Homogeneous First-Order Equations - Differential Equations

1. A student is solving the equation $x y' = 2x + y$. They correctly identify it as a homogeneous equation and apply the substitution $y=vx$. However, they make a common error in the substitution for $y'$, replacing it with $x\frac{dv}{dx}$ instead of the correct $v+x\frac{dv}{dx}$. Following this incorrect procedure, what general solution do they obtain?
   Answer: $y = Cx^2 - 2x$ This problem tests your understanding of the substitution method for homogeneous differential equations, specifically what happens when you make a common algebraic error during the substitution process. Starting with $xy' = 2x + y$ and using the substitution $y = vx$, you need to find $y'$. The correct derivative is $y' = v + x\frac{dv}{dx}$ using the product rule. However, the student incorrectly uses $y' = x\frac{dv}{dx}$, missing the $v$ term. With this incorrect substitution, the equation becomes: $x \cdot x\frac{dv}{dx} = 2x + vx$ $x^2\frac{dv}{dx} = x(2 + v)$ $x\frac{dv}{dx} = 2 + v$ $\frac{dv}{dx} = \frac{2 + v}{x}$ This separates to $\frac{dv}{2 + v} = \frac{dx}{x}$. Integrating both sides gives $\ln|2 + v| = \ln|x| + C_1$, which leads to $2 + v = Cx$, so $v = Cx - 2$. Since $y = vx$, the incorrect solution is $y = Cx^2 - 2x$. Answer D matches this flawed derivation. Answer A represents the correct solution to the original equation. Answer B looks similar to A but with terms reversed. Answer C suggests an exponential solution, which would arise from a completely different type of equation. Study tip: When using $y = vx$ substitution, always remember that $y' = v + x\frac{dv}{dx}$ by the product rule—the $v$ term is crucial and easily forgotten under exam pressure.
2. The general solution to the differential equation $2xy y' = y^2-x^2$ represents a family of curves. Which of the following provides the best geometric description of this family for $x > 0$?
   Answer: Circles passing through the origin with centers on the x-axis. When you encounter a differential equation that asks about the geometric description of its solution family, your goal is to solve the equation and identify the resulting curve type. This differential equation $2xy y' = y^2 - x^2$ can be solved using substitution. Rearranging gives $y' = \frac{y^2 - x^2}{2xy}$. This suggests trying the substitution $v = \frac{y}{x}$, so $y = vx$ and $y' = v + xv'$. Substituting: $v + xv' = \frac{(vx)^2 - x^2}{2x(vx)} = \frac{x^2(v^2 - 1)}{2x^2v} = \frac{v^2 - 1}{2v}$ This simplifies to $xv' = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = \frac{-v^2 - 1}{2v}$ Separating variables and integrating leads to the general solution: $x^2 + y^2 = 2Cx$, where $C$ is an arbitrary constant. Completing the square gives $(x - C)^2 + y^2 = C^2$, which represents circles with centers at $(C, 0)$ and radius $|C|$. Since the radius equals the x-coordinate of the center, these circles all pass through the origin. Choice A is wrong because these aren't parabolas—they're clearly circular equations. Choice B is incorrect because the centers are at $(C, 0)$, not at the origin. Choice D places centers on the y-axis, but our centers are at $(C, 0)$ on the x-axis. Study tip: When solving differential equations geometrically, always complete the square on the final implicit equation to identify the standard form and geometric properties of the curve family.
3. Consider the initial value problem $\frac{dy}{dx} = \frac{x-y}{x+y}$, with $y(1)=1$. What is the value of $y$ at $x=2$?
