hcju/proofwikips huggingface dataset:
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 1
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$. Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$. After a sufficiently long time, the electric current $I$ in $K$ is given by the equation: :$E_0  = R I$
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 2
Let the electric current flowing in $K$ at time $t = 0$ be $0$. Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$. The electric current $I$ in $K$ is given by the equation: :$I = \dfrac {E_0} R \paren {1 - e^{-R t / L} }$
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 3
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$. Let  EMF imposed upon $K$ be zero. The electric current $I$ in $K$ is given by the equation: :$I = I_0 e^{-R t / L}$
Current in Electric Circuit/L, R in Series/Condition for Ohm's Law
Ohm's Law is satisfied by $K$ whenever the current $I$ is at a maximum or a minimum.
Current in Electric Circuit/L, R in Series/Minimum Current implies Increasing EMF
Let the current $I$ be at a minimum. Then the EMF $E$ is increasing.
Current in Electric Circuit/L, R in Series/Maximum Current implies Decreasing EMF
Let the current $I$ be at a maximum. Then the EMF $E$ is decreasing.
Second Order ODE/y y'' = (y')^2
The second order ODE: :$(1): \quad y y'' = \paren {y'}^2$ has the general solution: :$y = C_2 e^{C_1 x}$
First Order ODE/(1 - x y) y' = y^2
The first order ODE: :$(1): \quad \paren {1 - x y} y' = y^2$ has the general solution: :$x y = \ln y + C$
First Order ODE/(2 x + 3 y + 1) dx + (2 y - 3 x + 5) dy = 0
The first order ODE: :$(1): \quad \paren {2 x + 3 y + 1} \rd x + \paren {2 y - 3 x + 5} \rd y = 0$ has the general solution: :$3 \, \map \arctan {\dfrac {y + 1} {x - 1} } = \map \ln {\paren {y + 1}^2 + \paren {x - 1}^2} + C$
First Order ODE/x y' = Root of (x^2 + y^2)
is a homogeneous differential equation with solution: :$3 x^2 \ln x = y \sqrt {x^2 + y^2} + x^2 \, \map \ln {y + \sqrt {x^2 + y^2} } + y^2 + C x^2$
Bernoulli's Equation/y^2 dx = (x^3 - x y) dy
The first order ODE: :$(1): \quad y^2 \rd x = \paren {x^3 - x y} \rd y$ has the general solution: :$3 y = 2 x^2 + C x^2 y^2$
First Order ODE/(x^2 y^3 + y) dx = (x^3 y^2 - x) dy
The first order ODE: :$(1): \quad \paren {x^2 y^3 + y} \rd x = \paren {x^3 y^2 - x} \rd y$ has the general solution: :$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
Second Order ODE/y y'' + (y')^2 - 2 y y' = 0
The second order ODE: :$y y'' + \paren {y'}^2 - 2 y y' = 0$ has the general solution: :$y^2 = C_2 e^{2 x} + C_1$
Rationals are Everywhere Dense in Sorgenfrey Line
$\Q$ is everywhere dense in the Sorgenfrey line.
Linear First Order ODE/x dy + y dx = x cosine x dx
The linear first order ODE: :$x \rd y + y \rd x = x \cos x \rd x$ has the general solution: :$x y = x \sin x + \cos x + C$
First Order ODE/x y dy = x^2 dy + y^2 dx
The first order ODE: :$(1): \quad x y \rd y = x^2 \rd y + y^2 \rd x$ has the general solution: :$y = x \ln y + C x$
Second Order ODE/y'' + 2 x (y')^2 = 0
The second order ODE: :$(1): \quad y'' + 2 x \paren {y'}^2 = 0$ has the general solution: :$C_1 \map \arctan {C_1 x} = y + C_2$
First Order ODE/(y over x^2) dx + (y - 1 over x) dy = 0
The first order ordinary differential equation: :$(1): \quad \dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$ is an exact differential equation with solution: :$\dfrac {y^2} 2 - \dfrac y x = C$
Focal Property of Parabola
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First Order ODE/(3 x^2 over y^4 - 1 over y^2) dy - 2 x over y^3 dx = 0
The first order ODE: :$(1): \quad \paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$ is an exact differential equation with general solution: :$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
First Order ODE/(y - 1 over x) dx + (x - y) dy = 0
The first order ODE: :$(1): \quad \paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$ is an exact differential equation with solution: :$x y - \ln x - \dfrac {y^2} 2 + C$
First Order ODE/1 over x^3 y^2 dx + (1 over x^2 y^3 + 3 y) dy = 0
The first order ODE: :$(1): \quad \dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$ is an exact differential equation with solution: :$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
Linear First Order ODE/y' - (y over x) = 3 x
The linear first order ODE: :$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = 3 x$ has the general solution: :$\dfrac y x = 3 x + C$ or: :$y = 3 x^2 + C x$
Characterization of Closure by Basis
Let $T = \left({S, \tau}\right)$ be a topological space. Let $\mathcal B \subseteq \tau$ be a basis. Let $A$ be a subset of $T$. Let $x$ be a point of $T$. Then $x \in A^-$ : :for every $U \in \mathcal B$: ::if $x \in U$ then $A \cap U \ne \varnothing$ where: :$A^-$ denotes the closure of $A$
Sorgenfrey Line is Separable
The Sorgenfrey line is separable.
