hcju/proofwikiqa huggingface dataset:
It can be seen that $(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = -\paren {\dfrac 1 x + 2 x^4}$ :$\map Q x = x^3$ :$n = 2$ and so is an example of Bernoulli's equation. By Solution to Bernoulli's Equation it has the general solution: :$(3): \quad \displaystyle \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \, \map \mu x \rd x + C$ where: :$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$ Thus $\map \mu x$ is evaluated: {{begin-eqn}} {{eqn | l = \paren {1 - n} \int \map P x \rd x       | r = \paren {1 - 2} \int -\paren {\dfrac 1 x + 2 x^4} \rd x       | c =  }} {{eqn | r = \ln x + \dfrac {2 x^5} 5       | c =  }} {{eqn | r = \map \ln {x \exp \dfrac {2 x^5} 5}       | c =  }} {{eqn | ll= \leadsto       | l = \map \mu x       | r = e^{\map \ln {x \exp \frac {2 x^5} 5} }       | c =  }} {{eqn | r = x \exp \dfrac {2 x^5} 5       | c =  }} {{end-eqn}} and so substituting into $(3)$: {{begin-eqn}} {{eqn | l = \frac x y \exp \dfrac {2 x^5} 5       | r = \paren {1 - 2} \int x^3 x \exp \dfrac {2 x^5} 5 \rd x       | c =  }} {{eqn | r = -\int x^4 \exp \dfrac {2 x^5} 5 \rd x       | c =  }} {{eqn | r = -\frac {\exp \dfrac {2 x^5} 5} 2 + \frac C 2       | c = by using the substitution $w = x^5$ }} {{eqn | ll= \leadsto       | l = \frac x y       | r = \frac C 2 \map \exp {-\dfrac {2 x^5} 5} - \frac 1 2       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 y       | r = \frac C {2 x} \map \exp {-\dfrac {2 x^5} 5} - \frac 1 {2 x}       | c =  }} {{eqn | r = \frac {C \, \map \exp {-\dfrac {2 x^5} 5} - 1} {2 x}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \frac {2 x} {C \, \map \exp {-\dfrac {2 x^5} 5} - 1}       | c =  }} {{end-eqn}} Hence the general solution to $(1)$ is: :$y = \dfrac {2 x} {C \, \map \exp {-\dfrac {2 x^5} 5} - 1}$ {{qed}} Category:Examples of Bernoulli's Equation 276qfq1i7tb9y18ikpfs6e09myqzajq
From Newton's Second Law of Motion: :$\mathbf F = \map {\dfrac \d {\d t} } {m \mathbf v}$ Then the added momentum being added to $B$ by the mass being added to it is given by: :$\paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t}$ Hence: {{begin-eqn}} {{eqn | l = \paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t} + \mathbf F       | r = \map {\dfrac \d {\d t} } {m \mathbf v}       | c =  }} {{eqn | r = m \dfrac {\d \mathbf v} {\d t} + \mathbf v \dfrac {\d m} {\d t}       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf w \dfrac {\d m} {\d t} + \mathbf F       | r = m \dfrac {\d \mathbf v} {\d t}       | c =  }} {{end-eqn}} {{qed}}
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - y = x$ $(2)$ is a linear first order ODE in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = -1$ :$\map Q x = x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = -\int \rd x       | c =  }} {{eqn | r = - x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = e^{-x}       | c =  }} {{eqn | r = \frac 1 {e^x}       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: {{begin-eqn}} {{eqn | l = \map {\dfrac {\d} {\d x} } {\dfrac y {e^x} }       | r = \frac x {e^x}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac y {e^x}       | r = \int \frac x {e^x} \rd x       | c =  }} {{eqn | r = -\frac 1 {e^x} \paren {x + 1} + C       | c = Primitive of $x e^{a x}$ }} {{eqn | ll= \leadsto       | l = y       | r = C e^x - x - 1       | c =  }} {{end-eqn}} {{qed}}
From Linear First Order ODE: $y' = x + y$, the general solution of $(1)$ is: :$y = C e^x - x - 1$ Setting $y = 1$ when $x = 0$ gives: :$1 = C + 1$ from which $C = 2$. Hence the result. {{qed}}
