nyuuzyou/wikis huggingface dataset:
Second Derivative of Locus of Cycloid

== Theorem ==
Consider a circle of radius $a$ rolling without slipping along the x-axis of a cartesian plane.
Consider the point $P$ on the circumference of this circle which is at the origin when its center is on the y-axis.
Consider the cycloid traced out by the point $P$.
Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.
The second derivative of the locus of $P$ is given by:
:$y = -\dfrac a {y^2}$
== Proof ==
From Equation of Cycloid:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
From Slope of Tangent to Cycloid:
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $2$
Category:Cycloids
Second Derivative of Natural Logarithm Function

== Theorem ==
Let $\ln x$ be the natural logarithm function.
Then:
:$\map {\dfrac {\d^2} {\d x^2} } {\ln x} = -\dfrac 1 {x^2}$
== Proof ==
From Derivative of Natural Logarithm Function:
:$\dfrac \d {\d x} \ln x = \dfrac 1 x$
From the Power Rule for Derivatives: Integer Index:
:$\dfrac {\d^2} {\d x^2} \ln x = \dfrac \d {\d x} \dfrac 1 x = -\dfrac 1 {x^2}$
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation
* : $\S 14.1$
Category:Derivatives involving Logarithm Functions
Category:Natural Logarithms
Second Derivative of PGF of Negative Binomial Distribution/Type 1

== Theorem ==
Let $X$ be a discrete random variable with the type $1$ negative binomial distribution with parameters $n$ and $p$.
Then the second derivative of the PGF of $X$ $s$ is:
:$\dfrac {\d^2} {\d s^2} \map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^{n + 2} \paren {\dfrac {n \paren {n - 1} + 2 n q s} {\paren {p s^2}^2} }$
== Proof ==
The Probability Generating Function of Negative Binomial Distribution (Type 1) is:
:$\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
We have that for a given negative binomial distribution, $n, p$ and $q$ are constant.
From First Derivative of PGF of Negative Binomial Distribution (Type 1):
Thus we have:
Category:Second Derivative of PGF of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 1)
Category:Derivatives of PGFs
Second Derivative of PGF of Negative Binomial Distribution/Type 2

== Theorem ==
Let $X$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $n$ and $p$.
Then the second derivative of the PGF of $X$ $s$ is:
:$\dfrac {\d^2} {\d s^2} \map {\Pi_X} s = \dfrac {n \paren {n + 1} p^2} {q^2} \paren {\dfrac q {1 - p s} }^{n + 2}$
where $q = 1 - p$.
== Proof ==
The Probability Generating Function of Negative Binomial Distribution (Type 2) is:
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^n$
From Derivatives of PGF of Negative Binomial Distribution: Type 2:
:$(1): \quad \dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \dfrac {n^{\overline k} p^k} {q^k} \paren {\dfrac q {1 - p s} }^{n + k}$
where:
:$n^{\overline k}$ is the rising factorial: $n^{\overline k} = n \paren {n + 1} \paren {n + 2} \cdots \paren {n + k - 1}$
:$q = 1 - p$
Putting $k = 2$ in $(1)$ above yields the required solution.
Category:Second Derivative of PGF of Negative Binomial Distribution
Category:Negative Binomial Distribution (Type 2)
Category:Derivatives of PGFs
Second Dual Operator Extends Operator

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $X^{\ast \ast}$ and $Y^{\ast \ast}$ be the second normed duals of $X$ and $Y$ respectively. 
Let $\iota_X : X \to X^{\ast \ast}$ be the evaluation linear transformation on $X$.
Let $\iota_Y : Y \to Y^{\ast \ast}$ be the evaluation linear transformation on $Y$.
Let $T^{\ast \ast} : X^{\ast \ast} \to Y^{\ast \ast}$ be the second dual operator of $T$.
Then $T^{\ast \ast}$ extends $T$ in the sense that: 
:$\map {T^{\ast \ast} } {\iota_X x} = \map {\iota_Y} {T x}$
== Proof ==
It is shown in Dual Operator is Weak-* to Weak-* Continuous that for each $x \in X$ we have: 
:$x^\wedge \circ T^\ast = \paren {T x}^\wedge$
where $x^\wedge = \iota_X x$ and $\paren {T x}^\wedge = \map {\iota_Y} {T x}$. 
From the definition of the second dual operator and the evaluation linear transformations, this is precisely the claim: 
:$\map {T^{\ast \ast} } {\iota_X x} = \map {\iota_Y} {T x}$
Category:Second Dual Operators
Second Hyperoperation is Multiplication Operation