First Order ODE/y dx + x dy = 0
The first order ODE: :$(1): \quad y \rd x + x \rd y = 0$ has the general solution: :$x y = C$
First Order ODE/y dy = k dx
Let $k \in \R$ be a real number. The first order ODE: :$y \, \dfrac {\d y} {\d x} = k$ has the general solution: :$y^2 = 2 k x + C$
First Order ODE/x dy = (y + y^3) dx
The first order ODE: :$(1): \quad x \dfrac {\d y} {\d x} = y + y^3$ has the general solution: :$y = \dfrac x {\sqrt {C^2 - x^2} }$
First Order ODE/y dy = k x dx
Let $k \in \R$ be a real number. The first order ODE: :$y \rd y = k x \rd x$ has the general solution: :$y^2 = k x^2 + C$
Primitive of Reciprocal of Root of x squared plus k
Let $k \in \R$. Then: :$\displaystyle \int \frac {\d x} {\sqrt {\size {x^2 + k} } } = \map \ln {x + \sqrt {\size {x^2 + k} } } + C$
Bernoulli's Equation/x^2 dy = (2 x y + y^2) dx
The first order ODE: :$(1): \quad x^2 \rd y = \paren {2 x y + y^2} \rd x$ has the general solution: :$y = - \dfrac {x^2} {x + C}$
Bernoulli's Equation/2 x y dx + (x^2 + 2 y) dy = 0
The first order ODE: :$(1): \quad 2 x y \rd x + \paren {x^2 + 2 y} \rd y = 0$ has the solution: :$y \paren {x^2 + y} = C$
Linear First Order ODE/y' - (y over x) = k x
Let $k \in \R$ be a real number. The linear first order ODE: :$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = k x$ has the general solution: :$\dfrac y x = k x + C$ or: :$y = k x^2 + C x$
First Order ODE/x dy = k y dx
The first order ODE: :$(1): \quad x \rd y = k y \rd x$ has the general solution: :$y = C x^k$
First Order ODE/dy = k y dx
Let $k \in \R$ be a real number. The first order ODE: :$\dfrac {\d y} {\d x} = k y$ has the general solution: :$y = C e^{k x}$
First Order ODE/(1 over x^3 y^2 + 1 over x) dx + (1 over x^2 y^3 - 1 over y) dy = 0
The first order ordinary differential equation: :$(1): \quad \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$ is an exact differential equation with solution: :$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
Linear First Order ODE/y' + (y over x) = k x^n
Let $k, n \in \R$ be real numbers. The linear first order ODE: :$(1): \quad \dfrac {\d y} {\d x} + \dfrac y x = k x^n$ has the general solution: :$y = \begin{cases} \dfrac {k x^{n + 1} } {n + 2} + \dfrac C x & : n \ne -2 \\ & \\ \dfrac {k \ln x} x + \dfrac C x & : n = -2 \end{cases}$
Linear First Order ODE/(x^2 + y) dx = x dy
The linear first order ODE: :$(1): \quad \paren {x^2 + y} \rd x = x \rd y$ has the general solution: :$y = x^2 + C x$
Linear First Order ODE/x y' + y = x^2 cosine x
The linear first order ODE: :$x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$ has the general solution: :$y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
First Order ODE/(6x + 4y + 3) dx + (3x + 2y + 2) dy = 0
The first order ODE: :$(1): \quad \paren {6 x + 4 y + 3} \rd x + \paren {3 x + 2 y + 2} \rd y = 0$ has the general solution: :$3 x + 2 y + \map \ln {\paren {3 x + 2 y}^2} + x = C$
First Order ODE/Cosine (x + y) dx = sine (x + y) dx + x sine (x + y) dy
The first order ordinary differential equation: :$\map \cos {x + y} \rd x = \map \sin {x + y} \rd x + x \map \sin {x + y} \rd y$ is an exact differential equation with solution: :$x \map \cos {x + y} = C$
Linear First Order ODE/x y' + y = f (x)
The linear first order ODE: :$(1): \quad x \, \dfrac {\d y} {\d x} + y = \map f x$ has the general solution: :$\displaystyle x y = \int \map f x \rd x + C$
Second Order ODE/x^2 y'' + x y' = 1
The second order ODE: :$x^2 y'' + x y' = 1$ has the general solution: :$y = \dfrac {\paren {\ln x}^2} 2 + C_1 \ln x + C_2$
Sorgenfrey Line is Lindelöf
The Sorgenfrey line is Lindelöf.
