## Quiz: Linearize ODEs

**1.** What is a method for linearizing a nonlinear function? Select all that apply.

**A.**Newton's Method

- Incorrect. Newton's method is used to find the root of an equation where it is equal to zero
*f(x)=0*.

**B.**Gauss Elimination

- Incorrect. Gauss elimination is an algorithm in linear algebra for solving a system of linear equations.

**C.**Taylor Series Expansion

- Correct. Linearization of the right-hand side of a differential equation is the first two terms of a Taylor series expansion with `f(x)=f(x_0)+f'(x_0)(x-x_0)`.

**D.**Power Series Expansion

- Correct. A power series is an example of a Taylor series. Linearization is using the first two terms of a Taylor series expansion with `f(x)=f(x_0)+f'(x_0)(x-x_0)`.

**2.** Determine the steady state value of *x* in the following equation, given a steady state input of *u = 2*.

$$\frac{dx}{dt}=\sqrt{x}-u^3$$

**A.**8

- Incorrect. Set the derivative equal to zero to find the steady state solution. Square both sides of the equation to solve for
*x*.

**B.**2

- Incorrect. Set the derivative equal to zero to find the steady state solution.

**C.**68

- Incorrect. Set the derivative equal to zero to find the steady state solution.

**D.**64

- Correct. `0=\sqrt{x}-2^3` is rearranged to find the solution `x=8^2`

**3.** Find the value `\alpha` for the function `f(y,u,d)` with `u_{ss}=4` and `y_{ss}=3`.

$$\frac{dy}{dt}=3y^3-(u^2-\sin(u))^{1/3}+ln(d)$$

The constant `\alpha` is the partial derivative of `f(y,u,d)` with respect to `y` and evaluated at steady state conditions.

$$\frac{dy'}{dt} = \alpha y' + \beta u' + \gamma d'$$

$$\alpha = \frac{\partial f}{\partial y}\bigg|_{\bar y,\bar u,\bar d}$$

**A.**81

- Correct. The partial derivative with respect to `y` is `9y^2`. Evaluating at `y_{ss}=3` gives the answer `\alpha=81`.

**B.**273

- Incorrect. Try again.

**C.**27

- Incorrect. Try again.

**D.**144

- Incorrect. You have swapped the values of `u_{ss}=4` and `y_{ss}=3`.

**4.** Find the value `\beta` for the function `f(y,u,d)` with `u_{ss}=4` and `y_{ss}=3`.

$$\frac{dy}{dt}=3y^3-(u^2-\sin(u))^{1/3}+ln(d)$$

The constant `\beta` is the partial derivative of `f(y,u,d)` with respect to `u` and evaluated at steady state conditions.

$$\frac{dy'}{dt} = \alpha y' + \beta u' + \gamma d'$$

$$\beta = \frac{\partial f}{\partial u}\bigg|_{\bar y,\bar u,\bar d}$$

Use the following Python source code to calculate the partial derivative, if needed.

import sympy as sp

# define symbols

y,u,d = sp.symbols(['y','u','d'])

# define equation

dydt = 3*y**3-(u**2-sp.sin(u))**(1/3)+sp.log(d)

# partial derivative with respect to u

beta = sp.diff(dydt,u)

# evaluate at steady state condition

print(beta.subs(u,4).evalf())

**A.**-0.4432

- Incorrect. Run the code to check your answer. Type
**pip install sympy**to install SymPy.

**B.**-0.4405

- Correct.

**C.**0.4402

- Incorrect. Run the code to check your answer. Type
**pip install sympy**to install SymPy.

**D.**-0.5442

- Incorrect. You have swapped the values of `u_{ss}=4` and `y_{ss}=3`.