Classifying ODEs - Differential Equations

1. A first-order differential equation is called homogeneous if it can be written in the form $\frac{dy}{dx} = F(\frac{y}{x})$. Which of the following equations has this property?
   Answer: $\frac{dy}{dx} = \frac{y^2 + xy}{x^2}$ To check if an equation is homogeneous, we try to write the right-hand side as a function of the ratio $v = y/x$. For choice B, we can divide the numerator and denominator by $x^2$: $\frac{dy}{dx} = \frac{y^2/x^2 + xy/x^2}{x^2/x^2} = \frac{(y/x)^2 + (y/x)}{1} = v^2 + v$. This is a function of $v=y/x$, so the equation is homogeneous. The other choices cannot be written solely as a function of $y/x$. For example, choice A becomes $x + y/x$, and choice D becomes $\sin(y/x)+x$; the presence of $x$ as a separate term violates the condition.
2. To solve the differential equation $(x+y)dx - x dy = 0$, a standard strategy is to use a substitution that transforms it into a separable equation. Which substitution is most appropriate for this purpose?
   Answer: The substitution $v = y/x$, because the equation is homogeneous. First, classify the equation. We can rewrite it as $\frac{dy}{dx} = \frac{x+y}{x} = 1 + \frac{y}{x}$. Since the right-hand side can be expressed as a function of the ratio $y/x$, the equation is homogeneous. The standard substitution for homogeneous equations is $v = y/x$ (or $y=vx$). This substitution is guaranteed to transform the equation into one that is separable in the variables $v$ and $x$.
3. A chemical reaction rate is modeled by the differential equation $\frac{dC}{dt} = k(C_0 - C)^2$, where $C(t)$ is the concentration of a product at time $t$, $C_0$ is the initial concentration of a reactant, and $k$ is a positive rate constant. A modification to the experiment introduces a catalyst that decays over time, making the 'rate constant' a function of time, $k(t) = k_0 e^{-\alpha t}$. Which classification accurately describes the modified differential equation?
   Answer: First-order, separable, and non-autonomous. The modified differential equation is $\frac{dC}{dt} = k_0 e^{-\alpha t}(C_0 - C)^2$. 1. Order: The highest derivative is $\frac{dC}{dt}$, so it is a first-order equation. 2. Linearity: The dependent variable $C$ appears in the term $(C_0 - C)^2 = C_0^2 - 2C_0C + C^2$. The presence of the $C^2$ term makes the equation non-linear. 3. Autonomy: The independent variable $t$ appears explicitly in the term $e^{-\alpha t}$. Therefore, the equation is non-autonomous. 4. Separability: The equation can be written as $\frac{dC}{(C_0-C)^2} = k_0 e^{-\alpha t} dt$. Since all terms involving $C$ are on one side and all terms involving $t$ are on the other, the equation is separable. Combining these, the best description is first-order, non-linear, non-autonomous, and separable. Choice B includes 'separable' and 'non-autonomous', which are key features. Choice D incorrectly omits that it is non-autonomous.
4. The differential equation $\frac{dy}{dx} = \frac{2y}{x}$ can be solved using several methods because it fits multiple classifications. Which of the following classifications does NOT apply to this equation?
   Answer: Exact Let's check each classification. (A) Separable: We can write $\frac{1}{y} dy = \frac{2}{x} dx$, so it is separable. (B) Linear: We can write $\frac{dy}{dx} - \frac{2}{x}y = 0$, which is in the form $y' + P(x)y = Q(x)$, so it is linear. (C) Homogeneous: The function $F(x,y) = \frac{2y}{x}$ satisfies $F(tx, ty) = \frac{2(ty)}{tx} = \frac{2y}{x} = F(x,y)$, so it is homogeneous. (D) Exact: Rewriting as $2y dx - x dy = 0$, we have $M=2y$ and $N=-x$. Then $\frac{\partial M}{\partial y} = 2$ and $\frac{\partial N}{\partial x} = -1$. Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact.
5. The equation $(x^2 + y^2)dx + k xy dy = 0$ is given, where $k$ is a constant. For what value of $k$ does this equation have a specific classification that allows for the most straightforward solution method?
