hcju/proofwikiqa huggingface dataset:
We differentiate $(1)$ partially {{WRT|Differentiation}} each of $x_j$ for $1 \le j \le m$: :$(2): \quad \dfrac {\partial f} {\partial x_j} + \dfrac {\partial f} {\partial z} \cdot \dfrac {\partial z} {\partial x_j} = 0$ Suppose $m \ge n$. Then there exist sufficient equations of the form of $2$ for the constants $c_1, c_2, \dotsc, c_n$ to be eliminated. Otherwise suppose $m > n$. Then we differentiate $(2)$ partially for each $j$ {{WRT|Differentiation}} each of $x_j$ for $1 \le j \le m$, to obtain: {{begin-eqn}} {{eqn | l = \dfrac {\partial^2 f} {\partial {x_j}^2} + 2 \dfrac {\partial^2 f} {\partial x_j \partial z} + \dfrac {\partial^2 f} {\partial z^2} \cdot \paren {\dfrac {\partial z} {\partial x_j} }^2 + \dfrac {\partial f} {\partial z} \cdot \dfrac {\partial^2 z} {\partial {x_j}^2}       | r = 0       | c = for $1 \le j \le m$ }} {{eqn | l = \dfrac {\partial^2 f} {\partial x_j \partial x_k} + \dfrac {\partial^2 f} {\partial x_j \partial z} \cdot \dfrac {\partial z} {\partial x_k} + \dfrac {\partial^2 f} {\partial x_k \partial z} \cdot \dfrac {\partial z} {\partial x_j} + \dfrac {\partial^2 f} {\partial z^2} \cdot \dfrac {\partial z} {\partial x_j} \cdot \dfrac {\partial z} {\partial x_k} + \dfrac {\partial f} {\partial z} \cdot \dfrac {\partial^2 f} {\partial x_j \partial x_k}       | r = 0       | c = for $1 \le j \le m$, $1 \le k \le m$, $j \ne k$ }} {{end-eqn}} and continue the operation until enough equations have been obtained so that the remaining constants can be eliminated. {{qed}}
Let $\CC$ be the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$. Let $\NN$ be the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$. Then $\Q_p$ is the quotient ring $\CC \, \big / \NN$ by definition of the $p$-adic numbers. By definition of the quotient ring $\CC \, \big / \NN$: :$\sequence {\alpha_n}$ and $\sequence {\beta_n}$ are representatives of the same equivalence class in $\CC \, \big / \NN$  {{iff}}: :$\sequence {\alpha_n}$ and $\sequence {\beta_n}$ are elements of the same left coset of $\NN$ in $\CC$. From Element in Left Coset iff Product with Inverse in Subgroup: :$\sequence {\alpha_n}$ and $\sequence {\beta_n}$ are elements of the same left coset of $\NN$ in $\CC$  {{iff}}:  :$\sequence {\alpha_n} - \sequence {\beta_n} = \sequence {\alpha_n - \beta_n} \in \NN$. The result follows. {{qed}} Category:P-adic Number Theory evmod0cug4kysj7xrof0tg0pkules93
From Equation of Plane, we have that the equation defining a general plane $P$ is: :$\alpha_1 x + \alpha_2 y + \alpha_3 z = \gamma$ which can be written as: :$z = a x + b y + c$ by setting: {{begin-eqn}} {{eqn | l = a       | r = \dfrac {-\alpha_1} {\alpha_3} }} {{eqn | l = b       | r = \dfrac {-\alpha_2} {\alpha_3} }} {{eqn | l = c       | r = \dfrac {-\gamma} {\alpha_3} }} {{end-eqn}} We use the technique of Elimination of Constants by Partial Differentiation. We see we have: :$1$ dependent variable, that is: $z$ :$2$ independent variables, that is: $x$ and $y$ :$3$ constants, that is: $a$, $b$ and $c$. Taking the partial first derivatives {{WRT|Differentiation}} $x$ and $y$, we get: {{begin-eqn}} {{eqn | l = \dfrac {\partial z} {\partial x}       | r = a }} {{eqn | l = \dfrac {\partial z} {\partial y}       | r = b }} {{end-eqn}} $2$ equations are insufficient to dispose of $3$ constants, so the process continues by taking the partial second derivatives {{WRT|Differentiation}} $x$ and $y$: {{begin-eqn}} {{eqn | l = \dfrac {\partial^2 z} {\partial x^2}       | r = 0 }} {{eqn | l = \dfrac {\partial^2 z} {\partial x \partial y}       | r = 0 }} {{eqn | l = \dfrac {\partial^2 z} {\partial y^2}       | r = 0 }} {{end-eqn}} and the system of partial differential equations has been established. {{qed}}
=== Lemma 1 === Given values $z_1, \ldots, z_{p + 1}$ and $1 \le m \le p$, then: {{begin-eqn}} {{eqn | n = 2       | l = \displaystyle \map {e_m} {z_1, \ldots, z_p, z_{p + 1} }       | r = z_{p+1} \map {e_{m - 1} } {z_1, \ldots, z_p} + \map {e_m} {z_1, \ldots, z_p}       | c = Elementary Symmetric Function/Examples/Recursion }} {{end-eqn}} {{qed|lemma}} === Lemma 2 === Let $X = \set {x_1, \ldots, x_n}$ and $\mathbf u = x_j$ for some $j = 1, \ldots, n$. Then: {{begin-eqn}} {{eqn | n = 3       | l = \displaystyle \sum_{k \mathop = 0}^{n - i} \paren {-1}^k \, \map {e_{n - i - k} } X \, \mathbf u^k       | r = \map {e_{n - i} } {X \setminus \set {\mathbf u} }       | c = Eisinberg (1981) }} {{end-eqn}} '''Proof of Lemma 2''': Let $S$ denote the {{LHS}} of $(3)$. Let $U = X \setminus \set {\mathbf u}$. Then: {{begin-eqn}} {{eqn | l = S       | r = \paren {-1}^{n - i} \mathbf u^{n - i} + \sum_{k \mathop = 0}^{n - i - 1} \paren {-1}^k \map {e_{n - i - k} } X {\mathbf u}^k       | c = splitting off the term for $k = n - i$ }} {{eqn | r = \paren {-1}^{n - i} \, {\mathbf u}^{n - i} + \sum_{k \mathop = 0}^{n - i - 1} \paren {-1}^k \, \paren {\mathbf u \, \map {e_{n - i - k - 1} } U + \map {e_{n - i - k} } U} \, \mathbf u^k       | c = by $(2)$ in Lemma 1 with $p = n - 1$ and $m = n - i - k$ }} {{eqn | r = \sum_{k \mathop = 0}^{n - i - 1} \paren {-1}^k \, \map {e_{n - i - k - 1} } U \mathbf u^{k + 1} + \sum_{k \mathop = 0}^{n - i} \paren {-1}^k \map {e_{n - i - k} } U \, \mathbf u^k       | c = reassembling summations }} {{eqn | r = \sum_{k \mathop = 1}^{n - i} \paren {-1}^{k - 1} \, \map {e_{n - i - k} } U \mathbf u^k + \sum_{k \mathop = 0}^{n - i} \paren {-1}^k \map {e_{n - i - k} } U \, \mathbf u^k       | c = reindexing the first sum }} {{eqn | r = \sum_{k \mathop = 1}^{n - i} \paren {-1}^k \paren {-\map {e_{n - i - k} } U + \map {e_{n - i - k} } U} \mathbf u^k + \paren {-1}^0 \map {e_{n - i} } U \mathbf u^0       | c = splitting off the term for $k = 0$ and collecting under one summation }} {{eqn | r = \map {e_{n-i} } U       | c =  }} {{end-eqn}} {{qed|lemma}} '''Proof of the Theorem''' {{begin-eqn}} {{eqn | l = d_{ij}       | r = \dfrac{\displaystyle \paren {-1}^{n - i} \map { e_{n - i} } {X \setminus \set {x_j} } } {\displaystyle \prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }       | c = transposing $W_n$, then applying corollary to Inverse of Vandermonde Matrix }} {{eqn | r = \dfrac {\displaystyle \sum_{k \mathop = 0}^{n - i} \paren {-1}^k \, \map {e_{n - i - k} } X \, x_j^k} {\displaystyle \prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }       | c = $(3)$ in Lemma 2 }}  {{eqn | r = \dfrac {\displaystyle \sum_{k \mathop = 0}^{n - i} a_{i + k} \, x_j^k} {\displaystyle \prod_{m \mathop = 1, m \mathop \ne j}^n \paren {x_j - x_m} }       | c = Viete's Formulas }}  {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x}       | r = \map f x       | c =  }} {{eqn | ll= \leadsto       | l = \int \d y       | r = \int \map f x \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = y       | r = \int \map f x \rd x       | c = Primitive of Constant }} {{end-eqn}} {{qed}}
It is seen that $(1)$ is an instance of the first order ordinary differential equation: :$\dfrac {\d y} {\d x} = \map f {x, y}$ which is: :subject to an initial condition: $\tuple {x_0, y_0}$ where: :$\map f {x, y}$ is actually $\map f x$ From Solution to First Order Initial Value Problem, this problem is equivalent to the integral equation: :$\displaystyle y = y_0 + \int_{x_0}^x \map f {\xi, \map y \xi} \rd \xi$ As $\map y \xi$ does not contribute towards $\map f x$, it can be ignored. Hence we have: :$\displaystyle y = y_0 + \int_{x_0}^x \map f \xi \rd \xi$ {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x}       | r = k y       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d y} {\d x} - k y       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = e^{-k x} \paren {\dfrac {\d y} {\d x} - k y}       | r = 0       | c = Solution to Linear First Order Ordinary Differential Equation: $e^{\int -k \rd x} = e^{-k x}$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {y e^{-k x} }       | r = 0       | c = Solution to Linear First Order Ordinary Differential Equation continued }} {{eqn | ll= \leadsto       | l = y e^{-k x}       | r = C       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = C e^{k x}       | c =  }} {{end-eqn}} {{qed}}
From the Product Rule for Derivatives: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \paren {e^{a x} y}       | r = e^{a x} \cdot \dfrac {\d y} {\d x} + y \cdot a e^{a x}       | c =  }} {{eqn | r = e^{a x} \paren {\dfrac {\d y} {\d x} + a y}       | c =  }} {{end-eqn}} Hence, multiplying $(1)$ all through by $e^{\int a \rd x}$: :$\dfrac \d {\d x} \left({e^{a x} y}\right) = e^{a x} \map Q x$ Integrating {{WRT|Integration}} $x$ now gives: :$\displaystyle e^{a x} y = \int e^{a x} \map Q x \rd x + C$ whence we get the result by dividing by $e^{a x}$. {{qed}}
This is a specific instance of Solution to Linear First Order Ordinary Differential Equation: {{:Solution to Linear First Order Ordinary Differential Equation}} In this instance, we have: :$\map P x = a$ Hence: :$\displaystyle \int \map P x = a x$ and the result follows. {{qed}}
From Solution to Linear First Order ODE with Constant Coefficients, the general solution to $(1)$ is: :$(2): \quad \displaystyle y = e^{-a x} \int e^{a x} \map Q x \rd x + C e^{-a x}$ Let $y = y_0$ when $x = x_0$. We have: :$(3): \quad y_0 = e^{-a x_0} \int e^{a x_0} \map Q {x_0} \rd x_0 + C e^{-a x_0}$ Thus: {{begin-eqn}} {{eqn | l = y e^{a x}       | r = \int e^{a x} \map Q x \rd x + C       | c = multiplying $(2)$ by $e^{a x}$ }} {{eqn | l = y_0 e^{a x_0}       | r = \int e^{a x_0} \map