== Theorem ==
The $2$nd hyperoperation is the multiplication operation restricted to the positive integers:
:$\forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = x \times y$
== Proof ==
By definition of the hyperoperation sequence:
:$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases}
y + 1 & : n = 0 \\
x & : n = 1, y = 0 \\
0 & : n = 2, y = 0 \\
1 & : n > 2, y = 0 \\
H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$
Thus the $2$nd hyperoperation is defined as:
:$\forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = \begin{cases}
0 & : y = 0 \\
H_1 \left({x, H_2 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$
From First Hyperoperation is Addition Operation:
:$(1): \quad \forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = \begin{cases}
0 & : y = 0 \\
x + H_2 \left({x, y - 1}\right) & : y > 0 \end{cases}$
The proof proceeds by induction.
For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition:
:$\forall x \in \Z_{\ge 0}: H_2 \left({x, y}\right) = x \times y$
=== Basis for the Induction ===
$P \left({0}\right)$ is the case:
Thus $P \left({0}\right)$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
:$\forall x \in \Z_{\ge 0}: H_2 \left({x, k}\right) = x \times k$
from which it is to be shown that:
:$\forall x \in \Z_{\ge 0}: H_2 \left({x, k + 1}\right) = x \times \left({k + 1}\right)$
=== Induction Step ===
This is the induction step:
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = x \times y$
Category:Hyperoperation
Second Inversion Formula for Stirling Numbers

== Theorem ==
For all $m, n \in \Z_{\ge 0}$:
:$\ds \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
where:
:$\ds {n \brace k}$ denotes a Stirling number of the second kind
:$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
:$\delta_{m n}$ denotes the Kronecker delta.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ has been shown to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true for all $r \ge 0$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {r \brace k} {k \brack m} \paren {-1}^{r - k} = \delta_{m r}$
from which it is to be shown that:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {r + 1 \brace k} {k \brack m} \paren {-1}^{r + 1 - k} = \delta_{m \paren {r + 1} }$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
== Also see ==
* First Inversion Formula for Stirling Numbers
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(47)$
Category:Stirling Numbers
Second Isomorphism Theorem

== Theorem ==
=== Groups ===
=== Rings ===
This result is also referred to by some sources as the first isomorphism theorem.
== Also known as ==
There is no standard numbering for the Isomorphism Theorems. Different authors use different labellings.
This particular result, for example, is also known as the first isomorphism theorem.
== Also see ==
* Isomorphism Theorems
Category:Named Theorems
Category:Isomorphisms (Abstract Algebra)
Category:Isomorphism Theorems
Second Isomorphism Theorem/Groups