   Answer: $-2 + \sqrt{10}$ The equation is homogeneous. Let $v=y/x$. Then $\frac{dy}{dx} = \frac{1-y/x}{1+y/x} = \frac{1-v}{1+v}$. Using $\frac{dy}{dx} = v+x\frac{dv}{dx}$, we get $v+x\frac{dv}{dx} = \frac{1-v}{1+v}$. This leads to $x\frac{dv}{dx} = \frac{1-v}{1+v} - v = \frac{1-v-v-v^2}{1+v} = -\frac{v^2+2v-1}{v+1}$. Separating variables: $\frac{v+1}{v^2+2v-1}dv = -\frac{dx}{x}$. Integrating gives $\frac{1}{2}\ln|v^2+2v-1| = -\ln|x|+C$. Substituting back $v=y/x$ and multiplying by $x^2$ leads to the implicit solution $y^2+2xy-x^2=k$. Using the initial condition $y(1)=1$, we find $1^2+2(1)(1)-1^2 = k$, so $k=2$. The particular solution is $y^2+2xy-x^2=2$. At $x=2$, we have $y^2+4y-4=2$, or $y^2+4y-6=0$. Using the quadratic formula, $y = \frac{-4 \pm \sqrt{16-4(-6)}}{2} = -2 \pm \sqrt{10}$. Since the solution starts at $y(1)=1$ (positive), and the function is continuous, we choose the positive root, $y(2) = -2+\sqrt{10} \approx 1.16$. - B is incorrect as it is the other root from the quadratic formula, which does not match the trajectory of the solution curve starting at $(1,1)$. - C is incorrect and could arise from an error in calculating the constant $k$, for example, finding $k=1$, which leads to the equation $y^2+4y-5=0$ at $x=2$. - D is incorrect and would result from a sign error during separation leading to the implicit solution $y^2+2xy+x^2=k$.
4. Consider the differential equation $\frac{dy}{dx} = \frac{x^2 + xy}{y^2 + xy}$. After applying the substitution $v = \frac{y}{x}$ and simplifying, which of the following represents the resulting separable equation in terms of $v$ and $x$?
   Answer: $\frac{dv}{dx} = \frac{1 + v - v^2}{x(v^2 + v)}$ Starting with $\frac{dy}{dx} = \frac{x^2 + xy}{y^2 + xy}$, divide numerator and denominator by $x^2$ to get $\frac{dy}{dx} = \frac{1 + \frac{y}{x}}{\frac{y^2}{x^2} + \frac{y}{x}}$. With $v = \frac{y}{x}$, this becomes $\frac{dy}{dx} = \frac{1 + v}{v^2 + v}$. Since $y = vx$, we have $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Setting equal: $v + x\frac{dv}{dx} = \frac{1 + v}{v^2 + v}$. Solving for $\frac{dv}{dx}$: $x\frac{dv}{dx} = \frac{1 + v}{v^2 + v} - v = \frac{1 + v - v(v^2 + v)}{v^2 + v} = \frac{1 + v - v^3 - v^2}{v^2 + v} = \frac{1 + v - v^2}{v^2 + v}$. Therefore $\frac{dv}{dx} = \frac{1 + v - v^2}{x(v^2 + v)}$. Choice B omits the $v$ term, choice C incorrectly separates terms, and choice D has the wrong sign.
5. Consider the differential equation $\frac{dy}{dx} = \frac{x^2 - y^2}{x^2 + xy}$. A student claims this equation is homogeneous and applies the substitution $y = vx$. After simplification, they obtain $\frac{dv}{dx} = \frac{1 - 3v^2}{x(1 + v)}$. What error, if any, did the student make?
   Answer: The student correctly identified the equation as homogeneous but made an algebraic error in the substitution First, check if the equation is homogeneous by dividing numerator and denominator by $x^2$: $\frac{dy}{dx} = \frac{1 - \left(\frac{y}{x}\right)^2}{1 + \frac{y}{x}}$. This is indeed a function of $\frac{y}{x}$ only, so it's homogeneous. With $v = \frac{y}{x}$, we have $\frac{dy}{dx} = \frac{1 - v^2}{1 + v}$ and $y = vx$ gives $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Setting equal: $v + x\frac{dv}{dx} = \frac{1 - v^2}{1 + v}$. Solving for $\frac{dv}{dx}$: $x\frac{dv}{dx} = \frac{1 - v^2}{1 + v} - v = \frac{1 - v^2 - v(1 + v)}{1 + v} = \frac{1 - v^2 - v - v^2}{1 + v} = \frac{1 - v - 2v^2}{1 + v}$. Therefore $\frac{dv}{dx} = \frac{1 - v - 2v^2}{x(1 + v)}$. The student wrote $\frac{1 - 3v^2}{x(1 + v)}$, missing the $-v$ term in the numerator. Choice A is wrong because the equation is homogeneous. Choice C is unnecessary since $y = vx$ is the standard substitution. Choice D is incorrect due to the algebraic error.
