nyuuzyou/wikis huggingface dataset:
Method of Truth Tables/Proof of Logical Implication

== Proof Technique ==
Suppose we wish to prove that $P \vdash Q$ for two propositional formulas $P$ and $Q$.
:Example: Let $P$ be $\neg p$ and $Q$ be $p \implies q$.
=== Express two statements as a single WFF ===
We express $P \vdash Q$ as a single WFF $\paren {P \implies Q}$ and perform the method of truth tables on that expression.
Demonstrating this with the example given:
$\begin{array}{cc||cccccc}
p & q & (\neg & p ) & \implies & (p & \implies & q) \\
\hline
\F & \F & \T & \F & \T & \F & \T & \F \\
\F & \T & \T & \F & \T & \F & \T & \T \\
\T & \F & \F & \T & \T & \T & \F & \F \\
\T & \T & \F & \T & \T & \T & \T & \T \\
\end{array}$
As can be seen, the column under the main connective $\implies$ of $\paren {P \implies Q}$ is all $\T$.
Hence $\paren {\paren {\neg p} \implies \paren {p \implies q} }$ is a tautology.
Hence from Equivalence of Logical Implication and Conditional, $\neg p \vdash p \implies q$.
=== Compare two WFFs in the same table ===
Alternatively, we can place the two WFFs side by side in the same truth table:
$\begin{array}{cc||cc||ccc}
p & q & \neg & p & p & \implies & q \\
\hline
\F & \F & \T & \F & \F & \T & \F \\
\F & \T & \T & \F & \F & \T & \T \\
\T & \F & \F & \T & \T & \F & \F \\
\T & \T & \F & \T & \T & \T & \T \\
\end{array}$
This time, we need to make sure that when the truth values in the columns under the first main connectives is $\T$, then it is also $\T$ under the second.
Note that this is exactly the same as putting a $\implies$ between the two and making a WFF out of the pair of them.
Category:Proof Techniques
Category:Truth Tables
Method of Truth Tables/Proof of Tautology

== Proof Technique ==
This is used to establish whether or not a given propositional formula is a tautology for boolean interpretations; that, is valid in all boolean interpretations.
Let $P$ be a propositional formula we wish to validate.
Subsequently, determine its truth table.
In the column under the main connective of $P$ itself can be found the truth value of $P$ for each boolean interpretation.
:If this contains nothing but $\T$, then $P$ is a tautology.
:If this contains nothing but $\F$, then $P$ is a contradiction.
:If this contains $\T$ for some boolean interpretations and $\F$ for others, then $P$ is a contingent statement.
== Examples ==
== Sources ==
* : $\S 3.3$: The Construction and Application of Truth-Tables
* : Chapter $2$: The Propositional Calculus $2$: $3$ Truth-Tables
* : $\S \text{I}: 4$: Using the Definitions
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables
* : $\S 1.6$: Truth Tables and Tautologies
Category:Proof Techniques
Category:Truth Tables
Method of Truth Tables/Proof of Tautology/Examples/Peirce's Law

== Examples of Proof of Tautology ==
Consider the truth table for Peirce's Law:
:$P = \paren {\paren {p \implies q} \implies p} \implies p$
which is:
$\begin{array}{cc||ccccccc}
p & q & ((p & \implies & q) & \implies & p) & \implies & p \\
\hline
\F & \F & \F & \T & \F & \F & \F & \T & \F \\
\F & \T & \F & \T & \T & \F & \F & \T & \F \\
\T & \F & \T & \F & \F & \T & \T & \T & \T \\
\T & \T & \T & \T & \T & \T & \T & \T & \T \\
\end{array}$
The main connective of $P$ is the rightmost instance of $\implies$.
The column beneath that connective is all $\T$, so $\paren {\paren {p \implies q} \implies p} \implies p$ is a tautology.
Category:Truth Tables
Method of Undetermined Coefficients

== Proof Technique ==
The Method of Undetermined Coefficients is a technique for solving a nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y' ' + p y' + q y = \map R x$
where $\map R x$ is one of the following types of expression:
:an exponential
:a sine or a cosine
:a polynomial
or a combination of such real functions.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y' ' + p y' + q y = 0$
From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically.
Let $\map {y_p} x$ be a particular solution to $(1)$.
Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
=== Exponential Function ===
=== Sine and Cosine Functions ===
=== Exponential of Sine and Cosine Functions ===
=== Polynomial Function ===
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.2$ The general equation
* : $\S 3.18$: The Method of Undetermined Coefficients
* 
Category:Method of Undetermined Coefficients
Category:Constant Coefficient LSOODEs
Category:Proof Techniques
Method of Undetermined Coefficients/Exponential