== Theorem ==
Let $G$ be a group, and let:
:$(1): \quad H$ be a subgroup of $G$
:$(2): \quad N$ be a normal subgroup of $G$.
Then:
:$\dfrac H {H \cap N} \cong \dfrac {H N} N$
where $\cong$ denotes group isomorphism.
== Proof ==
The fact that $N$ is normal, together with Intersection with Normal Subgroup is Normal, gives us that $N \cap H \lhd H$.
Also, $N \lhd N H = \gen {H, N}$ follows from Subset Product with Normal Subgroup as Generator.
Now we define a mapping $\phi: H \to H N / N$ by the rule:
:$\map \phi h = h N$
Note that $N$ need not be a subset of $H$.
Therefore, the coset $h N$ is an element of $H N / N$ rather than of $H / N$.
Then $\phi$ is a homomorphism, as:
:$\map \phi {x y} = x y N = \paren {x N} \paren {y N} = \map \phi x \map \phi y$
Then:
Then we see that $\phi$ is a surjection because $h n N = h N \in H N / N$ is $\map \phi h$.
The result follows from the First Isomorphism Theorem.
== Also known as ==
This result is also referred to by some sources as the first isomorphism theorem.
== Also see ==
* Isomorphism Theorems
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.16$
* : $1$: Subgroups: $1.\text{T}.5$
* : Chapter $\text{II}$: Groups: Problem $\text{HH}$
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 69$. The Second Isomorphism Theorem
* : Chapter $8$: The Homomorphism Theorem: Theorem $8.15$
Category:Group Isomorphisms
Category:Normal Subgroups
Category:Isomorphism Theorems
Second Isomorphism Theorem/Rings

== Theorem ==
Let $R$ be a ring, and let:
:$S$ be a subring of $R$
:$J$ be an ideal of $R$.
Then:
:$(1): \quad S + J$ is a subring of $R$
:$(2): \quad J$ is an ideal of $S + J$
:$(3): \quad S \cap J$ is an ideal of $S$
:$(4): \quad \dfrac S {S \cap J} \cong \dfrac {S + J} J$
where $\cong$ denotes group isomorphism.
== Proof ==
The relations being defined can be illustrated by this commutative diagram:
:
$(1): \quad S + J$ is a subring of $R$
From Sum of All Ring Products is Additive Subgroup, $S + J$ is an additive subgroup of $R$.
Suppose $s, s' \in S, j, j' \in J$.
Then:
:$\paren {s + j} \paren {s' + j'}$
so by the Subring Test $S + J$ is a subring of $R$.
$(2): \quad J$ is an ideal of $S + J$
Let $s + j \in S + J$ and let $j \in J$.
Then:
So $J$ is an ideal of $S + J$.
$(3): \quad S \cap J$ is an ideal of $S$
Let $\nu: R \to R / J$ be the quotient epimorphism.
Let $\nu'$ be the restriction of $\nu$ to $S$.
Then $\nu': S \to R / J$ is a homomorphism.
The image of $\nu'$ is the set of all cosets $s + J$ for $s \in S$:
:$\image {\nu'} = \dfrac {S + J} J$
Now, the kernel of $\nu'$ is the set of all elements of $S$ which are sent to $0_{S/J}$ by $\nu$.
That is, all the elements of $S$ which are also in $J$ itself, which is how the quotient ring behaves.
That is:
:$\ker \paren {\nu'} = S \cap J$
and so from Kernel of Ring Homomorphism is Ideal, $S \cap J$ is an ideal of $S$.
$(4): \quad \dfrac S {S \cap J} \cong \dfrac {S + J} J$
This follows directly from the First Ring Isomorphism Theorem.
== Also known as ==
== Also see ==
* Isomorphism Theorems
== Sources ==
* : $\S 2.2$: Homomorphisms: Theorem $2.10$
Category:Second Ring Isomorphism Theorem
Category:Ring Isomorphisms
Category:Ideal Theory
Category:Isomorphism Theorems
Second Order Fibonacci Number in terms of Fibonacci Numbers

== Theorem ==
The second order Fibonacci number $\FF_n$ can be expressed in terms of Fibonacci numbers as:
:$\dfrac {3 n + 3} 5 F_n - \dfrac n 5 F_{n + 1}$
== Proof ==
Let $\map \GG z = \ds \sum_{n \mathop \ge 0} \mathop F_n z^n$ be a generating function for $\FF_n$.
Then we have:
Thus:
Then from Summation over k to n of Product of kth with n-kth Fibonacci Numbers, the coefficient of $z^n$ in $\paren {\map G z}^2$ is:
:$\dfrac {\paren {n - 1} F_n + 2n F_{n - 1} } 5$
Thus the coefficient of $z^{n + 1}$ in $z \paren {\map G z}^2$ is likewise:
:$\dfrac {\paren {n - 1} F_n + 2n F_{n - 1} } 5$
and so the coefficient of $z^n$ in $\map G z + z \paren {\map G z}^2$ is:
:$F_n + \dfrac {\paren {n - 2} F_{n - 1} + 2 \paren {n - 1} F_{n - 2} } 5$
Hence:
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $12$
Category:Fibonacci Numbers
Second Order ODE/(x^2 + 2 y') y'' + 2 x y' = 0