   Answer: $k = 2$, making the equation exact and solvable by direct integration For the equation to be exact, we need $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ where $M = x^2 + y^2$ and $N = kxy$. Computing: $\frac{\partial M}{\partial y} = 2y$ and $\frac{\partial N}{\partial x} = ky$. Setting these equal: $2y = ky$, which gives $k = 2$. Choice B is incorrect because the equation is not homogeneous for any value of $k$ (it cannot be written as $f(y/x)$). Choice C is incorrect because $k = 1$ gives $\frac{\partial M}{\partial y} = 2y \neq y = \frac{\partial N}{\partial x}$. Choice D is incorrect because $k = 0$ makes $N = 0$, which doesn't lead to a separable form.
6. The initial value problem $y' = \sqrt{y^2 + x^2}$ with $y(1) = 2$ is given. How does the initial condition $y(1)=2$ influence the classification of the differential equation itself?
   Answer: The initial condition has no effect on the classification of the differential equation. The classification of a differential equation depends only on its algebraic form, involving the variables and their derivatives. Classifications like order, linearity, autonomy, and type (e.g., separable, exact, homogeneous) are properties of the equation $y' = f(x,y)$. An initial condition, such as $y(1)=2$, provides a specific point through which a particular solution must pass. It is used to find the value of the constant of integration after the general solution has been found. It does not alter the fundamental structure of the differential equation itself. Therefore, the initial condition has no effect on the classification.
7. The substitution $y = x^r$ is a key step in solving a specific class of second-order linear differential equations. For which of the following equations would this approach be most effective?
   Answer: $x^2 y'' + 5xy' + 4y = 0$ When you encounter the substitution $y = x^r$ in differential equations, you're dealing with Euler equations (also called Cauchy-Euler equations). These are second-order linear equations where each term has the same degree when you consider the order of differentiation plus the power of $x$. Let's examine why option D works perfectly. For $x^2 y'' + 5xy' + 4y = 0$, when you substitute $y = x^r$, you get $y' = rx^{r-1}$ and $y'' = r(r-1)x^{r-2}$. Substituting these: $x^2 \cdot r(r-1)x^{r-2} + 5x \cdot rx^{r-1} + 4x^r = 0$ This simplifies to $x^r[r(r-1) + 5r + 4] = 0$, giving you the characteristic equation $r^2 + 4r + 4 = 0$. Option A fails because it's a constant-coefficient equation with a non-homogeneous term $\ln(x)$ - you'd use variation of parameters, not the $x^r$ substitution. Option B has constant coefficients but includes an $xy'$ term that doesn't fit the Euler equation pattern. Option C is close but incorrect - the middle term should be $5xy'$, not just $5y'$, to maintain the proper degree structure. The key pattern to remember: Euler equations have the form $ax^2y'' + bxy' + cy = 0$ where each term's total degree (power of $x$ plus derivative order) equals 2. When you see this structure, immediately think "$y = x^r$ substitution."
8. A researcher encounters the equation $\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = xe^{2x}$. To solve this second-order linear ODE, which approach correctly addresses both the homogeneous and particular solution components?
   Answer: Find characteristic equation $r^2 - 4r + 4 = 0$; use method of undetermined coefficients with trial solution $y_p = Ax^2(Cx + D)e^{2x}$ The characteristic equation $r^2 - 4r + 4 = (r-2)^2 = 0$ has repeated root $r = 2$, giving homogeneous solution $y_h = (C_1 + C_2x)e^{2x}$. Since the forcing term $xe^{2x}$ has the same exponential factor as the homogeneous solution, and $x$ already appears in $y_h$, we need to multiply by $x^2$ (the multiplicity of the root) to avoid duplication. Thus $y_p = x^2(Ax + B)e^{2x} = (Ax^3 + Bx^2)e^{2x}$. Choice A fails because it doesn't account for the repeated root. Choice C would work but is unnecessarily complex. Choice D is inefficient for this type of problem.
9. A student attempts to solve $(3xy+y^2)dx + (x^2+xy)dy=0$ and finds it is not exact. They correctly determine that multiplying by an integrating factor $\mu(x)$ will make it exact. Which property of the equation's coefficients, $M(x,y)$ and $N(x,y)$, is the necessary and sufficient condition that justifies this specific approach?