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y' ' + p y' + q y = \map R x$
Let $\map R x$ be of the form of an exponential function:
:$\map R x = K e^{a x}$
The Method of Undetermined Coefficients can be used to solve $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y' ' + p y' + q y = 0$
From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically.
Let $\map {y_p} x$ be a particular solution to $(1)$.
Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Let $\map R x = K e^{a x}$.
Consider the auxiliary equation to $(1)$:
:$(2): \quad m^2 + p m + q = 0$
There are three cases to consider.
;$a$ is not a root of $(2)$
Assume that there is a particular solution to $(1)$ of the form:
:$y_p = A e^{a x}$
We have:
Inserting into $(1)$:
Hence:
:$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$
From Solution of Constant Coefficient Homogeneous LSOODE, $y_g$ depends on whether $(2)$ has equal or unequal roots.
Let $m_1$ and $m_2$ be the roots of $(2)$.
Then:
:$y = \begin{cases} C_1 e^{m_1 x} + C_2 e^{m_2 x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 \ne m_2: m_1, m_2 \in \R \\ C_1 e^{m_1 x} + C_2 x e^{m_1 x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 = m_2 \\ e^{r x} \paren {C_1 \cos s x + C_2 \sin s x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 = r + i s, m_2 = r - i s \end{cases}$
is the general solution to $(1)$.
;$a$ is a root of $(2)$
If $a$ is a root of $(2)$, then $a^2 + p a + q = 0$ and so $\dfrac {K e^{a x} } {a^2 + p a + q}$ is undefined.
Let the auxiliary equation to $(2)$ have two unequal real roots $a$ and $b$.
Assume that there is a particular solution to $(1)$ of the form:
:$y_p = A x e^{a x}$
We have:
Inserting into $(1)$:
Hence:
:$y_p = \dfrac {K x e^{a x} } {2 a + p}$
and so from Solution of Constant Coefficient Homogeneous LSOODE:
:$y = C_1 e^{a x} + C_2 e^{b x} + \dfrac {K x e^{a x} } {2 a + p}$
is the general solution to $(1)$.
;$a$ is a repeated root of $(2)$
If the auxiliary equation to $(2)$ has two equal real roots $a$, then:
:$a = - \dfrac p 2$
and so not only:
:$a^2 + p a + q = 0$
but also:
:$2 a + p = 0$
and so neither of the above expressions involving $e^{a x}$ and $x e^{a x}$ will work as particular solution to $(1)$.
So, assume that there is a particular solution to $(1)$ of the form:
:$y_p = A x^2 e^{a x}$
We have:
Inserting into $(1)$:
Hence:
:$y_p = \dfrac {K x^2 e^{a x} } 2$
and so:
:$y = y_g + \dfrac {K x^2 e^{a x} } 2$
and so from Solution of Constant Coefficient Homogeneous LSOODE:
:$y = C_1 e^{a x} + C_2 x e^{a x} + \dfrac {K x^2 e^{a x} } 2$
is the general solution to $(1)$.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.4$ Particular solution: exponential $\map f x$
* : $\S 3.18$: The Method of Undetermined Coefficients
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Exponential of Sine and Cosine

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y + p y' + q y = \map R x$
Let $\map R x$ be of the form:
:$\map R x = e^{a x} \paren {\alpha \sin b x + \beta \cos b x}$
The Method of Undetermined Coefficients can be used to solve $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$(2): \quad y + p y' + q y = 0$
From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically.
Let $\map {y_p} x$ be a particular solution to $(1)$.
Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Substitute a trial solution of similar form, either:
:$e^{a x} \paren {A \sin b x + B \cos b x}$
or replace the of $(1)$ by:
:$\paren {\alpha - i \beta} e^{i \paren {a + i b} x}$
find a solution, and take the real part.
If $e^{a x} \sin b x$ and $e^{a x} \cos b x$ appear in the general solution to $(2)$, then insert a factor of $x$:
:$x e^{a x} \paren {A \sin b x + B \cos b x}$
or:
:$x \paren {\alpha - i \beta} e^{i \paren {a + i b} x}$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(i)}$
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Polynomial

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y + p y' + q y = \map R x$
Let $\map R x$ be a polynomial in $x$:
:$\ds \map R x = \sum_{j \mathop = 0}^n a_j x^j$
The Method of Undetermined Coefficients can be used to solve $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y + p y' + q y = 0$
From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically.
Let $\map {y_p} x$ be a particular solution to $(1)$.
Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Let $\ds \map R x = \sum_{j \mathop = 0}^n a_j x^j$.
Assume that there is a particular solution to $(1)$ of the form:
:$\ds y_p = \sum_{j \mathop = 0}^n A_j x^j$
We have:
Inserting into $(1)$:
The coefficients $A_0$ to $A_n$ can hence be solved in terms of $a_0$ to $a_n$ using the techniques of simultaneous equations.
The general form is tedious and unenlightening to evaluate.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.3$ Particular solution: polynomial $\map f x$
* : $\S 3.18$: The Method of Undetermined Coefficients
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Product of Polynomial and Exponential

