MATH 20D cheat sheet

I don't know why it works. Just trust it.

Assumes x or t is the independent variable and y is dependent.

Separable equations

          \frac{dy}{dx} = g(x) \, p(y)

Solve

Multiply by dx, divide by p(y).

Integrate both sides.

                  \int \frac{1}{p(y)} \, dy = \int g(x) \, dx

Solve for y if possible.
y = D is a hidden solution when p(D) = 0.

Identify

Find g(x) and p(y).

Linear equations

          \frac{dy}{dx} + p(x) \, y = Q(x)

Solve

Calculate the integrating factor. Ignore C.

                  \mu(x) = e^{\int p(x) \, dx}

Multiply both sides by \mu(x). Note that \mu'(x) = \mu(x) \, p(x).

                  \mu y' + \mu' y = (\mu y)' = \mu(x) \, Q(x)

You need to show this step.

Integrate to solve for y.

                  y = \frac{1}{\mu} \int \mu(x) \, Q(x) \, dx

Identify and uniqueness

y and its derivatives can't be multiplied by each other or inside another function; they can only be multiplied by something in terms of x.

An initial condition has a unique solution in an interval where p and Q are continuous.

Exact equations

          M(x, y) + N(x, y) \cdot \frac{dy}{dx} = 0

Solve

                  F = \int M \, dx + g(y)

                  N = \frac{\partial F}{\partial y} = \text{(something)} + g'(y)

Find g'(y). You have two expressions for N.

                  g(y) = \int g'(y) \, dy + C

Plug g(y) into F.

F(x, y) = C is a solution.

Identify

Show that

              \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

If not, then multiply both M and N by a special integrating factor \mu, and find what \mu makes the equation exact. The form of \mu is either given (HW 2 Q3) or

a function only of x. Ignore C.

                  \mu = e^{\displaystyle \int \dfrac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} \, dx}

a function only of y. Ignore C.

                  \mu = e^{\displaystyle \int \dfrac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} \, dy}

\mu may cause gained or lost solutions. We aren't really told how, but it probably has to do with things like \mu's denominator being zero.

Homogeneous equations

          ay'' + by' + cy = 0

Solve

Find the roots of the characteristic equation.

                  ar^2 + br + c = 0

                  y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}

If there is a repeated root, multiply the repeated root's term by t.

                  y(t) = C_1 e^{rt} + C_2 t e^{rt} + C_3 t^2 e^{rt}

If the roots are complex r = \alpha \pm i\beta, they want a real solution.

                  e^{(\alpha + i\beta)t} = e^{\alpha t} e^{\beta i t} = e^{\alpha t} (\cos \beta t + i \sin \beta t)

You should show this use of Euler's formula.

                  y(t) = C_1 e^{\alpha t} \cos \beta t + C_2 e^{\alpha t} \sin \beta t

Uniqueness and springs

For second-order equations, a \neq 0. There is a unique solution.

This also works for higher order equations.

Spring word problems use this equation for the spring's motion, with mass m, damping constant b, and stiffness constant k.

              my'' + by' + ky  = F_{ext}(t)

Nonhomogeneous equations

          ay'' + by' + cy = f(t)

The following finds a particular solution y_p(t).

Undetermined coefficients

Find the solution y_c(t) for the homogenous part.
```
ay_c'' + by_c' + cy_c = 0
```
Guess y_p based on the terms in f(t) and its derivatives.

This sometimes doesn't work, so you'll have to use the other method.
Multiply terms by t until there is no repeated term between y_p and y_c.
Solve for the coefficients of y_p by plugging it into the original equation.
```
ay_p'' + by_p' + cy_p = f(t)
```

Variation of parameters

Find the complementary solution y_c(t) to find y_1 and y_2. Drop the constants.

                  ay_c'' + by_c' + cy_c = 0 \to y_c(t) = y_1 + y_2

Set up a system for v_1' and v_2'.

                  \begin{cases}
                  v_1' y_1 + v_2' y_2 = 0 \\
                  v_1' y_1' + v_2' y_2' = \dfrac{f(t)}{a}
                  \end{cases}

Find v_1 and v_2. Drop the integration constants.
```
v_1 = \int v_1'
```
```
v_2 = \int v_2'
```
```
y_p = v_1 y_1 + v_2 y_2
```

This also works with variable-coefficient equations.

