separates immediately
4. The equation cannot be separated because the exponential terms create non-linear coupling between $x$ and $y$ that cannot be factored into independent functions

Explanation: We have $e^{x+y} = e^x \cdot e^y$ and $e^{x-y} = e^x \cdot e^{-y}$. So the equation becomes $\frac{dy}{dx} = \frac{e^x e^y - e^x e^{-y}}{e^x e^y + e^x e^{-y}} = \frac{e^x(e^y - e^{-y})}{e^x(e^y + e^{-y})} = \frac{e^y - e^{-y}}{e^y + e^{-y}}$. Since $\frac{e^y - e^{-y}}{e^y + e^{-y}} = \tanh(y)$, we get $\frac{dy}{dx} = \tanh(y)$, which separates as $\frac{dy}{\tanh(y)} = dx$ or $\operatorname{coth}(y)\,dy = dx$. Choice B suggests an unnecessarily complex approach. Choice C has the right factorization but poor explanation. Choice D incorrectly concludes non-separability.

Question 19

1. The reasoning is correct; homogeneous equations of equal degree in numerator and denominator are always separable after appropriate algebraic manipulation
2. The reasoning is incorrect; this equation is homogeneous of degree 1, not 2, and requires the substitution $v = \frac{y}{x}$ to become separable (correct answer)
3. The reasoning is partially correct; the equation is homogeneous but becomes separable only after factoring as $\frac{dy}{dx} = \frac{xy(x+y)}{x^2+y^2}$ and making substitutions
4. The reasoning is incorrect; while both numerator and denominator involve degree-2 terms, the equation is not separable in the traditional sense due to mixed polynomial terms

Explanation: The equation $\frac{dy}{dx} = \frac{x^2y + xy^2}{x^2 + y^2}$ can be rewritten as $\frac{dy}{dx} = \frac{xy(x + y)}{x^2 + y^2}$. Dividing numerator and denominator by $x^2$ gives $\frac{dy}{dx} = \frac{\frac{y}{x}(1 + \frac{y}{x})}{1 + (\frac{y}{x})^2}$, which is homogeneous of degree 1 (since $\frac{dy}{dx}$ has degree 1). With $v = \frac{y}{x}$, so $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$, the equation becomes separable in $v$ and $x$. Choice A is wrong about the degree and the general principle. Choice C misunderstands the nature of homogeneous equations. Choice D incorrectly concludes non-separability.