   Answer: The expression $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$ simplifies to a function of $x$ only. When you encounter a differential equation that's not exact, you need to find an integrating factor to make it exact. The key is identifying which type of integrating factor exists based on specific conditions involving the coefficients. For the equation $(3xy+y^2)dx + (x^2+xy)dy=0$, we have $M(x,y) = 3xy+y^2$ and $N(x,y) = x^2+xy$. Let's check if an integrating factor $\mu(x)$ (depending only on $x$) exists. First, calculate the partial derivatives: $\frac{\partial M}{\partial y} = 3x+2y$ and $\frac{\partial N}{\partial x} = 2x+y$. Now compute $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{1}{x^2+xy}[(3x+2y)-(2x+y)] = \frac{x+y}{x(x+y)} = \frac{1}{x}$. Since this expression depends only on $x$, an integrating factor $\mu(x)$ exists, confirming answer C is correct. Answer A is wrong because while both functions are homogeneous of degree 2, this property alone doesn't guarantee the existence of an integrating factor depending only on $x$. Answer B incorrectly reverses the roles of $M$ and $N$ in the formula - this expression would indicate an integrating factor depending only on $y$. Answer D is incorrect because the equation cannot be easily rearranged into standard first-order linear form. Remember: when checking for integrating factors, if $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$ is a function of $x$ only, then $\mu(x)$ exists. If $\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$ is a function of $y$ only, then $\mu(y)$ exists.
10. A differential equation has the form $F\left(x, y, \frac{dy}{dx}\right) = 0$ where $F$ cannot be solved explicitly for $\frac{dy}{dx}$, but can be written as $p^2 + 2xyp - y^2 = 0$ where $p = \frac{dy}{dx}$. What classification and solution strategy should be used?
    Answer: Lagrange equation; solve the quadratic for $p$ in terms of $x$ and $y$, then solve the resulting first-order equations When you encounter a differential equation where $F(x, y, \frac{dy}{dx}) = 0$ cannot be solved explicitly for $\frac{dy}{dx}$, you need to classify the equation type to choose the right solution strategy. The key is recognizing the structure and applying the appropriate method. The equation $p^2 + 2xyp - y^2 = 0$ (where $p = \frac{dy}{dx}$) is quadratic in $p$. This is a Lagrange equation - a differential equation that's quadratic in the first derivative. The standard approach is to solve the quadratic equation for $p$ using the quadratic formula, giving you $p = \frac{-2xy \pm \sqrt{4x^2y^2 + 4y^2}}{2} = -xy \pm y\sqrt{x^2 + 1}$. This yields two expressions for $\frac{dy}{dx}$, each giving you a separable first-order differential equation to solve. Option B is incorrect because Clairaut equations have the specific form $y = px + f(p)$, which this equation doesn't match. Option C misses the point - while the equation is implicit, the systematic approach for quadratic-in-$p$ equations is to solve the quadratic first, not use repeated implicit differentiation. Option D incorrectly suggests introducing a parameter when we already have a workable quadratic structure. Study tip: When you see a differential equation that's quadratic in $p = \frac{dy}{dx}$, immediately think "Lagrange equation" and solve the quadratic for $p$. This transforms one complex equation into two simpler first-order equations you can handle with standard techniques.
11. To find the general solution for the ordinary differential equation $y'' - 4y' + 4y = xe^{2x}$, which combination of classifications and corresponding solution methods should be used?
    Answer: The equation is a non-homogeneous linear equation with constant coefficients, solved by finding a complementary function from a characteristic equation with a repeated root and a particular solution using undetermined coefficients. The equation is a second-order, linear, non-homogeneous ODE with constant coefficients. The solution is $y = y_c + y_p$. First, solve the associated homogeneous equation $y'' - 4y' + 4y = 0$. The characteristic equation is $r^2 - 4r + 4 = 0$, or $(r-2)^2 = 0$, which has a repeated real root $r=2$. Thus, the complementary function is $y_c = c_1e^{2x} + c_2xe^{2x}$. For the particular solution $y_p$, the method of undetermined coefficients is appropriate. Since the term $xe^{2x}$ on the right-hand side duplicates terms in $y_c$, the form of the guess for $y_p$ must be modified.
12. The differential equation $\sin(x)\frac{dy}{dx} + y\cos(x) = \tan(x)$ is given for $x \in (0, \pi/2)$. What is the most efficient classification and solution approach?