Q {x_0} \rd x + C       | c = multiplying $(3)$ by $e^{a x}$ }} {{eqn | ll= \leadsto       | l = y e^{a x}       | r = y_0 e^{a x_0} + \int e^{a x} \map Q x \rd x - \int e^{a x_0} \map Q {x_0} \rd x       | c = substituting for $C$ and rearranging }} {{eqn | r = y_0 e^{a x_0} + \int_{x_0}^x e^{a \xi} \map Q \xi \rd \xi       | c = Fundamental Theorem of Calculus }} {{eqn | ll= \leadsto       | l = y       | r = e^{a x} \int_{x_0}^x e^{a \xi} \map Q \xi \rd \xi + y_0 e^{-a \paren {x - x_0} }       | c = dividing by $e^{a x}$ and rearranging }} {{end-eqn}} {{qed}}
This is a linear first order ODE in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x y = -3 y$ :$\map Q x = \sin x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int -3 \rd x       | c =  }} {{eqn | r = -3 x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = e^{-3 x}       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor: {{begin-eqn}} {{eqn | l = \dfrac {\d} {\d x} \paren {e^{-3 x} y}       | r = e^{-3 x} \sin x       | c =  }} {{eqn | ll= \leadsto       | l = e^{-3 x} y       | r = \int e^{-3 x} \sin x \rd x + C       | c =  }} {{eqn | r = \frac {e^{3 x} \paren {3 \sin x - \cos x} } {3^2 + 1^2} + C       | c = Primitive of $e^{a x} \sin b x$ }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac 1 {10} \paren {3 \sin x - \cos x} + C e^{3 x}       | c =  }} {{end-eqn}} {{qed}}
This is a linear first order ODE with constant coefficents in the form: :$\dfrac {\d y} {\d x} + a y = \map Q x$ where: :$a = -3$ :$\map Q x = \sin x$ Thus from Solution to Linear First Order ODE with Constant Coefficients: {{begin-eqn}} {{eqn | l = y       | r = e^{-3 x} \int e^{3 x} \sin x \rd x + C e^{-3 x}       | c =  }} {{eqn | r = e^{-3 x} \cdot \frac {e^{3 x} \paren {3 \sin x - \cos x} } {3^2 + 1^2} + C       | c = Primitive of $e^{a x} \sin b x$ }} {{eqn | r = \dfrac 1 {10} \paren {3 \sin x - \cos x} + C e^{3 x}       | c =  }} {{end-eqn}} {{qed}}
This is a linear first order ODE with constant coefficents in the form: :$\dfrac {\d y} {\d x} + a y = \map Q x$ where: :$a = 1$ :$\map Q x = \dfrac 1 x$ with the initial condition $y = 0$ when $x = 1$. Thus from Solution to Linear First Order ODE with Constant Coefficients with Initial Condition: {{begin-eqn}} {{eqn | l = y       | r = e^{-x} \int_1^x \dfrac {e^\xi \rd \xi} \xi + 0 \cdot e^{x - 1}       | c =  }} {{eqn | r = e^{-x} \int_1^x \dfrac {e^\xi \rd \xi} \xi       | c =  }} {{end-eqn}} From Primitive of $\dfrac {e^x} x$ has no Solution in Elementary Functions, further work on this is not trivial. {{qed}}
From Solution to Linear First Order ODE with Constant Coefficients, $(1)$ has the general solution: :$\displaystyle y = e^{-a x} \int e^{a x} \map Q x \rd x + C e^{-a x}$ Setting $C = 0$ we get: :$\displaystyle y = e^{-a x} \int e^{a x} \map Q x \rd x$ which is a particular solution to $(1)$ By definition, the complementary function to $(1)$ is the general solution to the reduced equation: :$\dfrac {\d y} {\d x} + a y = 0$ From First Order ODE: $\dfrac {\d y} {\d x} = k y$, that general solution of $(1)$ is: :$y = C e^{-a x}$ Hence the result. {{qed}}