== Theorem ==
The second order ODE:
:$(1): \quad \paren {x^2 + 2 y'} y + 2 x y' = 0$
subject to the initial conditions:
:$y = 1$ and $y' = 0$ when $x = 0$
has the particular solution:
:$y = 1$
or:
:$3 y + x^3 = 3$
== Proof ==
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$ in $(1)$:
Consider the initial condition:
:$y' = p = 0$ when $x = 0$
Hence putting $p = x = 0$ in $(2)$ we get:
:$0 \cdot 0^2 + 0^2 = C_1$
:$C_1 = 0$
and so $(2)$ becomes:
There are two possibilities here:
From our initial condition:
:$y = 1$ when $x = 0$
gives us:
:$C_2 = 1$
and so the solution is obtained:
:$y = 1$
The other option is:
From our initial condition:
:$y = 1$ when $x = 0$
Hence putting $x = 0$ and $y = 1$ in $(3)$ we get:
:$1 = - \dfrac {0^3} 3 = C_2$
and so $C_2 = 1$.
Thus we have:
:$y + \dfrac {x^3} 3 = 1$
or:
:$3 y + x^3 = 3$
Hence the result.
== Sources ==
* : $\S 2.11$: Problem $2 \ \text{(a)}$
Category:Examples of Homogeneous LSOODEs
Second Order ODE/(x^2 - 1) y'' - 2 x y' + 2 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad \paren {x^2 - 1} y - 2 x y' + 2 y = 0$
has the general solution:
:$y = C_1 x + C_2 \paren {x^2 + 1}$
== Proof ==
Note that:
and so by inspection:
:$y_1 = x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y - \dfrac {2 x} {x^2 - 1} y' + \dfrac 2 {x^2 - 1} y = 0$
which is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = - \dfrac {2 x} {x^2 - 1}$
:$\map Q x = \dfrac 2 {x^2 - 1}$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
:$\map {y_2} x = \map v x \, \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 x + C_2 \paren {x^2 + 1}$
Category:Examples of Homogeneous LSOODEs
Second Order ODE/x^2 y'' + x y' = 1

== Theorem ==
The second order ODE:
:$x^2 y + x y' = 1$
has the general solution:
:$y = \dfrac {\paren {\ln x}^2} 2 + C_1 \ln x + C_2$
== Proof ==
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $16$
Category:Examples of Homogeneous LSOODEs
Second Order ODE/x^2 y'' = 2 x y' + (y')^2

== Theorem ==
The second order ODE:
:$x^2 y = 2 x y' + \paren {y'}^2$
has the general solution:
:$y = -\dfrac {x^2} 2 - C_1 x - {C_1}^2 \, \map \ln {x - C_1} + C_2$
== Proof ==
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$:
== Sources ==
* : $\S 2.11$: Problem $1 \ \text{(d)}$
Category:Examples of Second Order ODEs
Second Order ODE/x y'' + 3 y' = 0

== Theorem ==
The second order ODE:
:$(1): \quad x y + 3 y' = 0$
has the general solution:
:$y = C_1 + \dfrac {C_2} {x^2}$
== Proof ==
Note that:
and so by inspection:
:$y_1 = 1$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y + \dfrac 3 x y' = 0$
which is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = \dfrac 3 x$
:$\map Q x = 0$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
:$\map {y_2} x = \map v x \, \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 + k \paren {-\frac 1 {2 x^2} }$
where $k$ is arbitrary.
Setting $C_2 = -\dfrac k 2$ yields the result:
:$y = C_1 + \dfrac {C_2} {x^2}$
</onlyinclude>
== Sources ==
* : $\S 3.16$: Problem $3$
Category:Examples of Homogeneous LSOODEs
Second Order ODE/x y'' - (2 x + 1) y' + (x + 1) y = 0