6. The general solution to the homogeneous differential equation $x \frac{dy}{dx} = y + \sqrt{y^2-x^2}$ for $y>x>0$ is sought. Which of the following equations implicitly defines the solution family?
   Answer: $y + \sqrt{y^2-x^2} = Cx^2$ First, divide by $x$ to get $\frac{dy}{dx} = \frac{y}{x} + \sqrt{(\frac{y}{x})^2-1}$. Let $v=y/x$, so $\frac{dy}{dx}=v+x\frac{dv}{dx}$. The equation becomes $v+x\frac{dv}{dx} = v + \sqrt{v^2-1}$. This simplifies to $x\frac{dv}{dx} = \sqrt{v^2-1}$. Separating variables yields $\frac{dv}{\sqrt{v^2-1}} = \frac{dx}{x}$. Integrating both sides gives $\ln|v+\sqrt{v^2-1}| = \ln|x| + C_1$. Let $C_1 = \ln|C|$, so $\ln|v+\sqrt{v^2-1}| = \ln|Cx|$. This implies $v+\sqrt{v^2-1} = Cx$. Substituting $v=y/x$ back gives $\frac{y}{x} + \sqrt{(\frac{y}{x})^2-1} = Cx$, which simplifies to $\frac{y+\sqrt{y^2-x^2}}{x} = Cx$. Therefore, $y+\sqrt{y^2-x^2} = Cx^2$. - B is incorrect due to a common integration error, confusing the integral of $1/\sqrt{v^2-1}$ with that of $1/\sqrt{1-v^2}$. - C is incorrect and would result from a sign error, for instance starting with $x\frac{dv}{dx} = -\sqrt{v^2-1}$. - D is incorrect due to an algebraic error in the final step, failing to multiply $Cx$ by $x$.
7. The solution curve for the differential equation $(x+ye^{y/x})dx - xe^{y/x}dy=0$ passes through the point $(1, 0)$. What is the value of $y(e)$?
   Answer: $e\ln(2)$ This differential equation has the form $M(x,y)dx + N(x,y)dy = 0$ where you need to check if it's exact or can be made exact through an integrating factor. Here, $M = x + ye^{y/x}$ and $N = -xe^{y/x}$. Testing for exactness: $\frac{\partial M}{\partial y} = e^{y/x} + ye^{y/x} \cdot \frac{1}{x}$ and $\frac{\partial N}{\partial x} = -e^{y/x} - xe^{y/x} \cdot \frac{(-y)}{x^2} = -e^{y/x} + \frac{y}{x}e^{y/x}$. Since these aren't equal, the equation isn't exact. However, this equation is homogeneous since both terms have the same degree. Use the substitution $v = \frac{y}{x}$, so $y = vx$ and $dy = v dx + x dv$. Substituting: $(x + vxe^v)dx - xe^v(v dx + x dv) = 0$ Simplifying: $x dx + vxe^v dx - vxe^v dx - x^2e^v dv = 0$ This reduces to: $x dx - x^2e^v dv = 0$ Dividing by $x$: $dx - xe^v dv = 0$, or $\frac{dx}{x} = e^v dv$ Integrating: $\ln|x| = e^v + C$ Since $v = \frac{y}{x}$: $\ln|x| = e^{y/x} + C$ Using the initial condition $(1,0)$: $\ln(1) = e^{0/1} + C$, so $0 = 1 + C$, giving $C = -1$. The solution is $\ln|x| = e^{y/x} - 1$. For $x = e$: $\ln(e) = e^{y/e} - 1$, so $1 = e^{y/e} - 1$, meaning $e^{y/e} = 2$. Taking the natural log: $\frac{y}{e} = \ln(2)$, so $y = e\ln(2)$. Answer D is correct. Options A, B, and C result from algebraic errors in the integration process or incorrectly applying the initial condition. Strategy tip: For homogeneous differential equations, always try the substitution $v = y/x$ to separate variables effectively.
8. The general solution to $x\,dy - y\,dx = \sqrt{x^2-y^2}\,dx$ for $x>0$ is found using a substitution $y=vx$. The resulting separated equation can be written as $\frac{dv}{g(v)} = \frac{dx}{x}$. What is the function $g(v)$?