Let $z = \dfrac {\d y} {\d x}$. Then a solution to $(1)$ will yield a particular solution to: :$(2): \quad \begin{cases} \dfrac {\d y} {\d x} = z & , \map y {x_0} = y_0 \\ & \\ \dfrac {\d z} {\d x} = -\map P x \dfrac {\d y} {\d x} - \map Q x y + \map R x & , \map z {x_0} = {y_0}' \end{cases}$ From Lipschitz Condition on Linear ODE of Continuous Functions, $(2)$ satisfies the Lipschitz condition. Hence Picard's Existence Theorem applies. Hence the result. {{qed}}
Let $\map {y_g} {x, C_1, C_2}$ be a general solution of $(2)$. Note that $C_1$ and $C_2$ are the two arbitrary constants that are to be expected of a second order ODE. Let $\map {y_p} x$ be a certain fixed particular solution of $(1)$. Let $\map y x$ be an arbitrary particular solution of $(1)$. Then: {{begin-eqn}} {{eqn | o =        | r = \paren {y - y_p}'' + \map P x \paren {y - y_p}' + \map Q x \paren {y - y_p}       | c =  }} {{eqn | r = \paren {y'' + \map P x y' + \map Q x y} - \paren {y_p'' + \map P x y_p' + \map Q x y_p}       | c =  }} {{eqn | r = \map R x - \map R x       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} We have that $\map {y_g} {x, C_1, C_2}$ is a general solution of $(2)$. Thus: :$\map y x - \map {y_p} x = \map {y_g} {x, C_1, C_2}$ or: :$\map y x = \map {y_g} {x, C_1, C_2} + \map {y_p} x$ {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \paren {c_1 \, \map {y_1} x + c_2 \, \map {y_2} x}'' + \map P x \paren {c_1 \, \map {y_1} x + c_2 \, \map {y_2} x}' + \map Q x \paren {c_1 \, \map {y_1} x + c_2 \, \map {y_2} x}       | c =  }} {{eqn | r = \paren {c_1 \, \map {y_1''} x + c_2 \, \map {y_2''} x} + \map P x \paren {c_1 \, \map {y_1'} x + c_2 \, \map {y_2'} x} + \map Q x \paren {c_1 \, \map {y_1} x + c_2 \, \map {y_2} x}       | c = Linear Combination of Derivatives }} {{eqn | r = c_1 \paren {\map {y_1''} x + \map P x \, \map {y_1'} x + \map Q x \, \map {y_1} x} + c_2 \paren {\map {y_2''} x + \map P x \, \map {y_2'} x + \map Q x \, \map {y_2} x}       | c =  }} {{eqn | r = c_1 \cdot 0 + c_2 \cdot 0       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} Hence the result. {{qed}}
{{AimForCont}} $y_p$ is not the zero constant function. From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another, there exists another particular solution to $(1)$ such that $y_1$ and $y_2$ are linearly independent. At the point $\xi$: :$\map {y_p} \xi = 0$ :$\map { {y_p}'} \xi = 0$ Taking the Wronskian of $y_p$ and $y_2$: :$\map W {y_p, y_2} = y_p {y_2}' - {y_p}' y_2$ But at $\xi$ this works out as zero. It follows from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent that $y_p$ and $y_2$ cannot be linearly independent after all. From this contradiction, $y_p$ must the zero constant function. {{qed}}
Let $\xi \in \closedint a b$ be such that $\map {y_1} \xi = \map {y_2} \xi = 0$. Consider the Wronskian $\map W {y_1, y_2}$ at $\xi$: {{begin-eqn}} {{eqn | l = \map W {\map {y_1} \xi, \map {y_2} \xi}       | r = \map {y_1} \xi \map { {y_2}'} \xi - \map {y_2} \xi \map { {y_1}'} \xi       | c =  }} {{eqn | r = 0 \cdot \map { {y_2}'} \xi - 0 \cdot \map { {y_1}'} \xi       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE: :$\forall x \in \closedint a b: \map W {\map {y_1} \xi, \map {y_2} \xi} = 0$ and so from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent: :$y_1$ and $y_2$ are linearly dependent. {{qed}}
We have that: :$\forall x \in \left[{a \,.\,.\, b}\right]: g \left({x}\right) = 0 = 0 \times f \left({x}\right)$ and $0 \in \R$. Hence the result by definition of linearly dependent real functions.