== Theorem ==
The second order ODE:
:$(1): \quad x y - \paren {2 x + 1} y' + \paren {x + 1} y = 0$
has the general solution:
:$y = C_1 e^x + C_2 x^2 e^x$
== Proof ==
Note that:
:$x - \paren {2 x + 1} + \paren {x + 1} = 0$
so if $y = y' = y$ we find that $(1)$ is satisfied.
So:
and so:
:$y_1 = e^x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y - \dfrac {2 x + 1} x y' + \dfrac {x + 1} x y = 0$
which is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = -\dfrac {2 x + 1} x$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
:$\map {y_2} x = \map v x \, \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 e^x + k \dfrac {x^2} 2 e^x$
and so setting $C_2 = \dfrac k 2$:
:$y = C_1 e^x + C_2 x^2 e^x$
== Sources ==
* : $\S 3.16$: Problem $9$
Category:Examples of Homogeneous LSOODEs
Second Order ODE/x y'' - y' = 3 x^2

== Theorem ==
The second order ODE:
:$(1): \quad x y - y' = 3 x^2$
has the general solution:
:$y = x^3 + \dfrac {C_1 x^2} 2 + C^2$
== Proof ==
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$ in $(1)$:
:$x \dfrac {\d p} {\d x} - p = 3 x^2$
and divide through by $x$:
:$\dfrac {\d p} {\d x} - \dfrac p x = 3 x$
From:
:Linear First Order ODE: $y' - \dfrac y x = 3 x$
its solution is:
:$p = 3 x^2 + C_1 x$
Substituting back for $p$:
:$\dfrac {\d y} {\d x} = 3 x^2 + C_1 x$
which is separable, leading to:
:$y = x^3 + \dfrac {C_1 x^2} 2 + C^2$
== Sources ==
* : $\S 2.11$: Reduction of Order: Example $1$
Category:Examples of Homogeneous LSOODEs
Second Order ODE/x y'' = y' + (y')^3

== Theorem ==
The second order ODE:
:$(1): \quad x y = y' + \paren {y'}^3$
has the general solution:
:$x^2 + \paren {y - C_2}^2 = C_1^2$
== Proof ==
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$ in $(1)$:
== Sources ==
* : $\S 2.11$: Problem $1 \ \text{(b)}$
Category:Examples of Homogeneous LSOODEs
Second Order ODE/y'' + 2 x (y')^2 = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + 2 x \paren {y'}^2 = 0$
has the general solution:
:$C_1 \map \arctan {C_1 x} = y + C_2$
== Proof ==
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$ in $(1)$ and rearranging:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $11$
Category:Examples of Second Order ODEs
Second Order ODE/y'' - f(x) y' + (f(x) - 1) y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - \map f x y' + \paren {\map f x - 1} y = 0$
has the general solution:
:$\ds y = C_1 e^x + C_2 e^x \int e^{-2 x + \int \map f x \rd x} \rd x$
== Proof ==
Note that:
:$1 - \map f x + \paren {\map f x - 1} = 0$
so if $y = y' = y$ we find that $(1)$ is satisfied.
So:
and so:
:$y_1 = e^x$
is a particular solution of $(1)$.
$(1)$ is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = -\map f x$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
:$\map {y_2} x = \map v x \, \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$\ds y = C_1 e^x + C_2 e^x \int e^{-2 x + \int \map f x \rd x} \rd x$
== Sources ==
* : $\S 3.16$: Problem $10$
Category:Examples of Homogeneous LSOODEs
Second Order ODE/y'' - x f(x) y' + f(x) y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - x \, \map f x y' + \map f x y = 0$
has the general solution:
:$\ds y = C_1 x + C_2 x \int x^{-2} e^{\int x \, \map f x \rd x} \rd x$
== Proof ==
Note that:
Substituting into $(1)$:
and so it has been demonstrated that:
:$y_1 = x$
is a particular solution of $(1)$.
$(1)$ is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = -x \map f x$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
:$\map {y_2} x = \map v x \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$\ds y = C_1 x + C_2 x \int x^{-2} e^{\int x \map f x \rd x} \rd x$
== Sources ==
* : $\S 3.16$: Problem $8$
Category:Examples of Homogeneous LSOODEs
Second Order ODE/y'' = 1 + (y')^2