   Answer: $\sqrt{1-v^2}$ When you encounter a differential equation with terms like $x\,dy - y\,dx$ alongside expressions involving $\sqrt{x^2-y^2}$, this signals a homogeneous differential equation that's perfect for the substitution $y = vx$. Let's work through the substitution systematically. Starting with $y = vx$, we get $dy = v\,dx + x\,dv$. Substituting into the original equation: $x(v\,dx + x\,dv) - vx\,dx = \sqrt{x^2-(vx)^2}\,dx$ Expanding the left side: $xv\,dx + x^2\,dv - vx\,dx = \sqrt{x^2-v^2x^2}\,dx$ The $xv\,dx$ terms cancel, leaving: $x^2\,dv = \sqrt{x^2(1-v^2)}\,dx$ Since $x > 0$, we have $\sqrt{x^2(1-v^2)} = x\sqrt{1-v^2}$, so: $x^2\,dv = x\sqrt{1-v^2}\,dx$ Dividing both sides by $x$: $x\,dv = \sqrt{1-v^2}\,dx$ Separating variables: $\frac{dv}{\sqrt{1-v^2}} = \frac{dx}{x}$ Therefore, $g(v) = \sqrt{1-v^2}$, which is choice C. Looking at the wrong answers: A has an incorrect negative sign that would flip the direction of integration. B adds an extra $v$ term that doesn't appear in our derivation. D uses $\sqrt{v^2-1}$ instead of $\sqrt{1-v^2}$, which would only be valid if $|v| \geq 1$, contradicting our domain requirements. Study tip: When using $y = vx$ substitution, always remember that $dy = v\,dx + x\,dv$ and carefully track sign changes during algebraic manipulation.
9. A differential equation of the form $\frac{dy}{dx} = \frac{ax + by}{cx + dy}$ is homogeneous if and only if a certain condition is satisfied. Given the equation $\frac{dy}{dx} = \frac{3x + 2y}{x + ky}$ where $k$ is a parameter, for which value(s) of $k$ can this equation be solved using the substitution $v = \frac{y}{x}$?
   Answer: Any real value of $k$ including $k = 0$ A differential equation $\frac{dy}{dx} = \frac{ax + by}{cx + dy}$ is homogeneous when both the numerator and denominator are homogeneous functions of the same degree. For $\frac{dy}{dx} = \frac{3x + 2y}{x + ky}$, both the numerator $3x + 2y$ and denominator $x + ky$ are homogeneous functions of degree 1 (each term has total degree 1 in $x$ and $y$). This is true regardless of the value of $k$. We can verify by dividing numerator and denominator by $x$: $\frac{dy}{dx} = \frac{3 + 2\frac{y}{x}}{1 + k\frac{y}{x}}$, which is a function of $\frac{y}{x}$ alone for any value of $k$, including $k = 0$ (which gives $\frac{dy}{dx} = 3 + 2\frac{y}{x}$). The substitution $v = \frac{y}{x}$ will work for any value of $k$. Some students might think $k = 0$ causes problems because the denominator becomes just $x$, but this doesn't prevent the homogeneous substitution from working. Others might think there's a specific relationship needed between the coefficients, but homogeneity only requires that all terms have the same degree, which is satisfied here for any $k$.
10. Consider a solution curve $y=f(x)$ for a homogeneous first-order differential equation $\frac{dy}{dx} = G(y/x)$. If this curve passes through the point $(a, b)$, which of the following statements about other points and curves must be true?
    Answer: The curve defined by $y_c(x) = c f(x/c)$ for any constant $c > 0$ is also a solution curve. A key geometric property of homogeneous equations is that their direction fields are constant along any ray from the origin. This means that if the $(x,y)$ plane is scaled by a factor $c$, the solution curves are mapped onto other solution curves. If $y=f(x)$ is a solution, then the scaled function $y_c(x) = c f(x/c)$ is also a solution. This can be verified by differentiation: $\frac{dy_c}{dx} = c f'(x/c) \cdot (1/c) = f'(x/c)$. Since $f(x)$ is a solution, $f'(x) = G(f(x)/x)$. Thus, $\frac{dy_c}{dx} = G(f(x/c)/(x/c))$. Also, $G(y_c/x) = G(cf(x/c)/x) = G(f(x/c)/(x/c))$. So, $\frac{dy_c}{dx} = G(y_c/x)$, and $y_c(x)$ is a solution. - A is incorrect. This is only true for the simple case $y' = y/x$. - C is incorrect. The slope depends only on the ratio $y/x$. At $(ka, kb)$, the slope is $G(kb/ka) = G(b/a)$, which is the same as the slope at $(a,b)$. - D is incorrect. This describes vertical shifting, which is a property of solutions to equations of the form $y' = h(x)$, not homogeneous equations.
