Step-by-Step Guidance for Differential Equations and Calculus Concepts

Q1. For which integer values of n is $y(x) = x^n$ a solution to $x y'' - y' - 3y = 0$?

Background

Topic: Solving differential equations by seeking solutions of a specific form.

This question tests your ability to substitute a function of the form $x^n$ into a differential equation and determine for which values of $n$ it satisfies the equation.

Key Terms and Formulas:

• $y(x) = x^n$

• $y'(x) = n x^{n-1}$

• $y''(x) = n(n-1)x^{n-2}$

Step-by-Step Guidance

1. Compute $y'(x)$ and $y''(x)$ for $y(x) = x^n$.

2. Substitute $y(x)$, $y'(x)$, and $y''(x)$ into the given equation: $x y'' - y' - 3y = 0$.

3. Express the resulting equation in terms of $x$ and $n$.

4. Factor out $x^{n-1}$ and set the remaining expression in $n$ equal to zero.

Try solving on your own before revealing the answer!

Final Answer:

Substituting and simplifying, the equation becomes $[n(n-1) - n - 3x]x^{n-1} = 0$. Solving for $n$, the integer values are $n = 3$ and $n = 0$.

These are the values for which $y(x) = x^n$ solves the given differential equation.

Q2. For which integer values of n is $f(t) = t^n$ a solution to $\frac{d^2 f}{dt^2} + \frac{1}{t} \frac{df}{dt} = 0$?

Background

Topic: Solving differential equations by substitution.

This question asks you to substitute $f(t) = t^n$ into the ODE and find all integer $n$ for which it is a solution.

Key Terms and Formulas:

• $f(t) = t^n$

• $\frac{df}{dt} = n t^{n-1}$

• $\frac{d^2 f}{dt^2} = n(n-1) t^{n-2}$

Step-by-Step Guidance

1. Calculate $\frac{df}{dt}$ and $\frac{d^2 f}{dt^2}$ for $f(t) = t^n$.

2. Substitute these expressions into the ODE.

3. Combine like terms and factor out $t^{n-2}$.

4. Set the coefficient of $t^{n-2}$ equal to zero and solve for $n$.

Try solving on your own before revealing the answer!

Final Answer:

After substitution and simplification, the equation reduces to $n(n-1) + n = 0$, so $n = 0$ or $n = -1$.

These are the integer values for which $f(t) = t^n$ solves the ODE.

Q3. Find all equilibrium (constant) solutions to $y'' = e^{\cos t} y' - \sqrt{2t} y' + y^2 - t^2$ (i.e., find all values of $C$ for which $y(t) = C$ is a solution).

Background

Topic: Equilibrium solutions of differential equations.

This question tests your ability to find constant solutions (where $y(t) = C$) to a given ODE.

Key Terms and Formulas:

• Equilibrium solution: $y(t) = C$ (constant)

• $y'(t) = 0$, $y''(t) = 0$ for constant functions

Step-by-Step Guidance

1. Substitute $y(t) = C$ into the ODE. Recall that $y'(t) = 0$ and $y''(t) = 0$.

2. Simplify the equation, noting that all terms involving derivatives vanish.

3. Set the remaining expression equal to zero and solve for $C$.

Try solving on your own before revealing the answer!

Final Answer:

After substitution, the equation reduces to $C^2 - t^2 = 0$, so $C = t$ or $C = -t$. However, since $C$ must be constant, the only equilibrium solution is $C = 0$.

Thus, $y(t) = 0$ is the only constant solution.

Q4. Find all equilibrium (constant) solutions to $\frac{d\phi}{dq} = q^3 \phi - \frac{d\phi}{dq}$ (i.e., find all values of $C$ for which $\phi(q) = C$ is a solution).

Background

Topic: Equilibrium solutions of differential equations.

This question asks you to find constant solutions to a first-order ODE.

Key Terms and Formulas:

• Equilibrium solution: $\phi(q) = C$

• $\frac{d\phi}{dq} = 0$ for constant functions

Step-by-Step Guidance

1. Substitute $\phi(q) = C$ into the ODE, noting that $\frac{d\phi}{dq} = 0$.

2. Simplify the equation to find the values of $C$ that satisfy it.

Try solving on your own before revealing the answer!

Final Answer:

Substituting, the equation becomes $0 = q^3 C - 0$, so $q^3 C = 0$. For all $q$, this is only satisfied if $C = 0$.

Therefore, $\phi(q) = 0$ is the only constant solution.

Q5. For which values of $\omega$ is $y = \cos(\omega t)$ a solution to $y'' + 25y = 0$?

Background

Topic: Solving second-order linear ODEs with constant coefficients.

This question tests your ability to substitute a trigonometric function into a second-order ODE and find the parameter values that make it a solution.

Key Terms and Formulas:

• $y = \cos(\omega t)$

• $y'' = -\omega^2 \cos(\omega t)$

Step-by-Step Guidance

1. Compute $y''$ for $y = \cos(\omega t)$.

2. Substitute $y$ and $y''$ into the ODE: $y'' + 25y = 0$.

3. Factor out $\cos(\omega t)$ and set the coefficient equal to zero.

Try solving on your own before revealing the answer!

Final Answer:

Substituting, the equation becomes $-\omega^2 \cos(\omega t) + 25 \cos(\omega t) = 0$, so $\omega^2 = 25$.

Therefore, $\omega = 5$ or $\omega = -5$.

Q6. For which values of $r$ is $y = e^{rt}$ a solution to $y'' + 4y' + 3y = 0$?

Background

Topic: Solving second-order linear ODEs with constant coefficients.

This question tests your ability to substitute an exponential function into a second-order ODE and find the parameter values that make it a solution.

Key Terms and Formulas:

• $y = e^{rt}$

• $y' = r e^{rt}$

• $y'' = r^2 e^{rt}$

Step-by-Step Guidance

1. Compute $y'$, $y''$ for $y = e^{rt}$.

2. Substitute $y$, $y'$, and $y''$ into the ODE: $y'' + 4y' + 3y = 0$.

3. Factor out $e^{rt}$ and set the remaining expression in $r$ equal to zero.

Try solving on your own before revealing the answer!

Final Answer:

Substituting, the equation becomes $r^2 e^{rt} + 4 r e^{rt} + 3 e^{rt} = 0$, so $r^2 + 4r + 3 = 0$.

Solving, $r = -1$ and $r = -3$.

Q7. Find all equilibrium (constant) solutions to $y'(e^t + \cos(3t - 4)) - y^2 e^t - \frac{\sqrt{y} y'}{3 + \arctan(t^2 + 1)} = -e^t$ (i.e., find all values of $C$ for which $y(t) = C$ is a solution).

Background

Topic: Equilibrium solutions of nonlinear ODEs.

This question asks you to find constant solutions to a nonlinear ODE.

Key Terms and Formulas:

• Equilibrium solution: $y(t) = C$

• $y'(t) = 0$ for constant functions

Step-by-Step Guidance

1. Substitute $y(t) = C$ and $y'(t) = 0$ into the ODE.

2. Simplify the equation, noting that terms involving $y'$ vanish.

3. Set the remaining expression equal to zero and solve for $C$.

Try solving on your own before revealing the answer!

Final Answer:

After substitution, the equation reduces to $-C^2 e^t = -e^t$, so $C^2 = 1$.

Therefore, $C = 1$ or $C = -1$ are the equilibrium solutions.