Calculus

Limits and Continuity

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(Definitions) let a function \(f: I \longrightarrow \mathbb{R}\) be defined on an interval \(I\). We're interested in a point \(a\) in that interval, where \(f\) may or may not be defined.

When \(f(x)\) tends to a value \(l\) as \(x\) tends towards \(a\) from the left (negative side), we write the limit \(\lim_{x \rightarrow a-} f(x) = l\) Conversely, if \(f(x)\) tends to \(l\) as \(x\) to \(a\) from the other side we write \(\lim_{x \rightarrow a+} f(x) = l\) If \(\lim_{x \rightarrow a-} f(x) = l\) and \(\lim_{x \rightarrow a+} f(x) = l\) then \(\lim_{x \rightarrow a} f(x) = l.\) What this says that if every sequence \(x_n\) in \(I\) which converges to \(a\), \(x_n \neq a\), then \(\forall n\) the sequence \(f(x_n)\) converges to \(l\).

Floor and Ceiling functions

The \(\left \lfloor{\textrm{floor}}\right \rfloor\) and \(\left \lceil{\textrm{ceiling}}\right \rceil\) functions always round down and up respectively to the nearest integer.

\(\forall k \in \mathbb{Z}\):

Exercise

Prove that \(\lim_x \rightarrow 0 \frac{1}{x}\) does not exist.

Proof. We need to find a sequence \(x_n \rightarrow 0\) where \(f(x_n)\) does not converge. We can magic the sequence \(x_n = \frac{1}{n}\) which does converge to 0, but \(f(x_n) = n\) is an unbounded sequence which does not converge. Hence, the limit does not exist.

Combination rules for limits

If \(\lim_{x \rightarrow a} f(x) = l\)

and \(\lim_{x \rightarrow a} g(x) = m\) then

• Sum rule: \(\lim_{x \rightarrow a} (f(x) + g(x)) = l + m\)
• Multiple rule: \(\lim_{x \rightarrow a} \lambda f(x) = \lambda l\), \(\lambda \in \mathbb{R}\)
• Product rule: \(\lim_{x \rightarrow a} f(x)g(x) = lm\)
• Quotient rule: \(\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{l}{m}\) provided \(m \neq 0\)

Squeeze rule for limits

If \(f(x) \leq g(x) \leq h(x)\) for all \(x \neq a\),

Continuous functions

A continuous function is one with no "jumps" -

A function \(f : D \longrightarrow \mathbb{R}\) (\(D \subseteq \mathbb{R}\)) is continuous at a point \(a\) if \(\lim{x \rightarrow a} f(x)\) exists and equals \(f(a)\). \(f\) is continuous if it is continuous at every point in the interval.

Combination rules for continuous functions

If \(f, g\) are continuous at \(a\), then so are

• The sum \(f+g\)
• The scalar multiple \(\lambda f\) (\(\lambda \in \mathbb{R}\))
• The product \(fg\)
• The quotient \(\frac{f}{g}\) provided \(g(a) \neq 0\)

If \(f\) is continuous at \(a\) and \(g\) is continuous at \(f(a)\), then the composite \(g \circ f\) is also continuous at \(a\).

\(g \circ f \equiv g(f(x))\).

Basic continuous functions

• polynomials and rational functions*
• The modulus/absolute function
• The square root function, and nth root function where \(n \in \mathbb{Z}^+\)
• Trig functions
• Exponential functions
• Functions defined by power series

*The domain of rational functions exclude the divide by zero bit and so is continuous.

Intermediate Value Theorem

If \(f : [a, b] \longrightarrow \mathbb{R}\) is continuous, \(f(a), f(b)\) have opposite signs, then there is a point \(c\) between \(: f(c) = 0\).

Exercise

Show that there is a number \(x : x^{179} + \frac{163}{1 + x^2} = 119\).

Proof. Rearrange, let \(f(x) = x^{179} + \frac{163}{1 + x^2} - 119\).

\(f(0) > 0,\; f(1) < 0\) thus by intermediate value theorem \(\exists x \in [1, 0] : f(x) = 0\).

(You do have to conjure numbers to do this question)

Extreme Value Theorem

If \(f : [a, b] \longrightarrow \mathbb{R}\) is continuous, then \(\exists m, M \in [a, b]:\)

Basically, there's a maximum and minimum value of a function in a bound.

