Differential Equations Quiz: General Vs Particular Solutions

What this quiz covers

This quiz focuses on General Vs Particular Solutions, 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. 1, because all terms contain the function $e^{3x}$.
2. 2, because there are only two functionally independent solution forms, $e^{3x}$ and $x e^{3x}$. (correct answer)
3. 3, because there are three constants, $c_1$, $c_2$, and $c_3$, shown in the expression.
4. Cannot be determined without knowing the specific differential equation.

Explanation: The order of a linear differential equation is equal to the number of essential (linearly independent) arbitrary constants in its general solution. The given family of curves can be rewritten by combining the terms with $c_1$ and $c_3$: $y = (c_1 + c_3) e^{3x} + c_2 x e^{3x}$. If we define a new constant $C_1 = c_1 + c_3$, the solution becomes $y = C_1 e^{3x} + c_2 x e^{3x}$. This form reveals that there are only two essential arbitrary constants, $C_1$ and $c_2$. Therefore, the order of the corresponding differential equation is 2.

Question 2

1. The family of functions does not actually satisfy the differential equation for all $x$ and $C$.
2. The constant $C$ cannot be uniquely determined by a single initial condition like $y(x_0)=y_0$.
3. The family fails to include the singular solution $y(x) = 0$, which also satisfies the differential equation. (correct answer)
4. A general solution must be a single function, not an infinite family of functions.

Explanation: A general solution to a differential equation is a family of solutions that contains every possible solution, except possibly for some 'singular' solutions. In this case, the function $y(x) = 0$ for all $x$ is also a solution to $y' = 2\sqrt{y}$, since $0 = 2\sqrt{0}$. However, there is no value of the constant $C$ for which the family $y(x) = (x+C)^2$ yields the function $y(x) = 0$. Because this valid solution is omitted, the family is not the general solution. This missing solution is known as a singular solution.

Question 3

1. $3e^2$
2. $8e^2$
3. $5e^2$ (correct answer)
4. $11e^2$

Explanation: When you encounter a differential equation problem with given initial conditions, you need to use those conditions to find the specific values of the arbitrary constants in the general solution. Given the general solution $y(x) = c_1 e^{2x} + c_2 x e^{2x}$, start by applying the first initial condition. At $x = 0$: $y(0) = c_1 e^0 + c_2 \cdot 0 \cdot e^0 = c_1 = 3$. So $c_1 = 3$. For the second condition, you need the derivative. Using the product rule on the second term: $y'(x) = 2c_1 e^{2x} + c_2(e^{2x} + 2xe^{2x})$. At $x = 0$: $y'(0) = 2c_1 + c_2 = 2(3) + c_2 = 8$. Therefore $c_2 = 2$. The particular solution is $y(x) = 3e^{2x} + 2xe^{2x}$. At $x = 1$: $y(1) = 3e^2 + 2(1)e^2 = 3e^2 + 2e^2 = 5e^2$. Answer A ($3e^2$) represents what you'd get if you forgot the second term entirely—just evaluating $3e^{2x}$ at $x = 1$. Answer B ($8e^2$) might come from incorrectly using the initial condition $y'(0) = 8$ as if it were the coefficient. Answer D ($11e^2$) could result from arithmetic errors when combining terms or incorrectly calculating the derivative. Always remember: with initial value problems, methodically apply each condition one at a time to find your constants, then substitute your target $x$-value into the complete particular solution.