11. Given the initial value problem $\frac{dy}{dx} = \frac{y}{x} + \tan(\frac{y}{x})$ with $y(1) = \pi/6$, what is the value of $y(\sqrt{3})$?
    Answer: $\frac{\pi}{\sqrt{3}}$ When you encounter a differential equation of the form $\frac{dy}{dx} = f(\frac{y}{x})$, you're dealing with a homogeneous equation. The key strategy is to use the substitution $v = \frac{y}{x}$, which means $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Applying this substitution to $\frac{dy}{dx} = \frac{y}{x} + \tan(\frac{y}{x})$, we get: $v + x\frac{dv}{dx} = v + \tan(v)$ This simplifies to $x\frac{dv}{dx} = \tan(v)$, or $\frac{dv}{\tan(v)} = \frac{dx}{x}$. Since $\frac{1}{\tan(v)} = \cot(v)$, we have $\cot(v)dv = \frac{dx}{x}$. Integrating both sides: $\ln|\sin(v)| = \ln|x| + C$. This gives us $\sin(v) = Cx$, or $\sin(\frac{y}{x}) = Cx$. Using the initial condition $y(1) = \frac{\pi}{6}$: $\sin(\frac{\pi/6}{1}) = C \cdot 1$, so $C = \sin(\frac{\pi}{6}) = \frac{1}{2}$. Therefore, $\sin(\frac{y}{x}) = \frac{x}{2}$. At $x = \sqrt{3}$: $\sin(\frac{y}{\sqrt{3}}) = \frac{\sqrt{3}}{2}$. Since $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, we have $\frac{y}{\sqrt{3}} = \frac{\pi}{3}$, giving us $y = \frac{\pi}{\sqrt{3}}$. Choice A incorrectly uses the original substitution without proper integration. Choice B forgets to multiply back by $x$. Choice C appears to involve computational errors in the trigonometric evaluation. Remember: homogeneous differential equations always call for the substitution $v = \frac{y}{x}$ to separate variables effectively.
12. Find the particular solution to the differential equation $(x - y) dy = (x + y) dx$ that satisfies the initial condition $y(1) = 1$.
    Answer: $2\arctan(y/x) - \ln(x^2+y^2) = \pi/2 - \ln(2)$ Rewrite as $dy/dx = (x+y)/(x-y) = (1+y/x)/(1-y/x)$. Let $y=vx$, so $y' = v+xv'$. The equation becomes $v+xv' = (1+v)/(1-v)$. Then $xv' = (1+v)/(1-v) - v = (1+v - v + v^2)/(1-v) = (1+v^2)/(1-v)$. Separate variables: $((1-v)/(1+v^2)) dv = dx/x$. Integrate: $\int (1/(1+v^2) - v/(1+v^2)) dv = \int dx/x$. This gives $\arctan(v) - (1/2)\ln(1+v^2) = \ln|x| + C$. Substitute back $v=y/x$: $\arctan(y/x) - (1/2)\ln(1+(y/x)^2) = \ln|x| + C$. This simplifies to $\arctan(y/x) - (1/2)\ln(x^2+y^2) = C$. Using $y(1)=1$: $\arctan(1) - (1/2)\ln(1^2+1^2) = C$, so $C = \pi/4 - (1/2)\ln(2)$. The solution is $\arctan(y/x) - (1/2)\ln(x^2+y^2) = \pi/4 - (1/2)\ln(2)$. Multiplying by 2 gives $2\arctan(y/x) - \ln(x^2+y^2) = \pi/2 - \ln(2)$.
13. A solution to the differential equation $x dy = (y + \sqrt{x^2+y^2}) dx$ for $x>0$ passes through the point $(1, 0)$. What is the y-intercept of the line tangent to this solution curve at $x=\sqrt{3}$?