Differentiation

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If the limit of a real function \(f(x)\) at point \(a\) \(lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}\) exists, then \(f(x)\) is differentiable at \(a\), we get \(f'(a)\).

The \(\frac{d}{dx}\) thing is called Leibniz notation.

Theorem. If \(f\) is differentiable at \(a\), \(f\) is also continuous at \(a\). However, \(f\) may be continuous but not differentiable.

For example, the function \(f(x) = |x|\) is continuous at 0 but not differentiable, because at 0 the relevant limit does not exist (there's a crinkle).

Also though the notes is very from principles ""RIGOUR"" \(\frac{d(x^n)}{dx} = nx^{n-1}\) - in case you didn't know.

Combination Rules for Derivatives

If \(f, g\) are differentiable, the following are also differentiable and follow these rules:

• Sum rule (\(f + g\)): \((f +g)' = f' + g'\)
• Multiple rule (\(lambda f:\lambda \in \mathbb{R}\)): \((\lambda f)' = \lambda f'\)
• Product rule (\(fg\)): \((fg)' = fg' + gf'\)
• Quotient rule (\(f/g\)): \((\frac{f}{g})' = \frac{gf' - fg'}{g^2}\)

Trig derivatives

The Chain Rule

Providing the functions \(y = f(z), \; z = g(x)\) are differentiable

Differentiation of functions def' by power series

If \(\sum(a_nx^n\) is a power series with radius of convergence R, and \(f\) is defined \(f(x) = \sum_{n=0}^\infty a_nx^n : x \in (-R, R)\) then it is differentiable and \(f'(x) = \sum_{n=1}^\infty na_nx^{n-1}\)

Pay attention: \(n=0\) changes to \(n=1\)

The exponential function \(\exp(x)\) or \(e^x\) can be defined as a power series (its Maclaurin series) and thus differentiable to get... \(e^x\).

Exercise

Find \(\sum_{n=1}^\infty \frac{n}{2^n}\).

Solution. You've got to bear with me on this one, because the given solution is clearly written by someone who knew the answer already and have left out a lot of the logic leaps.

We start with a standard definition, for \(|x| = 1\) \(\sum_{n=0}^\infty x^n = \frac{1}{1-x}\) We differentiate the left to get \(\sum_{n=1}^\infty nx^{n-1} = \frac{d}{dx} (\frac{1}{1-x}).\) And the right resolves into \((1-x)^{-2}\).

This derivative form of the power set is very similar to the sum we want to find, so much so that if we take \(x = \frac{1}{2}\), \(\sum_{n=1}^\infty n(\frac{1}{2})^{n-1} = \sum_{n=1}^\infty \frac{n}{2^{n-1}}\) Which resolves down into \(4\) since we can just sub in \(x\) to the other side. Note that if we want to get to \(\frac{n}{2^n}\) from \(\frac{n}{2^n-1}\) we have to multiply by \(\frac{1}{2}\), so we multiply both sides by a half to get \(\sum_{n=1}^\infty \frac{n}{2^n} = 2.\)

Properties of Differentiable Functions

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(Things to be aware of) Turning points, of which are local maxima and minima.

Turning Point Theorem

If a differentiable function \(f\) has a turning point at \(a\), \(f'(a) = 0\).

All points \(a\) where \(f(a) = 0\) are called stationary points, which aren't necessarily turning points. A stationary point which is neither a local maximum nor minimum is a point of inflection

To locate maxima and minima of a continuous function in a range \([a, b]\), we need only consider values at

• The stationary points of \(f\) in \([a, b]\)
• The end points \(a, b\)

Rolle's Theorem

If \(f : [a, b] \longrightarrow \mathbb{R}\) is continuous, is differentiable on \(a, b\) and \(f(a) = f(b)\) then there is a point \(c \in a, b : f'(c) = 0\)

You have two end points with the same \(y\) value, so in the middle there must be some value which is a stationary point. (on horizontal lines, all the points work.)

Mean Value Theorem

If \(f\) is continuous and differentiable over \([a, b]\) then \(\exists c \in [a,b] :\) \(f'(c) = \frac{f(b) - f(a)}{b-a}.\)

Which is just Rolle's theorem but on a angled graph.