Question 4

1. $x^3 + 3xy^2 = C$ (correct answer)
2. $y^3 + 3yx^2 = C$
3. $x^2 + y^2 = Cxy$
4. $\ln(x^2+y^2) + 2y/x = C$

Explanation: This is a homogeneous equation. Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$. The equation becomes $v + x\frac{dv}{dx} = -\frac{x^2+(vx)^2}{2x(vx)} = -\frac{1+v^2}{2v}$. Rearranging gives $x\frac{dv}{dx} = -\frac{1+v^2}{2v} - v = -\frac{1+3v^2}{2v}$. This is separable: $\frac{2v}{1+3v^2} dv = -\frac{1}{x} dx$. Integrating both sides gives $\frac{1}{3}\ln(1+3v^2) = -\ln|x| + C_1$. Multiplying by 3 and exponentiating gives $1+3v^2 = K|x|^{-3}$. Substituting back $v=y/x$ gives $1+3(y/x)^2 = K/|x^3|$. Multiplying by $|x^3|$ gives $|x^3| + 3|x|y^2 = K$. Assuming $x>0$ for simplicity and renaming the constant, we have $x^3 + 3xy^2 = C$.

Question 5

1. $2$ (correct answer)
2. $-2$
3. $-2\sqrt{2}$
4. $0$

Explanation: First, use the initial position $x(0)=-2$: $x(0) = c_1\cos(0) + c_2\sin(0) = c_1$. So, $c_1=-2$. Next, find the velocity function: $v(t) = x'(t) = -3c_1\sin(3t) + 3c_2\cos(3t)$. Use the condition $v(\pi/12)=0$: $v(\pi/12) = -3c_1\sin(3\pi/12) + 3c_2\cos(3\pi/12) = -3c_1\sin(\pi/4) + 3c_2\cos(\pi/4) = 0$. This simplifies to $-3c_1(\sqrt{2}/2) + 3c_2(\sqrt{2}/2) = 0$, which implies $-c_1+c_2=0$, or $c_1=c_2$. Since $c_1=-2$, we have $c_2=-2$. The particular solution is $x(t) = -2\cos(3t) - 2\sin(3t)$. Finally, evaluate at $t=\pi/3$: $x(\pi/3) = -2\cos(3\pi/3) - 2\sin(3\pi/3) = -2\cos(\pi) - 2\sin(\pi) = -2(-1) - 2(0) = 2$.

Question 6

1. $y(x) = 0$
2. $y(x) = 5e^{2x} - 2e^{-3x}$
3. $y(x) = 4e^{-3x}$
4. $y(x) = e^{2x} + e^{-3x} + 1$ (correct answer)

Explanation: A particular solution is obtained by assigning specific values to the constants $c_1$ and $c_2$ in the general solution $y(x) = c_1 e^{2x} + c_2 e^{-3x}$. A) $y(x)=0$ corresponds to $c_1=0$ and $c_2=0$. B) $y(x) = 5e^{2x} - 2e^{-3x}$ corresponds to $c_1=5$ and $c_2=-2$. C) $y(x) = 4e^{-3x}$ corresponds to $c_1=0$ and $c_2=4$. D) $y(x) = e^{2x} + e^{-3x} + 1$ cannot be a solution because of the constant term '+1'. No choice of $c_1$ and $c_2$ in the general solution can produce this term. This function would be a solution to a non-homogeneous equation, not the given homogeneous one.

Question 7

1. The value of $c_1$ is always equal to the initial value $y(0)$ for any particular solution.
2. The value of $c_1$ is uniquely determined by the initial velocity $y'(0)$ at the origin. (correct answer)
3. The values of $c_1$ and $c_2$ can both be found using a single initial condition such as $y(1)=5$.
4. The values of $c_1$ and $c_2$ require initial conditions that must be specified away from $x=0$.

Explanation: Let's analyze the initial conditions at $x=0$. The general solution is $y(x) = c_1 x + c_2 x^2$. The derivative is $y'(x) = c_1 + 2c_2 x$. Evaluating at $x=0$: $y(0) = c_1(0) + c_2(0)^2 = 0$ and $y'(0) = c_1 + 2c_2(0) = c_1$. This shows that any particular solution from this family must pass through the origin ($y(0)=0$). The constant $c_1$ is uniquely determined by the initial velocity at the origin, $y'(0) = c_1$. The constant $c_2$ would require another condition, such as the value of $y$ or $y'$ at a point other than $x=0$.