    Answer: Linear first-order equation; recognize that $\frac{d}{dx}[y\sin(x)] = \sin(x)\frac{dy}{dx} + y\cos(x)$ and integrate directly The left side of the equation $\sin(x)\frac{dy}{dx} + y\cos(x)$ is exactly the derivative of $y\sin(x)$ by the product rule. Therefore, the equation becomes $\frac{d}{dx}[y\sin(x)] = \tan(x)$, which can be integrated directly: $y\sin(x) = \int \tan(x)dx = -\ln|\cos(x)| + C$. Choice A is incorrect because the equation cannot be separated. Choice B is incorrect because this approach is unnecessarily complex when direct integration is possible. Choice C is incorrect because while this would work, it's less efficient than recognizing the derivative pattern.
13. Consider the differential equation $(y \cos(x) + 2xe^y) dx + (\sin(x) + x^2e^y - 1) dy = 0$. Which of the following statements provides the correct classification and justification?
    Answer: The equation is exact, because $\frac{\partial}{\partial y}(y \cos(x) + 2xe^y) = \frac{\partial}{\partial x}(\sin(x) + x^2e^y - 1)$. To check if the equation $M(x, y)dx + N(x, y)dy = 0$ is exact, we must verify if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Here, $M(x, y) = y \cos(x) + 2xe^y$ and $N(x, y) = \sin(x) + x^2e^y - 1$. Calculating the partial derivatives: $\frac{\partial M}{\partial y} = \cos(x) + 2xe^y$ and $\frac{\partial N}{\partial x} = \cos(x) + 2xe^y$. Since the partial derivatives are equal, the equation is exact.
14. Consider the differential equation $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$. A student claims this can be solved using three different methods. Which combination of solution approaches is correct?
    Answer: Homogeneous substitution and exact equation after rearrangement, but separation is not possible The equation $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$ can be rewritten as $\frac{dy}{dx} = \frac{1}{2}\left(\frac{y}{x} - \frac{x}{y}\right)$, which is homogeneous (depends only on $y/x$). Rearranging to $(y^2 - x^2)dx - 2xy dy = 0$, we can verify exactness: $\frac{\partial M}{\partial y} = 2y = \frac{\partial N}{\partial x}$. However, the variables cannot be separated because of the mixed $xy$ terms. Choice A is incorrect because this is not a Bernoulli equation. Choice B is incorrect because separation is not possible. Choice C is incorrect because this is not a Riccati equation.
15. The differential equation $y'' - 6y' + 9y = e^{3x}\cos(2x)$ requires solution. What is the correct approach for finding the particular solution, given the nature of the homogeneous solution?
    Answer: Use trial solution $y_p = x^2e^{3x}(A\cos(2x) + B\sin(2x))$ due to repeated root conflict with exponential factor The characteristic equation $r^2 - 6r + 9 = (r-3)^2 = 0$ has repeated root $r = 3$, so $y_h = (C_1 + C_2x)e^{3x}$. The forcing term $e^{3x}\cos(2x)$ has the same exponential factor $e^{3x}$ as the homogeneous solution. Since $r = 3$ is a repeated root (multiplicity 2), we must multiply the usual trial solution by $x^2$ to ensure linear independence. Choice A fails to account for the conflict. Choice B uses $x$ instead of $x^2$, not accounting for the full multiplicity. Choice D is unnecessarily complex when undetermined coefficients can be applied with proper modification.
16. Consider the differential equation $y = x \frac{dy}{dx} - e^{dy/dx}$. This is an example of a Clairaut equation. Which of the following statements most accurately describes a key feature of its solution set, derived from its classification?
    Answer: As a non-linear equation, it possesses a general solution which is a family of straight lines, and may also have a singular solution that is not part of this family. The equation $y = xy' - e^{y'}$ is in the form of a Clairaut equation, $y = xy' + f(y')$. Clairaut equations are a special type of first-order, non-linear ODE. A key feature of their solutions is that the general solution is a family of straight lines, given by $y = Cx - e^C$, obtained by replacing $y'$ with a constant $C$. Additionally, Clairaut equations often have a singular solution that forms an envelope to the family of lines. This singular solution cannot be obtained by specifying a value for $C$ in the general solution. The other classifications are incorrect: it is non-linear, not exact, and not homogeneous.