Let the reduced equation of $(1)$ be expressed in the form: :$(3): \quad D^2 y + p D y + q y = 0$ where $D$ denotes the derivative operator {{WRT|Differentiation}} $x$: :$D := \dfrac \d {\d x}$ Thus: :$D^2 := \dfrac {\d^2} {\d x^2}$ We can express $(3)$ in the form: :$(4): \quad \paren {D^2 + p y + q} y = 0$ Consider the expression: :$(5): \quad \paren {D - k_1} \paren {D - k_2} y$ for constants $k_1$ and $k_2$ (not necessarily real). We have: {{begin-eqn}} {{eqn | l = \paren {D - k_1} \paren {D - k_2} y       | r = \paren {D - k_1} \paren {D y - k_2 y}       | c =  }} {{eqn | r = \paren {D - k_1} \paren {\dfrac {\d y} {\d x} - k_2 y}       | c =  }} {{eqn | r = D \paren {\dfrac {\d y} {\d x} - k_2 y} - k_1 \paren {\dfrac {\d y} {\d x} - k_2 y}       | c =  }} {{eqn | r = \map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} - k_2 y} - k_1 \paren {\dfrac {\d y} {\d x} - k_2 y}       | c =  }} {{eqn | r = \dfrac {\d^2 y} {\d x^2} - k_2 \dfrac {\d y} {\d x} - k_1 \dfrac {\d y} {\d x} + k_1 k_2 y       | c =  }} {{eqn | r = \dfrac {\d^2 y} {\d x^2} - \paren {k_1 + k_2} \dfrac {\d y} {\d x} + k_1 k_2 y       | c =  }} {{eqn | r = \dfrac {\d^2 y} {\d x^2} + p \dfrac {\d y} {\d x} + q y       | c = where $p = -\paren {k_1 + k_2}$ and $q = k_1 k_2$ }} {{end-eqn}} Thus $(3)$ can be written: :$(6): \quad \paren {D - k_1} \paren {D - k_2} y = 0$ From Sum of Roots of Quadratic Equation and Product of Roots of Quadratic Equation, we recognise that $k_1$ and $k_2$ are the solutions of $(2)$. Let $z := \paren {D - k_2} y$. Then from $(6)$ we have: :$\paren {D - k_1} z = 0$ That is: :$(7): \quad \dfrac {\d z} {\d x} - k_1 z = 0$ From Solution to Linear First Order ODE with Constant Coefficients, $(7)$ has the general solution: :$z = C e^{k_1 x}$ Thus we have: :$\dfrac {\d y} {\d x} - k_2 y = z = C_1 e^{k_1 x}$ From Solution to Linear First Order ODE with Constant Coefficients: :$(8): \quad y e^{-k_1 x} = C_2 \displaystyle \int e^{\paren {k_1 - k_2} x} \rd x + C_2$ Suppose $k_1 \ne k_2$. Then the {{RHS}} of $(8)$ evaluates to: :$\dfrac C {k_1 - k_2} e^{\paren {k_1 - k_2} x}$ If $k_1 = k_2$ then it is merely: :$\displaystyle \int C e^{0 \cdot x} \rd x = \int C \rd x = C x$ We can of course replace $\dfrac C {k_1 - k_2}$ with another constant. It follows that the general solution of $(3)$ can be expressed as a linear combination of : :$e^{k_1 x}$ and $e^{k_2 x}$ if $k_1 \ne k_2$, and: :$e^{k_1 x}$ and $x e^{k_1 x}$ if $k_1 = k_2$. If $k_1$ and $k_2$ are complex conjugates, we have that: :$k_1 = \alpha + i \omega$ :$k_2 = \alpha - i \omega$ for some real $\alpha$ and $\omega$. This leads to the corresponding solutions: :$e^{\paren {\alpha \pm i \omega} x} = e^{\alpha x} \paren {\cos \omega x \pm i \sin \omega x}$ Hence any linear combination of $e^{\alpha x} \paren {\cos \omega x \pm i \sin \omega x}$ can be expressed as: :$e^{\alpha x} \paren {A \cos \omega x + B \sin \omega x}$ and the task is complete. {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 + 2 m + 2 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = -1 + i$ :$m_2 = -1 - i$ These are complex conjugates. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = e^{-x} \paren {A \cos x + B \sin x}$ {{qed}}