Question 8

1. The two particular solutions are identical for all $x$.
2. The two particular solutions are different functions. (correct answer)
3. The solutions are the same, but the constants of integration are different.
4. The existence and uniqueness theorem does not guarantee a solution for the second problem.

Explanation: The general solution of $y' + 2y = 0$ is $y(x) = Ce^{-2x}$. For the first IVP, $y(0) = 3$, we find $3 = Ce^0$, so $C=3$. The solution is $y_1(x) = 3e^{-2x}$. For the second IVP, $y(1) = 3$, we find $3 = Ce^{-2}$, so $C = 3e^2$. The solution is $y_2(x) = (3e^2)e^{-2x} = 3e^{2-2x}$. Since $3e^{-2x}$ and $3e^{2-2x}$ are different functions, the two particular solutions are different. Choice C is contradictory; if the constants of integration are different, the particular solutions (which are specific functions) must also be different.

Question 9

1. $0$ (correct answer)
2. $\pi/6$
3. $\pi/3$
4. $\pi/2$

Explanation: First, use the initial condition $(0, \pi/6)$ to find the constant $C$ for the particular solution. Substitute $x=0$ and $y=\pi/6$ into the general solution: $\sin(\pi/6) - e^{-0^2} = C$, which simplifies to $1/2 - 1 = C$, so $C = -1/2$. The particular solution is defined by $\sin(y) - e^{-x^2} = -1/2$. Now, find the value of $y$ when $x = \sqrt{\ln(2)}$. Substitute this value of $x$ into the particular solution's equation: $\sin(y) - e^{-(\sqrt{\ln(2)})^2} = -1/2$. This simplifies to $\sin(y) - e^{-\ln(2)} = -1/2$. Since $e^{-\ln(2)} = e^{\ln(2^{-1})} = 1/2$, the equation becomes $\sin(y) - 1/2 = -1/2$, which gives $\sin(y) = 0$. The simplest solution for $y$ is $y=0$.

Question 10

1. 2, because of the power on the highest-order derivative.
2. 8, because this is the sum of the orders of all derivative terms.
3. 5, because of the power on the first-order derivative term.
4. 3, because the order of the equation is determined by the highest derivative. (correct answer)

Explanation: When you encounter a differential equation, the key concept to focus on is the order of the equation, which determines how many arbitrary constants appear in the general solution. The order is always determined by the highest derivative present, regardless of any powers, coefficients, or nonlinear terms. In this equation $\left(\frac{d^3y}{dt^3}\right)^2 - t^4 \left(\frac{dy}{dt}\right)^5 + y^3 = \cos(t)$, the highest derivative is $\frac{d^3y}{dt^3}$, making this a third-order differential equation. The fundamental theorem states that an nth-order differential equation has exactly n arbitrary constants in its general solution. Therefore, this third-order equation must have exactly 3 arbitrary constants. Let's examine why the other choices are incorrect. Choice A) suggests the answer is 2 because of the squared power on the third derivative, but powers on derivatives don't affect the number of constants—only the order matters. Choice B) incorrectly adds the orders of derivative terms (3 + 1 = 4, plus other terms to get 8), but this isn't how differential equation theory works. Choice C) focuses on the fifth power of the first derivative, but again, powers are irrelevant to determining the number of arbitrary constants. Study tip: Always identify the highest-order derivative first when analyzing any differential equation. The order of that derivative equals the number of arbitrary constants in the general solution, no matter how complex the equation appears with powers, products, or nonlinear terms.

Question 11

1. The radius of the circular solution curve.
2. The curvature of the circular solution curve.
3. The square of the radius of the circular solution curve. (correct answer)
4. The circumference of the circular solution curve.