First Order ODE/y' ln (x - y) = 1 + ln (x - y)
The first order ordinary differential equation: :$(1): \quad \dfrac {\d y} {\d x} \map \ln {x - y} \d x = 1 + \map \ln {x - y}$ is an exact differential equation with solution: :$\paren {x - y} \map \ln {x - y} = C - y$
Linear First Order ODE/y' + 2 x y = exp -x^2
The linear first order ODE: :$\dfrac {\d y} {\d x} + 2 x y = \map \exp {-x^2}$ has the general solution: :$y = \paren {x + C} \map \exp {-x^2}$
Linear First Order ODE/(1 + x^2) y' + 2 x y = 4 x^3
The linear first order ODE: :$\paren {1 + x^2} \dfrac {\d y} {\d x} + 2 x y = 4 x^3$ has the general solution: :$y = \dfrac {x^4} {1 + x^2} + \dfrac C {1 + x^2}$
First Order ODE/(1 + x^2) y' + x y = 0
The first order ODE: :$\paren {1 + x^2} y' + x y = 0$ has the general solution: :$y = \dfrac C {\sqrt {1 + x^2} }$
First Order ODE/(x exp y + y - x^2) dy = (2 x y - exp y - x) dx
The first order ordinary differential equation: :$(1): \quad \paren {x e^y + y - x^2} \rd y = \paren {2 x y - e^y - x} \rd x$ is an exact differential equation with solution: :$2 x e^y + x^2 + y^2 - 2 x^2 y = C$
First Order ODE/exp x (1 + x) dx = (x exp x - y exp y) dy
The first order ordinary differential equation: :$(1): \quad e^x \paren {1 + x} \rd x = \paren {x e^x - y e^y} \rd y$ has the solution: :$2 x e^x e^{-y} + y^2 + C$
Two Stage Radioactive Decay
Let $R_A$ be a radioactive isotope which decays into another radioactive isotope $R_B$ with the rate constant $k_1$. Let $R_B$ decay into a third element $R_C$ with the rate constant $k_2$. Let the initial quantity of $R_A$ be $x_0$. Let the amounts of $R_A$ and $R_B$ present at time $t$ be $x$ and $y$ respectively. Then: :$y = \begin{cases} \dfrac {k_1 x_0} {k_2 - k_1} \left({e^{-k_1 t} - e^{-k_2 t} }\right) & : k_1 \ne k_2 \\ & \\ k_1 x_0 t e^{-k_1 t} & : k_1 = k_2 \end{cases}$
General Solution of Riccati Equation from Particular Solution
Consider the Riccati equation: :$(1): \quad y' = \map p x + \map q x y + \map r x y^2$ Let $\map {y_1} x$ be a particular solution to $(1)$. Then the general solution to $(1)$ has the form: :$\map y x = \map {y_1} x + \map z x$ where $\map z x$ is the general solution to the Bernoulli equation: :$z' - \paren {q - 2 r y_1} z = r z^2$
Riccati Equation/y' = (y over x) + x^3 y^2 - x^5
The Riccati equation: :$(1): \quad y' = \dfrac y x + x^3 y^2 - x^5$ has the general solution: :$C \exp \dfrac {2 x^5} 5 = \dfrac {y - x} {y + x}$
Bernoulli's Equation/y' - (1 over x + 2 x^4) y = x^3 y^2
The first order ODE: :$(1): \quad y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$ has the general solution: :$y = \dfrac {2 x} {C \, \map \exp {-\dfrac {2 x^5} 5} - 1}$
Motion of Body with Variable Mass
Let $B$ be a body undergoing a force $\mathbf F$. Let $B$ be travelling at a velocity $\mathbf v$ at time $t$. Let mass travelling at a velocity $\mathbf v + \mathbf w$ be added to $B$ at a rate of $\dfrac {\d m} {\d t}$. Let $m$ be the mass of $B$ at time $t$. Then the equation of motion of $B$ is given by: :$\mathbf w \dfrac {\d m} {\d t} + \mathbf F = m \dfrac {\d \mathbf v} {\d t}$
Burnout Velocity of Upward Rocket under Constant Gravity