Consequences of the MVT

Suppose \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\),

1. • \(f'(x) = 0 \forall x \in (a, b) \implies f\) is constant on \([a, b]\)
   • \(f'(x) > 0 \forall x \in (a, b) \implies f\) is strictly increasing
   • \(f'(x) < 0 \forall x \in (a, b) \implies f\) is strictly decreasing
2. Second Derivative Test. Suppose \(f'(c) = 0\). \(f''(c) > 0 \implies f\) has local minimum; \(f''(c) < 0 \implies f\) has local maximum.

For curve sketching:

• Find stationary points, their co-ordinates, and their nature (max/min/inflection)
• Find where \(f(x) = 0\), i.e. intersects \(x\) axis
• Find \(f\) where \(x = 0\), i.e. y-intercept
• Determine what happens to \(f(x)\) as \(x \rightarrow \pm\infty\) (asymptotes?)
• Investigate near where (if) \(f(x) \rightarrow \infty\), where a divide by zero is found or some other asymptote.

L'Hopital's Rule and Implicit Differentiation

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L'Hopital's Rule

Suppose that \(f(a) = 0, g(a) = 0\). If

\(\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}\) exists then so does \(\lim_{x \rightarrow a} \frac{f(x)}{g(X)}\) also exists, \(\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}.\)

You can repeat this and keep on differentiating if you keep getting \(\frac{0}{0}\), however there is no guarantee that you will get anywhere.

Implicit differentiation

Implicit differentiation deals with functions not of the form \(y = f(x)\), such as \(x^2 + y^2 = 1\).

Two methods (that work the same):

• Either consider a \(y\) term as a function of \(x\), thus because of chain rule append a \(\frac{dy}{dx}\) to the end of each \(y\) term, differentiate \(x\) terms as normal.

• Or, consider both as functions and append a \(dx\) or \(dy\).

Exercise

Find \(\frac{dy}{dx}\) of \(y \sin x + \tan^{-1} y = 0\) in terms of \(y, x\).

Solution. First, differentiate \(y \sin x\). By product rule,

\(d(y \sin x) = \sin(x)dy + y\cos(x)dx\) Then \(\tan^{-1} y\) is a standard derivative, \(d(\tan^{-1} y) = \frac{1}{1+y^2} dy\) Thus we get \(\sin(x)dy + y\cos(x)dx + \frac{1}{1+y^2}dy = 0\) Collecting terms, \((\sin(x) + \frac{1}{1+y^2})dy = (-y\cos(x))dx\) Dividing through, \(\frac{dy}{dx} = \frac{-y \cos x}{\sin x + \frac{1}{1+ y^2}}\) Which one can simplify if they wish.

Differentiating Inverse Functions

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Recall from 130 Injection, Bijection, and Surjection. Recall also the range of a function \(f : A \longrightarrow B\) is \(\{y \in B : f(x) = y, x \in A \}\).

Let \(f : [a, b] \longrightarrow \mathbb{R}\) be a continuous function. If \(f\) differentiable on \((a, b)\) and \(f'(x) > 0 \lor f'(x) < 0 \; \forall x \in (a, b)\), then \(f\) has an inverse function \(f^{-1}\) which is differentiable. Let \(y = f(x);\)

Example

To find the derivative of \(f\) given \(f(x) = x^\frac{1}{n}\) for some fixed \(n > 0\). The inverse function is gotten by writing \(y = x^\frac{1}{n}\), such that \(x = y^n\), thus \(f^{-1}(x) = y^n\). So

\(y = x^\frac{1}{n}\) so \(\frac{dy}{dx} = y^{n-1} = x^\frac{n-1}{n}\), thus

We can differentiate any rational power of \(x\) by the following: \(\frac{d}{dx}x^\frac{p}{q} = ... = \frac{p}{q} x^{\frac{p}{q}-1}.\) same way for regular differentiation.

Trigonometric Inverse Derivatives

• \(\mathbf{sin}^{-1}\). The inverse for the function \(\sin : [\frac{-\pi}{2}, \frac{\pi}{2}] \longrightarrow [-1, 1]\);

• \(\mathbf{cos}^{-1}\). The inverse for the function \(\cos : [0, \pi] \longrightarrow [-1, 1]\);

• \(\mathbf{tan}^{-1}\). The inverse for the function \((\frac{-\pi}{2}, \frac{\pi}{2}) \longrightarrow \mathbb{R}\);

Integration

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Definition of integration

Let \(f:[a, b] \longrightarrow\mathbb{R}\) be a bounded function. A partition of \([a,b]\) is a set \(P = \{x_0, x_1, x_2,...,x_n\} :\) \(a = x_0 < x_1 < x_2 < ... < x_{n-1} < x_n = b.\) For such a partition P let

• \(m_r\) as the greatest lower bound \(\{f(x): x_{r-1} \leq x \leq x_r\}\)
• \(M_r\) as the least upper bound \(\{f(x): x_{r-1} \leq x \leq x_r\}\)

Also let the lower sum \(L(f, P\) and the upper sum \(U(f, P)\)

These are the lower and upper estimates of the area under the graph of \(f\). For more points in the partition, we get a better estimate of an area.