Then, to find the general solution,

y(t) = y_c(t) + y_p(t)

Cauchy–Euler equations

          at^2 y'' + bty' + cy = f(t)

Solve (homogeneous)

Find the roots of the characteristic equation.
```
ar^2 + (b - a)r + c = 0
```
```
y(t) = C_1 t^{r_1} + C_2 t^{r_2}
```

If r = \alpha + i\beta,

                  y(t) = C_1 t^\alpha \cos(\beta \ln t) + C_2 t^\alpha \sin(\beta \ln t)

If r is repeated,
```
y(t) = C_1 t^r + C_2 t^r \ln t
```
If the equation is nonhomogeneous, use variation of parameters.

Variable-coefficient equations' uniqueness

              y''(t) + p(t) \, y'(t) + q(t) \, y(t) = g(t)

An initial value problem of the above form has a unique solution if p, q, and g are continuous on (a, b).

Linear independence

Method 1

Let C_1 \vec{x_1} + \dots + C_n \vec{x_n} = 0.

Show C_1 = \dots = C_n = 0 for some t.

One way to do this is to use the matrix form, then determine whether its inverse exists by calculating its determinant. If it's nonzero for some t, then it's linearly independent. If it's always zero, you'll have to use a different approach.

                  \begin{bmatrix}
                    \vec{x_1} & \dots & \vec{x_n}
                  \end{bmatrix}
                  \begin{bmatrix}
                    C_1 \\
                    \vdots \\
                    C_2
                  \end{bmatrix} = 0
                  \to
                  \det \begin{bmatrix}
                    \vec{x_1} & \dots & \vec{x_n}
                  \end{bmatrix} \neq 0

Method 2

This only works if the problem says the vectors are solutions of \vec{x}' = A\vec{x}.

Show that for every t,

                  w(\vec{x_1}, \dots, \vec{x_n}) = \det \begin{bmatrix}
                    x_{11}(t) & \dots & x_{n1}(t) \\
                    \vdots & & \vdots \\
                    x_{1n}(t) & \dots & x_{nn}(t)
                  \end{bmatrix} \neq 0

Linear systems with a constant-coefficient matrix

          \vec{x}' = A\vec{x}

Solve

Solve for the eigenvalues r.

                  \det(A - rI) = 0

If r_1 \neq r_2,

For each eigenvalue r, find its eigenvector \vec{u}.

                  (A - rI) \vec{u} = 0

                  \vec{x}(t) = C_1 e^{r_1 t} \vec{u_1} + \dots + C_n e^{r_n t} \vec{u_n}

For complex eigenvalues r_1 = \alpha + i\beta,

You only need to find one eigenvector \vec{u_1} = \vec{a} + i\vec{b} because \vec{u_2} = \vec{a} - i\vec{b}.

Use Euler's formula and collect the real and imaginary terms.

                  \begin{aligned}
                    e^{(\alpha + i\beta) t} \vec{u_1} &= e^{\alpha t} (\cos \beta t + i \sin \beta t) \vec{u_1} \\
                    &= e^{\alpha t} \left( \underbrace{\left(\cos(\beta t) \vec{a} - \sin(\beta t) \vec{b} \right)}_{\vec{x_1}} + i \underbrace{\left(\cos(\beta t) \vec{b} + \sin(\beta t) \vec{a} \right)}_{\vec{x_2}} \right)
                  \end{aligned}

                  \vec{x}(t) = C_1 e^{\alpha t} \vec{x_1} + C_2 e^{\alpha t} \vec{x_2}

If there is a repeated eigenvalue on the homework or midterm, you are probably wrong. But for your information,

Solve for \vec{v}.