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y + p y' + q y = \map R x$
Let $\map R x$ be of the form:
:$\map R x = e^{a x} \paren {\map f x}$
where $\map f x$ is a real polynomial function.
The Method of Undetermined Coefficients can be used to solve $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$(2): \quad y + p y' + q y = 0$
From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically.
Let $\map {y_p} x$ be a particular solution to $(1)$.
Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Substitute a trial solution of similar form:
:$e^{a x} \paren {\map g x}$
where $\map g x$ is a real polynomial function with undetermined coefficients of as high a degree as $f$.
Then:
:differentiate twice $x$
:establish a set of simultaneous equations by equating powers
:solve these equations for the coefficients.
If $e^{a x} \paren {\map g x}$ appears in the general solution to $(2)$, then add a further degree to $g$.
The last step may need to be repeated if that last polynomial also appears as a general solution to $(2)$.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(ii)}$
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Product of Polynomial and Function of Sine and Cosine

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y' ' + p y' + q y = \map R x$
Let $\map R x$ be of the form:
:$\map R x = \paren {\alpha \cos b x + \beta \sin b x} \paren {\map f x}$
where $\map f x$ is a real polynomial function.
The Method of Undetermined Coefficients can be used to solve $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$(2): \quad y' ' + p y' + q y = 0$
From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically.
Let $\map {y_p} x$ be a particular solution to $(1)$.
Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Substitute a trial solution of similar form, either:
:$\paren {\alpha \cos b x + \beta \sin b x} \paren {\map g x}$
or replace the of $(1)$ by:
:$\paren {\alpha - i \beta} e^{i \paren {a + i b} x} \paren {\map g x}$
find a solution, and take the real part.
In the above, $\map g x$ is a real polynomial function with undetermined coefficients of as high a degree as $f$.
Then:
:differentiate twice $x$
:establish a set of simultaneous equations by equating powers
:solve these equations for the coefficients.
If $\paren {\alpha \cos b x + \beta \sin b x} \paren {\map g x}$ appears in the general solution to $(2)$, then add a further degree to $g$.
The last step may need to be repeated if that last polynomial also appears as a general solution to $(2)$.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(iii)}$
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Sine and Cosine

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y' ' + p y' + q y = \map R x$
Let $\map R x$ be a linear combination of sine and cosine:
:$\map R x = \alpha \sin b x + \beta \cos b x$
The Method of Undetermined Coefficients can be used to solve $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y' ' + p y' + q y = 0$
From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically.
Let $\map {y_p} x$ be a particular solution to $(1)$.
Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Let $\map R x = \alpha \sin b x + \beta \cos b x$.
Consider the auxiliary equation to $(1)$:
:$(2): \quad m^2 + p m + q = 0$
There are two cases which may apply.
=== $i b$ is not Root of Auxiliary Equation ===
First we investigate the case where $i b$ is not a root of the auxiliary equation to $(1)$.
=== $i b$ is Root of Auxiliary Equation ===
Now suppose that $(1)$ is of the form $y' ' + b^2 y = A \sin b x + B \cos b x$.
Thus one of the $i b$ is one of the roots of the auxiliary equation to $(1)$.
From Linear Second Order ODE: $y + k^2 y = 0$ the general solution to $(2)$ is:
:$y = C_1 \sin b x + C_2 \cos b x$
and it can be seen that an expression of the form $A \sin b x + B \cos b x$ is already a particular solution of $(2)$.
Thus we have:
:$\paren {q - b^2}^2 + b^2 p^2 = 0$.
But using the Method of Undetermined Coefficients in the above manner, this would result in an attempt to calculate:
:$\dfrac {\alpha \paren {q - b^2} + \beta b p} {\paren {q - b^2}^2 + b^2 p^2}$
and:
:$\dfrac {\beta \paren {q - b^2} - \alpha b p} {\paren {q - b^2}^2 + b^2 p^2}$
both of which are are undefined.
== Sources ==
* : $\S 3.18$: The Method of Undetermined Coefficients
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Sine and Cosine/Particular Solution

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y' ' + p y' + q y = \map R x$
Let $\map R x$ be a linear combination of sine and cosine:
:$\map R x = \alpha \sin b x + \beta \cos b x$
The Method of Undetermined Coefficients can be used to find a particular solution to $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y' ' + p y' + q y = 0$
From General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
<onlyinclude>
Let $\map R x = \alpha \sin b x + \beta \cos b x$.
Consider the auxiliary equation to $(1)$:
:$(2): \quad m^2 + p m + q = 0$
There are two cases which may apply.
=== $i b$ is not Root of Auxiliary Equation ===
=== $i b$ is Root of Auxiliary Equation ===
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is Root of Auxiliary Equation