If there is a unique number A: \(L(f,P) \leq A \leq U(f,P)\) for every partition P of \([a, b]\) then we say that \(f\) is integrable over \([a, b]\). A is the definite integral of \(f\) between \(a, b\). \(\vartriangleright\)

Some basic integrable functions (referring to the range \([a, b]\)):

1. Any continuous function.
2. Any function which is increasing.
3. Any function which is decreasing.

After this section are basic rules for integrals. However, I'm going to put collapsible sections in here for more integration techniques one can use.

Integration by Substitution

Like reverse-engineering a chain rule, this methods relies on the integrand being able to be written in the following form: \(\int f(g(x))g'(x)dx\) Where the function \(g(x)\) is contained within a bigger function, and its derivative is outside as a separate factor.

We replace \(g(x)\) with a new variable, such as \(u\) and \(g'(x)dx\) with \(du\), and integrate over just the function of \(u\), \(\int f(u) du\), replacing in what \(u\) is meant to be afterwards.

Why it works...

Let \(u = g(x)\). If we differentiate \(g(x)\) we get \(\frac{du}{dx} = g'(x)\) Now, look at the integral, \(\int f(g(x))g'(x)dx\). Replace every \(g\) instance with the equivalent \(u\) form to get

Try working out \(\int 2x\cos(x^2) dx\) this way, and \(\int \frac{9x^2 + 2}{3x^3+2x+7} dx\).

Answers...

1. \[-\sin(x^2)\]
2. \[\ln(3x^3 + 2x + 7)\]

Tangentially related is the reverse chain rule itself. Say we have an integral, \(\int \cos(3x^2+2x) dx\), we can note that \(\sin (3x^2 + 2x)\) differentiates into \((6x+2) \cos(3x^2+2x) dx\) by chain rule, so integrating the original integrand back we divide by this extra factor, to get \(\frac{\sin(3x^2+2x)}{6x + 2}\).

In certain cases (i.e. like \(\int \frac{1}{1 + x^2} dx\)) we would have to substitute trig functions - in this case \(x = \tan \theta\). (\(dx = \sec^2 (\theta)d\theta\).) Then when we substitute, using said example we get \(\int \frac{1}{\sec^2 \theta} \sec^2 \theta \: d\theta\). This is just \(\int 1 d\theta\) and is \(\theta + c\), or \(\tan^{-1} x + c\).

Integration by Parts

Integration by parts is just a formula:

It can alternatively be written \(\int u \: dv = uv + \int v \: du\)

Try work out \(\int x \cos (x )dx\).

Answer...

Properties of definite intervals

1. Sum rule: \(f, g\) are integrable then so is \(f+g\), and \(\int_{a}^{b} (f(x) + g(x))dx = \int_{a}^{b} f(x)dx + \int_{a}^{b} g(x)dx.\)
2. Multiple (scalar) rule: \(f\) integrable, \(\lambda \in \mathbb{R}\), then \(\lambda f\) integrable, and \(\int_a^b \lambda f(x)dx = \lambda \int_a^b f(x)dx.\)
3. \(f\) integrable over \([a,c]\) and \([c,b]\), then \(f\) also integrable over \([a,b]\), and \(\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx.\)
4. If \(f(x) \leq g(x) \; \forall x \in [a, b]\) then provided both integrals exist, \(\int_a^b f(x)dx \leq \int_a^b g(x)dx.\)

Fundamental theorems of calculus

First Fundamental Theorem of Calculus

Let \(f : [a,b] \longrightarrow \mathbb{R}\) be integrable, and define \(F : [a,b] \longrightarrow \mathbb{R}\) as \(F(x) = \int_a^x f(t)dt.\) If \(f\) continuous at \(c \in (a, b)\) then \(F\) is differentiable at \(c\), and \(F'(c) = f(c).\)

Second Fundamental Theorem of Calculus

Let \(f : [a,b] \longrightarrow \mathbb{R}\) be continuous and suppose \(F\) is a differentiable function : \(F' = f\), then \(\int_a^b f(x)dx = [F(x)]^b_a\) Where \([F(x)]^b_a\) denotes \(F(b) - F(a)\)

Alternatively, \(\frac{d}{dx}F(x) = f(x) \; \forall x\), and \(f\) continuous, then \(\int f(x)dx = F(x) + c\) for some c.