    Answer: $-2$ When you encounter a differential equation like this one, recognize it as a homogeneous equation that can be solved using the substitution method. The presence of $\sqrt{x^2+y^2}$ is a key indicator that the substitution $v = \frac{y}{x}$ (so $y = vx$) will simplify the equation. Making this substitution and using $\frac{dy}{dx} = v + x\frac{dv}{dx}$, the equation becomes: $x(v + x\frac{dv}{dx}) = v + \sqrt{1+v^2}$ This simplifies to $x\frac{dv}{dx} = \sqrt{1+v^2}$, which separates as: $\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$ Integrating both sides gives $\sinh^{-1}(v) = \ln|x| + C$. Using the initial condition $(1,0)$ where $v = 0$, we find $C = 0$. Therefore $v = \sinh(\ln x) = \frac{x^2-1}{2x}$, so $y = \frac{x^2-1}{2}$. At $x = \sqrt{3}$, we have $y = \frac{3-1}{2} = 1$ and $\frac{dy}{dx} = x = \sqrt{3}$. The tangent line at $(\sqrt{3}, 1)$ with slope $\sqrt{3}$ is: $y - 1 = \sqrt{3}(x - \sqrt{3})$ $y = \sqrt{3}x - 2$ The y-intercept is $-2$, which is choice C. Choice A ($0$) might come from incorrectly using the initial point. Choice B ($-1$) could result from calculation errors in finding the slope or intercept. Choice D ($1-2\sqrt{3}$) might arise from sign errors or incorrect substitution of the point coordinates. Always verify your solution satisfies both the differential equation and initial condition before finding tangent lines.
14. The differential equation $x\frac{dy}{dx} = y + \sqrt{x^2 + y^2}$ can be solved using a homogeneous substitution. After making the appropriate substitution and simplification, the resulting equation in terms of the new variable requires which integration technique?
    Answer: Trigonometric substitution using $v = \tan(\theta)$ to handle the square root Dividing by $x$: $\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2}$. With $v = \frac{y}{x}$, this becomes $v + x\frac{dv}{dx} = v + \sqrt{1 + v^2}$, so $x\frac{dv}{dx} = \sqrt{1 + v^2}$. Separating: $\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$. The left side integral $\int \frac{dv}{\sqrt{1 + v^2}}$ is a standard form that requires the substitution $v = \tan(\theta)$, $dv = \sec^2(\theta)d\theta$, and $\sqrt{1 + v^2} = \sqrt{1 + \tan^2(\theta)} = \sec(\theta)$. This gives $\int \frac{\sec^2(\theta)d\theta}{\sec(\theta)} = \int \sec(\theta)d\theta = \ln|\sec(\theta) + \tan(\theta)| + C$. Choice A is incorrect because direct integration isn't possible with the square root. Choice C is wrong because integration by parts isn't needed. Choice D is incorrect because partial fractions don't apply to this radical expression.
15. The differential equation $\frac{dy}{dx} = \frac{2xy}{y^2-x^2}$ can be solved as a homogeneous equation. While both standard substitutions ($y=vx$ and $x=vy$) are valid, one leads to a significantly simpler integration process. Why is the substitution $x=vy$ preferable in this case?
    Answer: Because the resulting separable equation for $v$ can be solved with a basic $u$-substitution, whereas the alternative requires partial fractions. Let's analyze both substitutions. With $y=vx$, the equation becomes $v+x\frac{dv}{dx} = \frac{2v}{v^2-1}$, which leads to $x\frac{dv}{dx} = \frac{3v-v^3}{v^2-1}$. The separated form is $\frac{v^2-1}{v(3-v^2)}dv = \frac{dx}{x}$, which requires integration using partial fractions. Now consider $x=vy$. This implies $\frac{dx}{dy} = \frac{y^2-x^2}{2xy} = \frac{1-(x/y)^2}{2(x/y)}$. Let $v=x/y$, so $\frac{dx}{dy} = v+y\frac{dv}{dy}$. The equation becomes $v+y\frac{dv}{dy} = \frac{1-v^2}{2v}$, which leads to $y\frac{dv}{dy} = \frac{1-3v^2}{2v}$. The separated form is $\frac{2v}{1-3v^2}dv = \frac{dy}{y}$. The integral on the left can be solved with a simple $u$-substitution ($u=1-3v^2$). Thus, the substitution $x=vy$ is preferable because it leads to a much simpler integral. - A is incorrect as this reasoning is heuristic and not mathematically precise. - B is incorrect because the substitution leads to a separable equation, not a linear one. - D is incorrect because while domain issues can be a concern, they are not the primary reason for preferring one substitution over the other based on computational efficiency.