                  (A - rI) \vec{v} = \vec{u}

                  \vec{x}(t) = te^{rt} \vec{u} + e^{rt} \vec{v}

Fundamentals

With a general solution

              \vec{x}(t) = C_1 \vec{x_1} + \dots + C_n \vec{x_n}

the fundamental set is

              
              \mathscr{S} = \{ \vec{x_1}, \dots, \vec{x_n} \}

and the fundamental matrix is

              X(t) = \begin{bmatrix}
                \vec{x_1} (t) & \dots & \vec{x_n} (t)
              \end{bmatrix}

Laplace transform

          \mathscr{L}\{f\}(s) = F(s) = \int_0^\infty e^{-st} f(t) \,dt = \lim_{N \to \infty} \int_0^N e^{-st} f(t) \,dt

Examples

              \begin{aligned}
                f(t) &\longleftrightarrow F(s) \\
                1 &\longleftrightarrow \frac{1}{s} && \text{for} ~ s > 0 \\
                e^{at} &\longleftrightarrow \frac{1}{s - a} && \text{for} ~ s > a \\
                \sin bt &\longleftrightarrow \frac{b}{s^2 + b^2} && \text{for} ~ s > 0 \\
                \cos bt &\longleftrightarrow \frac{s}{s^2 + b^2} \\
              \end{aligned}

Properties

              \begin{aligned}
                f(t) &\longleftrightarrow F(s) && \text{for} ~ s > \alpha \\
                af_1(t) + bf_2(t) &\longleftrightarrow aF_1(s) + bF_2(s) \\
                e^{at}f(t) &\longleftrightarrow F(s - a) && \text{for} ~ s > \alpha + a \\
                f'(t) &\longleftrightarrow sF(s) - f(0) && \text{for} ~ s > \alpha \\
                f^{(n)}(t) &\longleftrightarrow s^nF(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \dots - f^{(n-1)}(0) \\
                t^n f(t) &\longleftrightarrow (-1)^n \frac{d^n F(s)}{ds^n} \\
              \end{aligned}

Partial fractions

For \dfrac{p(s)}{Q(s)},

Nonrepeated linear factors. If the roots are all distinct,

                  \frac{p(s)}{Q(s)} = \frac{A_1}{s - r_1} + \dots + \frac{A_n}{s - r_n}

Repeated linear factors. If Q(s) = (s - r)^m,

                  \frac{p(s)}{Q(s)} = \frac{A_1}{s - r} + \dots + \frac{A_m}{(s - r)^m}

Quadratic factors. If Q(s) = \left((s - \alpha)^2 + \beta^2\right)^m,

                  \frac{p(s)}{Q(s)} = \frac{A_1(s - \alpha) + B_1 \beta}{(s-\alpha)^2 \beta^2} + \dots + \frac{A_m(s - \alpha) + B_m \beta}{((s-\alpha)^2 + \beta^2)^m}

The test will not ask for more than these three cases.

Method

If the initial value condition is w(k) where k \neq 0, then let

                  \begin{cases}
                    y(t) = w(t + k) \\
                    y'(t) = w'(t + k) \\
                    y''(t) = w''(t + k) \\
                  \end{cases}

Take the Laplace transform of both sides.

                  a\left(s^2 Y - sy(0) - y'(0)\right) + b(sY - y(0)) + cY = \mathscr{L}\{f\}

Apply properties to solve for Y(s).
Apply an inverse transform to find y(t).

Don't forget w(t) = y(t - k) if necessary.

Power series

Ratio test

For

              \sum_{n=0}^\infty a_n (x - x_0)^n

the radius of convergence

              L = \lim_{n \to \infty} \left| \frac{a_n}{a_{n + 1}} \right|

Taylor series

At x = 0:

              f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n

Singular points

              y'' + p(x) y' + q(x) y = 0

The singular points are when p and q aren't defined.

Method

Plug in

                \begin{cases}
                  y(x) = \displaystyle\sum\limits_{n=0}^\infty a_n x^n \\
                  y'(x) = \displaystyle\sum\limits_{n=1}^\infty n a_n x^{n-1} \\
                  y''(x) = \displaystyle\sum\limits_{n=2}^\infty n(n-1) a_n x^{n-2} \\
                \end{cases}

Simplify by moving all the terms inside summations.
Shift indices so that all summations are in terms of x^n.
Merge all summations into a single summation in terms of x^n. Set this all equal to the right hand side.

Solve for a_n.

Kisun Lee wants us to collect terms into the following form:

                y(x) = a_0 \underbrace{(\dots)}_{y_1} + a_1 \underbrace{(\dots)}_{y_2}