Logs and Exponentials

Link to the PDF.

In the notes \(\log x\) means the natural log of \(x\), but here I will use \(\ln x\). It is always useful to specify.

For each \(x > 0\); \(\ln x = \int_1^x \frac{1}{t}dt.\)

Properties of Logarithms

1. \[\ln(1) = 0.\]
2. log is a strictly increasing function.
3. log is a differentiable function, \(\frac{d(\ln x)}{dx}\)
4. \[\ln(xy) = \ln x + \ln y\]
5. \[\ln(\frac{x}{y}) = \ln x - \ln y\]
6. \(\ln : (0, \infty) \longrightarrow \mathbb{R}\) is bijective

Since \(\ln : (0, \infty) \longrightarrow \mathbb{R}\) is bijective, it has an inverse function \(\exp : \mathbb{R} \longrightarrow (0, \infty)\). \(e = \exp(1); e^x = \exp(x)\).

Properties of Exponentials

For any \(x, y \in \mathbb{R}\),

1. \(\exp(x+y) = \exp(x)\exp(y)\) but likewise \(e^{x+y} = e^xe^y\)
2. \(\frac{d(\exp(x))}{dx} = \exp(x)\), \(\exp(0) = 1\)
3. \[\exp(x) = \lim_{n \rightarrow \infty} (1 + (\frac{x}{n})^n\]
4. \[\exp(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}\]

For \(a > 0, x \in \mathbb{R}\), any exponential can be defined by \(a^x = e^{x \ln a}\) Then for \(a, b > 0; x, y \in \mathbb{R}\) the following properties hold:

We can change bases of logarithms (provided \(b \neq 1\)); \(\log_b x = \frac{\ln x}{\ln b}\)

Taylor's Theorem

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Taylor's Theorem

Let \(f\) be an \(n+1\) times differentiable function on an open interval containing some points \(a, x\). Then \(f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + ... + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x)\) Where \(R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\) For some number \(c\) between \(a, x\).

The resulting polynomial from evaluating some \(n\) terms is called the Taylor polynomial of degree \(n\) of \(f\) at \(a\). This is however an approximation, and so the end term \(R_n(x)\) is the error in the approximation.

Taylor series

If we can show \(R_n (x) \rightarrow 0\) as \(n \rightarrow \infty\) then we get a sequence of better and better approximations until we get the Taylor series, \(f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n\) This will only converge for values of \(x\) within the radius of convergence of the power series. They can be used to approximate functions.

When \(n = 0\), Taylor's reduces down to the mean value theorem. (Proof of Taylor's is omitted)

Nth derivative test for the nature of stationary points

Suppose a function \(f\) has a stationary point at \(a\) however \(f'(a) = f''(a) = ... = f^{(n-1)} = 0\), while \(f^{n}(a) \neq 0\). If \(f^{n}(a)\) is continuous,

1. \(n\) even, \(f^{(n)}(a) > 0\); \(f\) has a local maximum
2. \(n\) even, \(f^{(n)}(a) < 0\); \(f\) has a local minimum
3. \(n\) odd; \(f\) has a point of inflection at \(a\)

Maclaurin Series

When \(a = 0\), we get the Maclaurin's series, which is a special case of Taylor's series \(f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n\)

Important Maclaurin Series

These are all expressible as sums using the nth term, though bear in mind the last two start from \(n=1\).

First Order ODEs

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ODEs, Ordinary Differential Equations, are equations including derivatives, such as \(y' = 4x\) or \(y'' + 4y = x\), etc.

The order of a DE represents the highest derivative contained within it. This section looks at 1st orders.

First order equations of the form \(y' = f(x)\) can be easily solved via integration. This will give us a general solution with the \(+ c\), representing a family of curves. If we are given constraints, we can get a particular solution.