16. The general solution to the homogeneous differential equation $x \frac{dy}{dx} = y + x e^{y/x}$ can be written in the implicit form $f(x, y) = C$. Which of the following is a valid expression for $f(x, y)$?
    Answer: $ln|x| + e^{-y/x}$ First, rewrite the equation as $\frac{dy}{dx} = \frac{y}{x} + e^{y/x}$. This is a homogeneous equation. Use the substitution $y = vx$, which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$. The equation becomes $v + x \frac{dv}{dx} = v + e^v$. This simplifies to $x \frac{dv}{dx} = e^v$. Separating variables, we get $e^{-v} dv = \frac{dx}{x}$. Integrating both sides yields $\int e^{-v} dv = \int \frac{dx}{x}$, which gives $-e^{-v} = ln|x| + C_1$. Substituting back $v = y/x$, we have $-e^{-y/x} = ln|x| + C_1$. Rearranging to the form $f(x, y) = C$, we get $ln|x| + e^{-y/x} = -C_1$. Letting $C = -C_1$, the solution is $ln|x| + e^{-y/x} = C$.
17. A homogeneous differential equation of the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ is given by $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$. After applying the substitution $v = \frac{y}{x}$ and reaching the separated form, the student obtains $\int \frac{2v}{v^2 - 1} dv = \int \frac{dx}{x}$. Which of the following represents the correct evaluation of the left-hand integral?
    Answer: $\ln|v^2 - 1| + C$ When you encounter a homogeneous differential equation and need to evaluate an integral after substitution, the key is recognizing the correct integration technique and applying it carefully. To evaluate $\int \frac{2v}{v^2 - 1} dv$, you should use substitution. Let $u = v^2 - 1$, so $du = 2v \, dv$. This transforms the integral to $\int \frac{1}{u} du = \ln|u| + C$. Substituting back gives $\ln|v^2 - 1| + C$. Answer C is correct because it represents the proper application of the natural logarithm integration rule after the appropriate substitution. Answer A, $\ln|v^2 - 1|^{1/2} + C$, incorrectly applies logarithm properties. While $\ln|v^2 - 1|^{1/2} = \frac{1}{2}\ln|v^2 - 1|$, this suggests the student might have confused the coefficient placement. Answer B, $\frac{1}{2}\ln|v^2 - 1| + C$, represents a common error where students incorrectly handle the coefficient 2 in the numerator. This would be the result if you mistakenly treated the integral as $\frac{1}{2} \int \frac{2v}{v^2 - 1} dv$ instead of recognizing the perfect substitution. Answer D uses partial fraction decomposition, which is unnecessary here. While $\frac{2v}{v^2 - 1}$ could theoretically be decomposed, the direct substitution method is more efficient and less error-prone. Study tip: When you see a rational function where the numerator is the derivative (or a constant multiple) of the denominator, always try $u$-substitution first before attempting partial fractions. This pattern appears frequently in differential equations.
18. A particular solution to the differential equation $(x^2 + 3y^2) dx - 2xy dy = 0$ passes through the point $(1, 1)$. What is the value of $y$ when $x=2$?
    Answer: $2\sqrt{3}$ The equation can be rewritten as $dy/dx = (x^2 + 3y^2) / (2xy)$. Dividing the numerator and denominator by $x^2$ gives $dy/dx = (1 + 3(y/x)^2) / (2(y/x))$, which is a homogeneous equation. Let $v = y/x$, so $y = vx$ and $dy/dx = v + x(dv/dx)$. The equation becomes $v + x(dv/dx) = (1 + 3v^2)/(2v)$. Rearranging gives $x(dv/dx) = (1 + 3v^2)/(2v) - v = (1 + v^2)/(2v)$. Separating variables yields $(2v / (1 + v^2)) dv = dx/x$. Integrating both sides gives $ln(1 + v^2) = ln|x| + C$. Using the initial condition $y(1)=1$, we have $x=1$ and $v=y/x=1$. So, $ln(1+1^2) = ln(1) + C$, which implies $C = ln(2)$. The solution is $ln(1+v^2) = ln|x| + ln(2) = ln(2|x|)$. For $x>0$, this is $1+v^2 = 2x$. Substituting back $v=y/x$, we get $1 + (y/x)^2 = 2x$, which simplifies to $x^2 + y^2 = 2x^3$. To find $y$ when $x=2$, we substitute $x=2$ into the solution: $2^2 + y^2 = 2(2^3)$, which is $4 + y^2 = 16$. Thus, $y^2 = 12$, and since the initial condition has $y>0$, we take the positive root, $y = \sqrt{12} = 2\sqrt{3}$.