Differentiating $(1)$ twice {{WRT|Differentiation}} $x$ gives: {{begin-eqn}} {{eqn | l = y'       | r = -e^{-x} \paren {A \cos x + B \sin x} + e^{-x} \paren {-A \sin x + B \cos x}       | c =  }} {{eqn | r = e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x}       | c = rearranging }} {{eqn | ll= \leadsto       | l = y''       | r = -e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x} + e^{-x} \paren {-\paren {B - A} \sin x - \paren {A + B} \cos x}       | c =  }} {{eqn | r = e^{-x} \paren {-2 B \cos x + 2 A \sin x}       | c = rearranging }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = y'' + 2 y' + 2 y       | r = e^{-x} \paren {-2 B \cos x + 2 A \sin x} + 2 \paren {e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x} } + 2 e^{-x} \paren {A \cos x + B \sin x}       | c =  }} {{eqn | r = e^{-x} \paren {\paren {-2 B + 2 \paren {B - A} + 2 A} \cos x - \paren {2 A - 2 \paren {A + B} + 2 B} \sin x}       | c = rearranging }} {{eqn | r = 0       | c = simplifying }} {{end-eqn}} thus demonstrating that $(1)$ is indeed a solution to the second order ODE given. {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = -7$ :$q = -5$ :$\map R x = x^3 - 1$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' - 7 y' - 5 y = 0$ From Linear Second Order ODE: $y'' - 7 y' - 5 y = 0$, this has the general solution: :$y_g = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x}$ We have that: :$\map R x = x^3 - 1$ and it is noted that $x^3 - 1$ is not itself a particular solution of $(2)$. So from the Method of Undetermined Coefficients for Polynomials: :$y_p = A_0 + A_1 x + A_2 x^2 + A_3 x^3$ for $A_n$ to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A_0 + A_1 x + A_2 x^2 + A_3 x^3       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = A_1 + 2 A_2 x + 3 A_3 x^2       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = 2 A_2 + 6 A_3 x       | c = Power Rule for Derivatives }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = 2 A_2 + 6 A_3 x - 7 \paren {A_1 + 2 A_2 x + 3 A_3 x^2} - 5 \paren {A_0 + A_1 x + A_2 x^2 + A_3 x^3}       | r = x^3 - 1       | c =  }} {{eqn | ll= \leadsto       | l = -5 A_3 x^3       | r = x^3       | c = equating powers }} {{eqn | l = \paren {-21 A_3 - 5 A_2} x^2       | r = 0       | c =  }} {{eqn | l = \paren {6 A_3 - 14 A_2 - 5 A_1} x       | r = 0       | c =  }} {{eqn | l = 2 A_2 - 7 A_1 - 5 A_0       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = A_3       | r = -\dfrac 1 5       | c =  }} {{eqn | l = A_2       | r = -\dfrac {21} 5 A_3       | c =  }} {{eqn | r = -\dfrac {21} {25}       | c =  }} {{eqn | l = 5 A_1       | r = -\dfrac 6 5 - \dfrac {294} {25}       | c =  }} {{eqn | ll= \leadsto       | l = A_1       | r = -\dfrac {324} {125}       | c =  }} {{eqn | l = 5 A_0       | r = \dfrac {42} {25} + \dfrac {2268} {125} + 1       | c =  }} {{eqn | ll= \leadsto       | l = A_0       | r = \dfrac {2603} {625}       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x} + \dfrac 1 {625} \paren {-125 x^3 + 525 x^2 - 1620 x + 2603}$ {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 - 7 m - 5 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: {{begin-eqn}} {{eqn | l = m       | r = \dfrac {7 \pm \sqrt {7^2 - 4 \times 1 \times \paren {-5} } } 2       | c =  }} {{eqn | r = \dfrac 7 2 \pm \dfrac {\sqrt {69} } 2       | c =  }} {{end-eqn}} These are real and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x}$ {{qed}} Category:Examples of Constant Coefficient Homogeneous LSOODEs batruzmvs9x9u6pyqw4b7b8jbtpm31k