17. The substitution $v = y^{-2}$ is applied to the Bernoulli differential equation $y' + p(x)y = q(x)y^3$. Which of the following accurately describes the resulting differential equation in the variable $v$ and its properties?
    Answer: The resulting equation is a first-order linear equation in $v$, which is solvable using an integrating factor. The substitution for a Bernoulli equation of the form $y' + p(x)y = q(x)y^n$ is $v = y^{1-n}$. In this case, $n=3$, so the substitution is $v = y^{1-3} = y^{-2}$. We differentiate $v$ with respect to $x$: $v' = -2y^{-3}y'$. We can write $y' = -\frac{1}{2}y^3 v'$. Substitute this into the original equation: $(-\frac{1}{2}y^3 v') + p(x)y = q(x)y^3$. Now, divide by $-\frac{1}{2}y^3$: $v' - 2p(x)y^{-2} = -2q(x)$. Since $v = y^{-2}$, the equation becomes $v' - 2p(x)v = -2q(x)$. This is a first-order linear differential equation in the variable $v$, which can be solved using an integrating factor. The general form is $v' + P(x)v = Q(x)$.
18. Which of the following first-order differential equations is exact but is NOT separable?
    Answer: $(2x+y)dx + (x-1)dy = 0$ When you encounter questions about exact versus separable differential equations, you need to test each equation using the specific criteria for both properties. An exact equation has the form $M(x,y)dx + N(x,y)dy = 0$ where $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. A separable equation can be written as $\frac{dy}{dx} = f(x)g(y)$ where variables can be completely separated. Let's examine option C: $(2x+y)dx + (x-1)dy = 0$. Here, $M = 2x+y$ and $N = x-1$. Testing for exactness: $\frac{\partial M}{\partial y} = 1$ and $\frac{\partial N}{\partial x} = 1$. Since these partial derivatives are equal, the equation is exact. However, when we try to separate variables by rewriting as $\frac{dy}{dx} = -\frac{2x+y}{x-1}$, we cannot factor this into a product $f(x) \cdot g(y)$ because $y$ appears in the numerator mixed with $x$. Therefore, C is exact but not separable. Option A fails because both $M = x+y$ and $N = x+y$, making $\frac{\partial M}{\partial y} = 1$ and $\frac{\partial N}{\partial x} = 1$, so it's exact, but it's also separable when rewritten properly. Option B is clearly separable ($y \, dy = x^2 \, dx$) but not exact. Option D is separable ($dy = (xy+1)dx$) but checking exactness shows $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$. Remember: always check both conditions systematically. Exact equations often "look mixed" with both variables appearing together, while separable equations can be factored cleanly.
19. The differential equation $(3x^2y - 2y^3)dx + (x^3 - 6xy^2 + 2y)dy = 0$ is being analyzed for solution methods. Which approach should be used first?
    Answer: Solve directly as an exact equation since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ throughout the domain Computing the partial derivatives: $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2y - 2y^3) = 3x^2 - 6y^2$ and $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^3 - 6xy^2 + 2y) = 3x^2 - 6y^2$. Since these are equal, the equation is exact and can be solved directly. Choice A is incorrect because an integrating factor is not needed when the equation is already exact. Choice C is incorrect because the equation cannot be separated into functions of $x$ and $y$ independently. Choice D is incorrect because no integrating factor is required.
20. A student encounters the differential equation $y' + \frac{2}{x}y = x^2y^{-2}$. To solve this equation efficiently, what classification and solution strategy should be employed?
    Answer: Bernoulli equation; substitute $v = y^3$, then solve the resulting linear equation in $v$ This equation has the form $y' + P(x)y = Q(x)y^n$ where $P(x) = \frac{2}{x}$, $Q(x) = x^2$, and $n = -2$. This is a Bernoulli equation. The standard substitution is $v = y^{1-n} = y^{1-(-2)} = y^3$. Then $\frac{dv}{dx} = 3y^2\frac{dy}{dx}$, and the equation becomes linear in $v$. Choice A is incorrect because the $y^{-2}$ term prevents this from being linear in $y$. Choice C is incorrect because this is not homogeneous. Choice D is incorrect because this is not a Ricatti equation.

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