Explanation: When you encounter a differential equation whose general solution represents a geometric family like circles, you need to understand what the arbitrary constant represents geometrically, not just algebraically. To find the particular solution through $(3, 4)$, substitute these coordinates into the general solution $x^2 + y^2 = C$: $3^2 + 4^2 = C$ $9 + 16 = 25 = C$ So the particular solution is $x^2 + y^2 = 25$, which represents a circle centered at the origin. Now consider what $C = 25$ means geometrically. The standard form of a circle centered at the origin is $x^2 + y^2 = r^2$, where $r$ is the radius. Comparing this with our solution $x^2 + y^2 = 25$, we see that $r^2 = 25$, so the radius is $r = 5$. Therefore, $C$ represents the square of the radius, making (C) correct. Let's examine why the other choices fail: (A) claims $C$ is the radius itself, but $C = 25$ while the radius is only $5$. (B) suggests $C$ represents curvature, but curvature for a circle equals $1/r = 1/5$, not $25$. (D) proposes $C$ is the circumference, but circumference equals $2\pi r = 10\pi$, which doesn't match $25$. Study tip: When working with geometric solution families, always compare the given equation with the standard form to identify what each parameter represents. The constant in $x^2 + y^2 = C$ will always be $r^2$, not $r$ itself.

Question 12

1. Both students are correct; the particular solutions differ only by a constant, confirming the general solution structure (correct answer)
2. The first student is correct, but the second student is wrong because general solutions require homogeneous equations
3. The first student is wrong; two different particular solutions cannot satisfy the same differential equation simultaneously
4. Both students are wrong; the given functions are solutions to different differential equations entirely

Explanation: Both particular solutions have the same derivative structure: $y' = 4x + 3$ and $y'' = 4$. They satisfy the same second-order differential equation $y'' = 4$. The general solution to $\frac{d^2y}{dx^2} = 4$ is $y = 2x^2 + C_1x + C_2$, and with appropriate initial conditions, this reduces to the form $y = 2x^2 + 3x + C$. The two particular solutions differ by a constant ($5 - (-2) = 7$), which is exactly what we expect for different particular solutions of the same differential equation.

Question 13

1. The solution is unique because fourth-order differential equations always have unique solutions when boundary conditions are given
2. The solution is unique because the differential equation is linear and the initial conditions are specified at the same point
3. The solution is unique because the initial conditions provide enough information to solve for all four unknown constants simultaneously
4. The solution is unique because the number of initial conditions equals the number of arbitrary constants in the general solution (correct answer)

Explanation: When you encounter a differential equation with a general solution containing arbitrary constants, the key to uniqueness lies in having sufficient conditions to determine all those constants. The general solution $y = C_1 x^3 + C_2 x^2 + C_3 x + C_4$ contains four arbitrary constants: $C_1$, $C_2$, $C_3$, and $C_4$. To find a unique particular solution, you need exactly four independent conditions to create a system of four equations in four unknowns. The given conditions $y(1) = 2$, $y'(1) = 6$, $y''(1) = 12$, and $y'''(1) = 18$ provide precisely this—four independent conditions that allow you to solve for all four constants simultaneously, yielding a unique solution. Option A is incorrect because uniqueness isn't guaranteed simply by having a fourth-order equation with boundary conditions—you need the right number and type of conditions. Option B misses the fundamental point: while linearity helps ensure solutions exist and behave nicely, and having conditions at the same point matters for initial value problems, neither directly explains why this particular solution is unique. Option C is partially correct but imprecise—it's not just that you have "enough information," but specifically that you have exactly the right amount. Option D captures this precisely: the number of conditions (four) equals the number of arbitrary constants (four). Study tip: Always count the arbitrary constants in a general solution, then verify you have that many independent conditions for uniqueness. This one-to-one correspondence is the fundamental principle for unique solutions in linear differential equations.

Question 14

1. They obtain different particular solutions because they used different forms of the general solution with different constants
2. They obtain the same particular solution despite using different constant labels, demonstrating equivalent general solution forms (correct answer)
3. Student A's solution is correct while Student B's contains an error in the constant relationship
4. Both solutions are incorrect because the general solution should not contain the term $-xe^x$ for this differential equation

Explanation: Both students have equivalent forms of the general solution. For Student A: $3 = Ce^0 - 0 \cdot e^0 = C$, so the particular solution is $y = 3e^x - xe^x$. For Student B: $3 = De^0 - 0 \cdot e^0 = D$, so the particular solution is $y = 3e^x - xe^x$. The different constant labels ($C$ vs $D$) don't affect the final answer. This demonstrates that general solutions can be written with different parameter names but still represent the same family of solutions.