Let $\sequence {x_n}$ be a sequence in $X$ that converges to the limit $L \in X$. Let $\epsilon > 0$.  Then also $\dfrac \epsilon 2 > 0$. Because $\sequence {x_n}$ converges to $L$, we have: :$\exists N: \forall n > N: \norm {x_n - L} < \dfrac \epsilon 2$ So if $m > N$ and $n > N$, then: {{begin-eqn}} {{eqn | l = \norm {x_n - x_m}       | r = \norm {x_n - L + L - x_m} }} {{eqn | o = \le       | r = \norm {x_n - L} + \norm {x_m - L}       | c = Triangle Inequality Axiom of Norm }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c = (by choice of $N$) }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {x_n}$ is a Cauchy sequence. {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = -3$ :$q = 2$ :$\map R x = 5 e^{3 x}$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - 3 y' + 2 y = 0$ From Linear Second Order ODE: $y'' - 3 y' + 2 y = 0$, this has the general solution: :$y_g = C_1 e^x + C_2 e^{2 x}$ We have that: :$\map R x = 5 e^{3 x}$ and so from the Method of Undetermined Coefficients for the Exponential function: :$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$ where: :$K = 5$ :$a = 3$ :$p = -3$ :$q = 2$ Hence: {{begin-eqn}} {{eqn | l = y_p       | r = \dfrac {5 e^{3 x} } {3^2 - 3 \times 3 + 2}       | c =  }} {{eqn | r = \dfrac {5 e^{3 x} } {9 - 9 + 2}       | c =  }} {{eqn | r = \dfrac {5 e^{3 x} } 2       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^x + C_2 e^{2 x} + \dfrac {5 e^{3 x} } 2$ {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = -1$ :$\map R x = 3 e^{-x}$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - y = 0$ From Linear Second Order ODE: $y'' - y = 0$, this has the general solution: :$y_g = C_1 e^x + C_2 e^{-x}$ We have that: :$\map R x = 3 e^{-x}$ and so from the Method of Undetermined Coefficients for the Exponential function: :$y_p = \dfrac {K x e^{a x} } {2 a + p}$ where: :$K = 3$ :$a = -1$ :$p = 0$ Hence: {{begin-eqn}} {{eqn | l = y_p       | r = \dfrac {3 x e^{-x} } {2 \times \paren {-1} + 0}       | c =  }} {{eqn | r = \dfrac {3 x e^{-x} } {-2}       | c =  }} {{eqn | r = -\dfrac {3 x e^{-x} } 2       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^x + C_2 e^{-x} - \dfrac {3 x e^{-x} } 2$ {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = -2$ :$q = 1$ :$\map R x = e^x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - 2 y' + y = 0$ From Linear Second Order ODE: $y'' - 2 y' + y = 0$, this has the general solution: :$y_g = C_1 e^x + C_2 x e^x$ We have that: :$\map R x = e^x$ and so from the Method of Undetermined Coefficients for the Exponential function: :$y_p = \dfrac {K x^2 e^{a x} } 2$ where: :$K = 1$ :$a = 1$ Hence: {{begin-eqn}} {{eqn | l = y_p       | r = \dfrac {x^2 e^x} 2       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^x + C_2 x e^x + \dfrac {x^2 e^x} 2$ {{qed}}