Separable Equations

A [1st order] ODE is called a **separable equation if it can be written in the form \(\frac{dy}{dx} = f(x)g(y)\) Since this can be rewritten as \(\frac{1}{g(y)} \frac{dy}{dx} = f(x) \textrm{ or } \frac{1}{g(y)}dy = f(x)dx\) Which can be directly integrated, \(\int \frac{1}{g(y)} dy = \int f(x)dx.\)

An example of a separable equation is \(\frac{dy}{dx} + xy = 0\).

Homogenous Equations

An ODE is called homogenous if it can be written in the form \(\frac{dy}{dx} = f(\frac{y}{x})\) We let some variable \(v = \frac{y}{x} : y = xv\), thus \(\frac{d(xv)}{dx} = f(v) \Longleftrightarrow x\frac{dv}{dx} + v = f(v) \Longleftrightarrow \frac{dv}{dx} = \frac{f(v) -v}{x}\) This brings it into a separable form, into

And replace \(v\) by \(\frac{y}{x}\) to get the original general equation.

The following exercise may help.

Exercise

Solve \(2xy \frac{dy}{dx} = y^2 - x^2.\)

Solution. We can rearrange; \(\implies \frac{dy}{dx} = \frac{y^2-x^2}{2xy} = \frac{1}{2}(\frac{y^2}{xy} - \frac{x^2}{xy}) = \frac{1}{2}(\frac{y}{x}-\frac{x}{y}).\) This is homogenous, so we apply the above principle to this equation;

Which is now in separable form. Now, skipping straight to \(\ln x = \dots\) form,

Let the constant \(k = e^c\) and replacing \(v\) by \(\frac{y}{x}\). Thus \(x(1 + \frac{y^2}{x^2}) = k \Longleftrightarrow x^2 + y^2 = kx.\)

Linear Equations

An ODE is called linear if it has the form \(\frac{dy}{dx} + P(x)y = Q(x).\)

Related to this is the integrating factor, \(I(x) = e^{\int P(x) dx}\). This can be derived if \(Q(x) = 0\).

We multiply both sides of the equation by \(I(x)\), to get \(\frac{dy}{dx}I(x) + yI(x)P(x) = Q(x)I(x)\) Note the LHS is the result of the implicit differentiation of \(yI(x)\), thus

Exercise

Find the general solution to \((1+x^2)\frac{dy}{dx} + xy = x\sqrt{1+x^2},\) and the particular solution which satisfies \(y = \frac{1}{2}, x=0\).

Solution. We can rearrange the equation to \(\frac{dy}{dx} + \frac{xy}{1+x^2} = \frac{x}{\sqrt{1+x^2}}\) The integrating factor is thus \(e^{\int \frac{x}{1+x^2}dx} = e^{\frac{1}{2} \ln (1+x^2)} = (1 + x^2)^{\frac{1}{2}}.\) Multiplying by this integrating factor,

Hence the general solution is \(\sqrt{1+x^2}y = x^2 / 2 + c.\) The particular solution is when you substitute the values for \(y, x\) into the equation to find c, and \(c = \frac{1}{2}\). That gives you the final specific curve.

Second Order ODEs

Link to the PDF.

These are where there is a \(y''\) term, and are dealt with very similarly to recurrences. They take the form \(ay'' + by' + cy = f(x)\)

If \(f(x) = 0\), then the equation is homogenous. To find the general solution, find the general solution of \(ay'' + by' + cy = 0\) find the particular solution of \(ay'' + by' + cy = f(x)\) and add them together.

Solving homogenous equations

For the homogenous equation \(ay'' + by' + cy = 0\)

There's a lot of waffle, but we have an auxiliary equation that is relevant, which is \(a\lambda^2 + b\lambda + c = 0\)

The roots of this equation determine the form of the general solution.

• Roots are real and distinct. The general solution is \(y = Ae^{\lambda_1 x} + Be^{\lambda_2 x}\).
• Roots are equal. The general solution is \(y = Ae^{\lambda x} + Bxe^{\lambda x} = (A + Bx) e^{\lambda x}\).
• Complex Roots. Let \(\lambda_1 = \alpha + i\beta\) and \(\lambda_2 = \alpha - i\beta\). Then, the general equation is \(y = e^{\alpha x} (A \cos \beta x + B \sin \beta x)\).

Finding particular solutions

For the equation \(ay'' + by' + cy = f(x)\) Like in recurrences, we want like functions.