19. Find the explicit solution to the initial value problem $y' = \frac{y}{x} + \cos^2(\frac{y}{x})$, with $y(1) = \pi/4$.
    Answer: $y = x \arctan(\ln x + 1)$ When you encounter a differential equation of the form $y' = f\left(\frac{y}{x}\right)$, this signals a homogeneous equation that requires the substitution $v = \frac{y}{x}$, making $y = vx$ and $y' = v'x + v$. Applying this substitution to $y' = \frac{y}{x} + \cos^2\left(\frac{y}{x}\right)$: $v'x + v = v + \cos^2(v)$ $v'x = \cos^2(v)$ $\frac{dv}{\cos^2(v)} = \frac{dx}{x}$ Integrating both sides: $\int \sec^2(v) \, dv = \int \frac{dx}{x}$ This gives us $\tan(v) = \ln|x| + C$, so $v = \arctan(\ln|x| + C)$. Since $y = vx$, we have $y = x \arctan(\ln|x| + C)$. Using the initial condition $y(1) = \pi/4$: $\frac{\pi}{4} = 1 \cdot \arctan(\ln(1) + C) = \arctan(0 + C) = \arctan(C)$ Therefore $C = \tan(\pi/4) = 1$, giving us $y = x \arctan(\ln x + 1)$, which is choice B. Choice A omits the constant of integration. Choice C incorrectly assumes a linear relationship without proper integration of the trigonometric term. Choice D uses arccotangent instead of arctangent and has the wrong sign pattern. For homogeneous differential equations, always remember the key substitution $v = \frac{y}{x}$ and don't forget to apply your initial conditions to find the integration constant.
20. A homogeneous differential equation has the form $dy/dx = F(y/x)$. When the substitution $y=vx$ is made, the equation transforms into a separable equation of the form $x \frac{dv}{dx} = f(v)$. If $F(v) = \frac{1-v}{1+v}$, which of the following expressions represents the integral $\int \frac{1}{f(v)} dv$?
    Answer: $-\frac{1}{2}\ln|1-2v-v^2| + C$ When you encounter a homogeneous differential equation problem involving the substitution $y = vx$, you need to carefully track how the transformation affects the equation structure. Starting with $\frac{dy}{dx} = F(y/x) = \frac{1-v}{1+v}$ where $v = y/x$, the substitution $y = vx$ gives us $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Setting this equal to $F(v)$: $v + x\frac{dv}{dx} = \frac{1-v}{1+v}$ Solving for $x\frac{dv}{dx}$: $x\frac{dv}{dx} = \frac{1-v}{1+v} - v = \frac{1-v-v(1+v)}{1+v} = \frac{1-v-v-v^2}{1+v} = \frac{1-2v-v^2}{1+v}$ Therefore, $f(v) = \frac{1-2v-v^2}{1+v}$, and we need $\int \frac{1}{f(v)} dv = \int \frac{1+v}{1-2v-v^2} dv$. To evaluate this integral, notice that the derivative of the denominator is $-2-2v$, which is $-2(1+v)$. This means: $\int \frac{1+v}{1-2v-v^2} dv = -\frac{1}{2}\int \frac{-2(1+v)}{1-2v-v^2} dv = -\frac{1}{2}\ln|1-2v-v^2| + C$ This confirms answer B is correct. Answer A has the wrong sign on the coefficient. Answer C appears to come from incorrectly attempting partial fractions or making algebraic errors in the substitution. Answer D oversimplifies the denominator structure. Remember: in homogeneous equation problems, always double-check your algebra when computing $f(v) = F(v) - v$, as small errors here cascade through the entire solution.

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