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y' ' + b^2 y = \alpha \sin b x + \beta \cos b x$
The Method of Undetermined Coefficients can be used to find a particular solution to $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y' ' + b^2 y = 0$
From General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Consider the auxiliary equation to $(1)$:
:$(2): \quad m^2 + b^2 = 0$
We have that $i b$ is a root of $(2)$.
=== Trigonometric Form ===
=== Exponential Form ===
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.5$ Particular solution: trigonometric $\map f x$
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is Root of Auxiliary Equation/Exponential Form

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y' ' + b^2 y = \alpha \sin b x + \beta \cos b x$
The Method of Undetermined Coefficients can be used to find a particular solution to $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y' ' + b^2 y = 0$
From General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Assume that there is a particular solution to $(1)$ of the form:
:$y_p = x \paren {A \sin b x + B \cos b x}$
From Euler's Formula:
:$\cos b x + i \sin b x = e^{i b x}$
and so:
:$x \paren {A \sin b x + B \cos b x}$ is the real part of $x \paren {A - i B} \paren {\cos b x + i \sin b x} = x \paren {A - i B} e^{i b x}$
It is assumed that $A$, $B$, $p$ and $q$ are all real numbers.
Suppose we have found a solution $y$ of $(1)$ where:
:$\map f x = \map {f_1} x + i \, \map {f_2} x$
where $\map y x$ and $\map f x$ are complex-valued.
Letting $\map y x = \map {y_1} x + \map {y_2} x$, where $y_1$ and $y_2$ are the real and imaginary parts of $\map y x$, we have:
:${y_1}' ' + p {y_1}' + q y_1 + i \paren { {y_2}' ' + p {y_2}' + q y_2} = \map {f_1} x + i \, \map {f_2} x$
Equating real parts:
:${y_1}' ' + p {y_1}' + q y_1 = \map {f_1} x$
Equating imaginary parts:
:${y_2}' ' + p {y_2}' + q y_2 = \map {f_2} x$
Thus if $y$ is a particular solution to $(1)$ when the is $\map f x$:
:$\map \Re y$ is a particular solution to $(1)$ when the is $\map \Re {\map f x}$
:$\map \Im y$ is a particular solution to $(1)$ when the is $\map \Im {\map f x}$
So to find a particular solution when the is $K \cos x$ or $K \sin x$, we can first find a particular solution when the is $K e^{i b x}$ and then take its real part or imaginary part as necessary.
Hence, when we have $A \cos b x + B \sin b x$ on the :
:replace it with $x \paren {A - i B} e^{i b x}$
:use the Method of Undetermined Coefficients for Exponential functions
and then take its real part.
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is Root of Auxiliary Equation/Trigonometric Form

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y' ' + b^2 y = \alpha \sin b x + \beta \cos b x$
The Method of Undetermined Coefficients can be used to find a particular solution to $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y' ' + b^2 y = 0$
From General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Assume that there is a particular solution to $(1)$ of the form:
:$y_p = x \paren {A \sin b x + B \cos b x}$
We have:
Inserting into $(1)$:
Hence $A$ and $B$ can be expressed in terms of $\alpha$ and $\beta$:
Hence:
:$y_p = \dfrac {\beta x \sin b x} {2 b} - \dfrac {\alpha x \cos b x} {2 b}$
== Sources ==
* : $\S 3.18$: The Method of Undetermined Coefficients
Category:Method of Undetermined Coefficients
Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is not Root of Auxiliary Equation

== Proof Technique ==
Consider the nonhomogeneous linear second order ODE with constant coefficients:
:$(1): \quad y + p y' + q y = \map R x$
Let $\map R x$ be a linear combination of sine and cosine:
:$\map R x = \alpha \sin b x + \beta \cos b x$
such that $i b$ is not a root of the auxiliary equation to $(1)$.
The Method of Undetermined Coefficients can be used to find a particular solution to $(1)$ in the following manner.
== Method and Proof ==
Let $\map {y_g} x$ be the general solution to:
:$y + p y' + q y = 0$
From General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$\map {y_g} x + \map {y_p} x$
is the general solution to $(1)$.
It remains to find $\map {y_p} x$.
Let $\map R x = \alpha \sin b x + \beta \cos b x$.
Consider the auxiliary equation to $(1)$:
:$(2): \quad m^2 + p m + q = 0$
We have that $i b$ is not a root of $(2)$.
=== Trigonometric Form ===
=== Exponential Form ===
Category:Method of Undetermined Coefficients