Let $R$ be a rocket whose structural mass is $m_1$. Let $R$ contain fuel of initial mass $m_2$. Let $R$ be fired straight up from the surface of a planet whose gravitational field exerts an Acceleration Due to Gravity of $g$, assumed constant. Let $R$ burn fuel at a constant rate $a$, producing a constant exhaust velocity $b$ relative to $R$. Let all forces on $R$ except that due to the gravitational field be neglected. Then the burnout velocity of $R$ is given by: :$v_b = b \ln \left({1 + \dfrac {m_1} {m_2} }\right) - \dfrac {g m_2} a$
Burnout Height of Upward Rocket under Constant Gravity
Let $R$ be a rocket whose structural mass is $m_1$. Let $R$ contain fuel of initial mass $m_2$. Let $R$ be fired straight up from the surface of a planet whose gravitational field exerts an Acceleration Due to Gravity of $g$, assumed constant. Let $R$ burn fuel at a constant rate $a$, with a constant exhaust velocity $b$ relative to $R$. Let all forces on $R$ except that due to  the gravitational field be neglected. Then the burnout height of $R$ is given by: :$h_b = -\dfrac {g m_2^2} {2 a^2} + \dfrac {b m_2} a + \dfrac {b m_1} a \ln \dfrac {m_1} {m_1 + m_2}$
Linear First Order ODE/y' = x + y
The linear first order ODE: :$(1): \quad \dfrac {\d y} {\d x} = x + y$ has the general solution: :$y = C e^x - x - 1$
Linear First Order ODE/y' = x + y/y(0) = 1
The linear first order ODE: :$(1): \quad \dfrac {\d y} {\d x} = x + y$ with initial condition: :$\map y 0 = 1$ has the particular solution: :$y = 2 e^x - x - 1$
Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions
Let $\map P x$, $\map Q x$ and $\map R x$ be continuous real functions on a closed real interval $\closedint a b$. Let $x_0$ be any point in $\closedint a b$. Let $y_0$ and ${y_0}'$ be real numbers. Then the linear second order ordinary differential equation: :$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = \map R x$ has a unique particular solution $\map y x$ on $\closedint a b$ such that: :$\map y {x_0} = y_0$ and: :$\map {y'} {x_0} = {y_0}'$
General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution
Consider the nonhomogeneous linear second order ODE: :$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = \map R x$ Let $\map {y_g} x$ be the general solution of the homogeneous linear second order ODE: :$(2): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$ Let $\map {y_p} x$ be a particular solution of $(1)$. Then $\map {y_g} x + \map {y_p} x$ is the general solution of $(1)$.
Trivial Solution of Homogeneous Linear 2nd Order ODE
The homogeneous linear second order ODE: :$\dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$ has the particular solution: :$\map y x = 0$ that is, the zero constant function. This particular solution is referred to as the '''trivial solution'''.
Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE
Let $c_1$ and $c_2$ be real numbers. Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE: :$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$ Then: :$c_1 \, \map {y_1} x + c_2 \, \map {y_2} x$ is also a particular solution to $(1)$. That is, a linear combination of particular solutions to a homogeneous linear second order ODE is also a particular solution to that ODE.
Solutions of Linear 2nd Order ODE have Common Zero iff Linearly Dependent
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE: :$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$ on a closed interval $\closedint a b$. Let $y_1$ and $y_2$ both have a zero for the same value of $x$ in $\closedint a b$. Then $y_1$ and $y_2$ are constant multiples of each other. That is, $y_1$ and $y_2$ are linearly dependent.