Question 15

1. The argument is incorrect; the phase-shifted form $A\sin(3x + \phi)$ represents fewer solutions than the original general form
2. The argument is incorrect; the trigonometric identity changes the fundamental structure of the differential equation's solution
3. The argument is correct for the homogeneous part but fails to account for the particular solution $\frac{1}{4}$ properly
4. The argument is correct; both forms span the same solution space and contain the same number of arbitrary constants (correct answer)

Explanation: When you encounter questions about general solutions to differential equations, focus on whether different forms represent the same solution space with the same degrees of freedom. The key insight is recognizing that $C_1 \sin(3x) + C_2 \cos(3x)$ and $A\sin(3x + \phi)$ are mathematically equivalent expressions for the homogeneous solution. Using the angle addition formula, $A\sin(3x + \phi) = A\sin(3x)\cos(\phi) + A\cos(3x)\sin(\phi)$. If we set $C_1 = A\cos(\phi)$ and $C_2 = A\sin(\phi)$, we can convert between the forms. Both expressions contain exactly two arbitrary constants and span the same two-dimensional solution space. The complete general solution $y = A\sin(3x + \phi) + \frac{1}{4}$ is therefore equivalent to the original form, making answer D correct. Answer A is wrong because both forms represent exactly the same number of solutions—the phase-shifted form doesn't reduce the solution space. Answer B incorrectly suggests that trigonometric identities somehow alter the differential equation's structure, when they simply provide alternative representations of the same mathematical object. Answer C misunderstands the relationship between homogeneous and particular solutions—the particular solution $\frac{1}{4}$ is handled identically in both forms and doesn't affect the equivalence of the homogeneous parts. Remember: different algebraic forms of the same expression are equivalent as long as they contain the same number of independent arbitrary constants and span the same solution space.

Question 16

1. Verify that the three solution components $e^{2x}$, $xe^{2x}$, and $e^{-x}$ are linearly independent functions
2. Check that the number of initial conditions matches the order of the equation, ensuring automatic existence of a unique solution
3. Calculate $y'(x)$ and $y''(x)$, then solve the resulting $3 \times 3$ linear system for $C_1$, $C_2$, and $C_3$ (correct answer)
4. Substitute the initial conditions directly into the general solution without computing derivatives, then solve for the constants

Explanation: When dealing with initial value problems for linear differential equations, the key question is whether you can uniquely determine all arbitrary constants using the given conditions. This requires setting up and solving a system of equations. The correct approach (C) involves finding the derivatives $y'(x)$ and $y''(x)$ of the general solution, then applying all three initial conditions to create a system of three equations in three unknowns. From $y = C_1 e^{2x} + C_2 xe^{2x} + C_3 e^{-x}$, you'd compute $y' = 2C_1 e^{2x} + C_2(e^{2x} + 2xe^{2x}) + C_3(-e^{-x})$ and $y''$, then substitute $x = 0$ into each expression along with the given values. This gives you three equations to solve for $C_1$, $C_2$, and $C_3$. If this system has a unique solution, then a unique particular solution exists. Option A is insufficient because linear independence of the solution components tells you the general solution is complete, but doesn't guarantee the initial value problem is solvable. Option B makes a false assumption—having the right number of conditions is necessary but not sufficient for uniqueness. The coefficient matrix of your system could still be singular. Option D fails because you need all three conditions (including those involving derivatives) to determine all three constants; using only $y(0) = 1$ leaves you with one equation and three unknowns. Remember: for initial value problems, always set up the complete linear system using all given conditions before concluding anything about existence or uniqueness.