== Theorem ==
The second order ODE:
:$(1): \quad y = 1 + \paren {y'}^2$
has the general solution:
:$y = \map {\ln \sec} {x + C_1} + C_2$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 2.11$: Problem $3$
Category:Examples of Second Order ODEs
Category:Second Order ODE/y = 1 + (y')^2
Second Order ODE/y'' = 1 + (y')^2/Proof 1

== Theorem ==
== Proof ==
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
Making the subtitution $u = A_1 e^y$:
:$\d u = A_1 e^y \rd y = u \rd y$
Thus $(2)$ transforms into:
== Sources ==
* : $\S 2.11$: Problem $3$
Category:Second Order ODE/y = 1 + (y')^2
Second Order ODE/y'' = 1 + (y')^2/Proof 2

== Theorem ==
== Proof ==
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$ in $(1)$:
== Sources ==
* : $\S 2.11$: Problem $3$
Category:Second Order ODE/y = 1 + (y')^2
Second Order ODE/y'' = f(x)

== Theorem ==
The second order ODE:
:$\dfrac {\d^2 y} {\d x^2} = \map f x$
has the general solution:
:$\ds y = \iint \map f x \rd x \rd x + C_1 x + C_2$
where $C_1$ and $C_2$ are arbitrary constants.
== Proof ==
== Sources ==
* : : $(1)$ Equations of the form $\dfrac {\d^2 y} {\d x^2} = \map f x$
* : : $(1)$ Equations of the form $\dfrac {\d^2 y} {\d x^2} = \map f x$
Category:Examples of Second Order ODEs
Second Order ODE/y'' = f(y)

== Theorem ==
The second order ODE:
:$\dfrac {\d^2 y} {\d x^2} = \map f y$
has the general solution:
:$\dfrac {\sqrt 2} 2 \ds \int \dfrac {\d y} {\sqrt {\ds \int \map f y \rd y + C_1} } = x + C_2$
where $C_1$ and $C_2$ are arbitrary constants.
== Proof ==
== Sources ==
* : : $(2)$ Equations of the form $\dfrac {\d^2 y} {\d x^2} = \map f y$
* : : $(2)$ Equations of the form $\dfrac {\d^2 y} {\d x^2} = \map f y$
Category:Examples of Second Order ODEs
Second Order ODE/y y'' + (y')^2 - 2 y y' = 0

== Theorem ==
The second order ODE:
:$y y + \paren {y'}^2 - 2 y y' = 0$
has the general solution:
:$y^2 = C_2 e^{2 x} + C_1$
== Proof ==
Using Solution of Second Order Differential Equation with Missing Independent Variable:
After algebra, and reassigning constants:
:$y^2 = C_2 e^{2 x} + C_1$
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $7$
Category:Examples of Second Order ODEs
Second Order ODE/y y'' + (y')^2 = 0

== Theorem ==
The second order ODE:
:$(1): \quad y y + \paren {y'}^2 = 0$
has the general solution:
:$y^2 = C_1 x + C_2$
== Proof ==
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
== Sources ==
* : $\S 2.11$: Problem $1 \ \text{(a)}$
Category:Examples of Second Order ODEs
Second Order ODE/y y'' = (y')^2

== Theorem ==
The second order ODE:
:$(1): \quad y y = \paren {y'}^2$
has the general solution:
:$y = C_2 e^{C_1 x}$
== Proof ==
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $1$
Category:Examples of Second Order ODEs