Real Function is Linearly Dependent with Zero Function
Let $f \left({x}\right)$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$. Let $g \left({x}\right)$ be the constant zero function on $\left[{a \,.\,.\, b}\right]$: :$\forall x \in \left[{a \,.\,.\, b}\right]: g \left({x}\right) = 0$ Then $f$ and $g$ are linearly dependent on $\left[{a \,.\,.\, b}\right]$.
Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE: :$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$ on a closed interval $\closedint a b$. Let $y_1$ and $y_2$ be linearly independent. Then the general solution to $(1)$ is: :$y = C_1 \map {y_1} x + C_2 \map {y_2} x$ where $C_1 \in \R$ and $C_2 \in \R$ are arbitrary constants.
Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE: :$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$ on a closed interval $\closedint a b$. Let $y_1$ and $y_2$ be linearly independent. Then their Wronskian is either never zero, or zero everywhere on $\closedint a b$.
Real Number Line is Lindelöf
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology. Then $\struct {\R, \tau_d}$ is Lindelöf.
Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE: :$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$ on a closed interval $\closedint a b$. Then: :$y_1$ and $y_2$ are linearly dependent : :the Wronskian $\map W {y_1, y_2}$ of $y_1$ and $y_2$ is zero everywhere on $\closedint a b$.
Linear Second Order ODE/y'' + y = 0
The second order ODE: :$(1): \quad y'' + y = 0$ has the general solution: :$y = C_1 \sin x + C_2 \cos x$
Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0
The second order ODE: :$(1): \quad x^2 y'' - 2 x y' + 2 y = 0$ has the general solution: :$y = C_1 x + C_2 x^2$ on any closed real interval which does not contain $0$.
Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0/y(1) = 3, y'(1) = 5
The second order ODE: :$(1): \quad x^2 y'' - 2 x y' + 2 y = 0$ with initial conditions: :$\map y 1 = 3$ :$\map {y'} 1 = 5$ has the particular solution: :$y = x + 2 x^2$
Linear Second Order ODE/y'' - 3 y' + 2 y = 0
The second order ODE: :$(1): \quad y'' - 3 y' + 2 y = 0$ has the general solution: :$y = C_1 e^x + C_2 e^{2 x}$
Linear Second Order ODE/y'' - 3 y' + 2 y = 0/y(0) = -1, y'(0) = 1
The second order ODE: :$(1): \quad y'' - 3 y' + 2 y = 0$ with initial conditions: :$\map y 0 = -1$ :$\map {y'} 0 = 1$ has the particular solution: :$y = -3 e^x + 2 e^{2 x}$
Linear Second Order ODE/y'' - 4 y' + 4 y = 0
The second order ODE: :$(1): \quad y'' - 4 y' + 4 y = 0$ has the general solution: :$y = C_1 e^{2 x} + C_2 x e^{2 x}$
Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another
Let $\map {y_1} x$ be a particular solution to the homogeneous linear second order ODE: :$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$ such that $y_1$ is not the trivial solution. Then there exists a standard procedure to determine another particular solution $\map {y_2} x$ of $(1)$ such that $y_1$ and $y_2$ are linearly independent.
Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution
Let $\map {y_1} x$ be a particular solution to the homogeneous linear second order ODE: :$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$ such that $y_1$ is not the trivial solution. Let $\map {y_2} x$ be the real function defined as: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ Then $y_2$ and $y_1$ are linearly independent.
Linear Second Order ODE/x^2 y'' + x y' - y = 0
The second order ODE: :$(1): \quad x^2 y'' + x y' - y = 0$ has the general solution: :$y = C_1 x + \dfrac {C_2} x$
Linear Second Order ODE/y'' - y = 0
The second order ODE: :$(1): \quad y'' - y = 0$ has the general solution: :$y = C_1 e^x + C_2 e^{-x}$
Second Order ODE/x y'' + 3 y' = 0
The second order ODE: :$(1): \quad x y'' + 3 y' = 0$ has the general solution: :$y = C_1 + \dfrac {C_2} {x^2}$
Linear Second Order ODE/x^2 y'' + x y' - 4 y = 0
The second order ODE: :$(1): \quad x^2 y'' + x y' - 4 y = 0$ has the general solution: :$y = C_1 x^2 + \dfrac {C_2} {x^2}$
Topological Subspace of Real Number Line is Lindelöf
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology. Let $W$ be a non-empty subset of $\R$. Then $R_W$ is Lindelöf where $R_W$ denotes the topological subspace of $R$ on $W$.