Question 17

1. $y_3$ is a particular solution only for specific values of $a$ and $b$ that must be determined by substitution
2. $y_3$ is a particular solution only when $a$ and $b$ are both positive and $a + b = 1$
3. $y_3$ is always a particular solution to the same differential equation, regardless of the specific values of $a$ and $b$ (correct answer)
4. $y_3$ represents a linear combination that may not satisfy the original differential equation due to nonlinearity

Explanation: When you encounter questions about linear combinations of solutions to differential equations, you're dealing with the fundamental principle of superposition. This principle states that if you have a linear differential equation, any linear combination of its solutions is also a solution. Since both $y_1 = 2x^2 + 3x - 1$ and $y_2 = x^2 - x + 4$ are particular solutions to the same differential equation (they can both be written in the form $y = C_1x^2 + C_2x + C_3$), their linear combination $y_3 = ay_1 + by_2$ will automatically satisfy the original equation. The constraint $a + b = 1$ doesn't affect this property—it's simply given information that doesn't change the fundamental fact that linear combinations of solutions remain solutions. Let's examine why the other options miss the mark: Option A suggests you need specific values of $a$ and $b$, but superposition works for any constants, not just particular ones. Option B incorrectly claims the constants must be positive—the sign of $a$ and $b$ is irrelevant to whether $y_3$ satisfies the equation. Option D mentions "nonlinearity," but we're dealing with a linear differential equation here, as evidenced by the polynomial form of the general solution. Study tip: Remember that for linear differential equations, the superposition principle always holds. Any linear combination of solutions (particular or homogeneous) will be a solution to the same equation. This is a powerful tool for constructing new solutions from known ones.

Question 18

1. $A = 3$ and $B = -2$ represent one possible assignment that makes the general solution equal the particular solution (correct answer)
2. $A$ and $B$ must always equal $3$ and $-2$ respectively for any solution to this differential equation
3. $A = 3$ and $B = -2$ are the only values that satisfy the differential equation's boundary conditions
4. The values $3$ and $-2$ are incorrect because they don't satisfy the characteristic equation of the differential equation

Explanation: The general solution $y_1 = A\cos(2x) + B\sin(2x)$ represents all possible solutions to the differential equation, with $A$ and $B$ being arbitrary constants determined by initial conditions. The particular solution $y_2 = 3\cos(2x) - 2\sin(2x)$ is obtained from the general solution by setting $A = 3$ and $B = -2$. This represents just one specific member of the infinite family of solutions. Option B is wrong because $A$ and $B$ can take any real values. Option C is wrong because no boundary conditions are given. Option D is wrong because both $3\cos(2x)$ and $-2\sin(2x)$ satisfy the differential equation (the characteristic equation $r^2 + 4 = 0$ gives $r = \pm 2i$, leading to the cosine and sine solutions).

Question 19

1. The claim is correct because the homogeneous and particular solutions address different aspects of the problem independently
2. The claim is correct because linear superposition allows the homogeneous solution constants to be determined separately from the particular solution (correct answer)
3. The claim is incorrect because the particular solution $y_p = 2$ constrains the possible values of $C_1$ and $C_2$
4. The claim is incorrect because the constant term $2$ should be absorbed into one of the arbitrary constants $C_1$ or $C_2$

Explanation: The student's claim is mathematically correct. For linear nonhomogeneous differential equations, the general solution is $y = y_h + y_p$ where $y_h$ contains all arbitrary constants and $y_p$ is any particular solution to the nonhomogeneous equation. The particular solution $y_p = 2$ is fixed and satisfies the nonhomogeneous equation, while the homogeneous solution $y_h = C_1 e^{2x} + C_2 e^{3x}$ provides the freedom to satisfy initial conditions. The constants $C_1$ and $C_2$ can indeed be chosen independently because the exponential functions $e^{2x}$ and $e^{3x}$ are linearly independent. Option A is less precise about the mathematical principle involved. Option C incorrectly suggests constraint where none exists. Option D misunderstands the structure of the general solution.