Bessel's Equation/x^2 y'' + x y' + (x^2 - (1 over 4)) y = 0/Particular Solution
The special case of Bessel's equation: :$(1): \quad x^2 y'' + x y' + \left({x^2 - \dfrac 1 4}\right) y = 0$ has a particular solution: :$y = \dfrac {\sin x} {\sqrt x}$
Linear Second Order ODE/(x - 1) y'' - y' + y = 0
The second order ODE: :$(1): \quad \paren {x - 1} y'' - x y' + y = 0$ has the general solution: :$y = C_1 x + C_2 e^x$
Linear Second Order ODE/x^2 y'' + 2 x y' - 2 y = 0
The second order ODE: :$(1): \quad x^2 y'' + 2 x y' - 2 y = 0$ has the general solution: :$y = C_1 x + \dfrac {C_2} {x^2}$
Second Order ODE/x y'' - (2 x + 1) y' + (x + 1) y = 0
The second order ODE: :$(1): \quad x y'' - \paren {2 x + 1} y' + \paren {x + 1} y = 0$ has the general solution: :$y = C_1 e^x + C_2 x^2 e^x$
Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation
Let: :$(1): \quad y'' + p y' + q y = 0$ be a constant coefficient homogeneous linear second order ODE. Then: :$y = e^{m_1 x}$ is a solution to $(1)$  :$m_1$ is a root of the auxiliary equation $m^2 + p m + q = 0$
Solution of Constant Coefficient Homogeneous LSOODE
Let: :$(1): \quad y'' + p y' + q y = 0$ be a constant coefficient homogeneous linear second order ODE. Let $m_1$ and $m_2$ be the roots of the auxiliary equation $m^2 + p m + q = 0$.
Solution of Constant Coefficient Homogeneous LSOODE/Real Roots of Auxiliary Equation
{{:Solution of Constant Coefficient Homogeneous LSOODE}} Let $p^2 > 4 q$. Then $(1)$ has the general solution: :$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation
{{:Solution of Constant Coefficient Homogeneous LSOODE}} Let $p^2 < 4 q$. Then $(1)$ has the general solution: :$y = e^{a x} \paren {C_1 \cos b x + C_2 \sin b x}$ where: :$m_1 = a + i b$ :$m_2 = a - i b$
Solution of Constant Coefficient Homogeneous LSOODE/Equal Real Roots of Auxiliary Equation
{{:Solution of Constant Coefficient Homogeneous LSOODE}} Let $p^2 = 4 q$. Then $(1)$ has the general solution: :$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
Linear Second Order ODE/y'' + y' - 6 y = 0
The second order ODE: :$(1): \quad y'' + y' - 6 y = 0$ has the general solution: :$y = C_1 e^{2 x} + C_2 e^{-3 x}$
Linear Second Order ODE/y'' + 2 y' + y = 0
The second order ODE: :$(1): \quad y'' + 2 y' + y = 0$ has the general solution: :$y = C_1 e^{-x} + C_2 x e^{-x}$
Linear Second Order ODE/2 y'' - 4 y' + 8 y = 0
The second order ODE: :$(1): \quad 2 y'' - 4 y + 8 y = 0$ has the general solution: :$y = e^x \paren {C_1 \cos \sqrt 3 x + C_2 \sin \sqrt 3 x}$
Linear Second Order ODE/y'' - 9 y' + 20 y = 0
The second order ODE: :$(1): \quad y'' - 9 y' + 20 y = 0$ has the general solution: :$y = C_1 e^{4 x} + C_2 e^{5 x}$
Condition for Solutions to Constant Coefficient Homogeneous LSOODE to tend to Zero
Let: :$(1): \quad y'' + p y' + q y = 0$ be a constant coefficient homogeneous linear second order ODE. Let the general solution to $(1)$ be $\map y {x, C_1, C_2}$. Then: :$\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$  :$p$ and $q$ are both strictly positive.
Derivative of Solution to Constant Coefficient Homogeneous LSOODE is also Solution
Let: :$(1): \quad y'' + p y' + q y = 0$ be a constant coefficient homogeneous linear second order ODE. Let $\map y x$ be a particular solution of $(1)$. Then its derivative $\map {y'} x$ is also a particular solution of $(1)$.