Let $\map y x$ be any particular solution to $(1)$ on $\closedint a b$. It is to be shown that constants $C_1$ and $C_2$ can be found such that: :$\map y x = C_1 \map {y_1} x + C_2 \map {y_2} x$ for all $x \in \closedint a b$. By Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions: :a particular solution to $(1)$ over $\closedint a b$ is completely determined by: ::its value :and: ::the value of its derivative at a single point. From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE: :$C_1 \map {y_1} x + C_2 \map {y_2} x$ is a particular solution to $(1)$ over $\closedint a b$ We also have: :$\map y x$ is a particular solution to $(1)$ over $\closedint a b$ Thus it is sufficient to prove that: :$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that: ::$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$ :and: ::$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$ For this system to be solvable for $C_1$ and $C_2$ it is necessary that: :$\begin{vmatrix} \map {y_1} x & \map {y_2} x \\ \map { {y_1}'} x & \map { {y_2}'} x \\ \end{vmatrix} = \map {y_1} x \map { {y_2}'} x - \map {y_2} x \map { {y_1}'} x \ne 0$ That is, that the Wronskian $\map W {y_1, y_2} \ne 0$ at $x_0$. From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE: :if $\map W {y_1, y_2} \ne 0$ at $x_0$, then $\map W {y_1, y_2} \ne 0$ for all $x \in \closedint a b$. Hence it does not matter what point is taken for $x_0$; if the Wronskian is non-zero at one such point, it will be non-zero for all such points. From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent: :$W \left({y_1, y_2}\right) = 0$ for all $x \in \closedint a b$ {{iff}} $y_1$ and $y_2$ are linearly dependent. But we have that $y_1$ and $y_2$ are linearly independent. Hence: :$\forall x \in \closedint a b: \map W {y_1, y_2} \ne 0$ and so: :$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that: ::$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$ :and: ::$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$ The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = y_1 {y_2}' - y_2 {y_1}'       | c =  }} {{eqn | ll= \leadsto       | l = \map {W'} {y_1, y_2}       | r = \paren {y_1 {y_2}'' + {y_1}' {y_2}'} - \paren {y_2 {y_1}'' + {y_2}' {y_1}'}       | c = Product Rule }} {{eqn | r = y_1 {y_2}'' - y_2 {y_1}''       | c =  }} {{end-eqn}} Because $y_1$ and $y_2$ are both particular solutions of $(1)$: {{begin-eqn}} {{eqn | n = 2       | l = {y_1}'' + \map P x {y_1}' + \map Q x y_1       | r = 0       | c =  }} {{eqn | n = 3       | l = {y_2}'' + \map P x {y_2}' + \map Q x y_2       | r = 0       | c =  }} {{eqn | n = 4       | ll= \leadsto       | l = y_2 {y_1}'' + \map P x y_2 {y_1}' + \map Q x y_2 y_1       | r = 0       | c = $(2)$ multiplied by $y_2$ }} {{eqn | n = 5       | l = y_1 {y_2}'' + \map P x y_1 {y_2}' + \map Q x y_1 y_2       | r = 0       | c = $(3)$ multiplied by $y_1$ }} {{eqn | n = 6       | ll= \leadsto       | l = \paren {y_1 {y_2}'' - y_2 {y_1}''} + \map P x \paren {y_1 {y_2}' - y_2 {y_1}'}       | r = 0       | c = $(5)$ subtracted from $(6)$ }} {{end-eqn}} That is: :$\dfrac {\d P} {\d W} + P W = 0$ This is a linear first order ODE. From Solution to Linear First Order Ordinary Differential Equation: :$W = C e^{-\int P \rd x}$ The exponential function is never zero: Therefore: :$W = 0 \iff C = 0$ and the result follows. {{qed}}
From Real Number Line is Second-Countable we have that $\struct {\R, \tau_d}$ is a second-countable space. The result follows from Second-Countable Space is Lindelöf. {{qed}}
=== Sufficient Condition === Let $y_1$ and $y_2$ are linearly dependent. Suppose either $y_1$ or $y_2$ is zero everywhere on $\closedint a b$. Then either ${y_1}'$ or ${y_2}'$ is also zero everywhere on $\closedint a b$. Thus: :$y_1 {y_2}' - y_2 {y_1}' = 0$ and so by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$. Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$. {{WLOG}}, it follows by definition of linearly dependent that: :$y_2 = C y_1$ for some $C \in \R$. Thus by Derivative of Constant Multiple: :${y_2}' = C {y_1}'$ Hence: {{begin-eqn}} {{eqn | l = C       | r = \frac {y_2} {y_1}       | c =  }} {{eqn | ll= \leadsto       | l = {y_2}'       | r = \frac {y_2} {y_1} {y_1}'       | c =  }} {{eqn | ll= \leadsto       | l = y_1 {y_2}'       | r = y_2 {y_1}'       | c =  }} {{eqn | ll= \leadsto       | l = y_1 {y_2}' - y_2 {y_1}'       | r = 0       | c =  }} {{end-eqn}} Hence by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$. {{qed|lemma}} === Necessary Condition === Let $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$. One way for this to happen is for either $y_1$ or $y_2$ to be zero everywhere on $\closedint a b$. {{WLOG}}, suppose $y_1$ is zero everywhere on $\closedint a b$. Then by Real Function is Linearly Dependent with Zero Function, $y_1$ and $y_2$ are linearly dependent. Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$. As $y_1$ is continuous, there is some closed interval: :$\closedint c d \subseteq \closedint a b$ on which: :$\forall x \in \closedint c d: \map {y_1} x \ne 0$ Since $\map W {y_1, y_2} = 0$: :$\dfrac {y_1 {y_2}' - y_2 {y_1}'} { {y_1}^2} = 0$ on $\closedint c d$. By Quotient Rule for Derivatives, this is the same as: :$\paren {\dfrac {y_2} {y_1} }' = 0$ Integrating {{WRT|Integration}} $x$, this gives: :$\dfrac {y_2} {y_1} = k$ Explicitly: :$\forall x \in \closedint c d: \map {y_2} x = k \, \map {y_1} x$ As $\map {y_2} x$ and $k \, \map {y_1} x$ have equal values on $\closedint c d$, their derivatives are equal there as well. From Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions, it follows that: :$\forall x \in \closedint a b: \map {y_2} x = k \, \map {y_1} x$. {{qed}}
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: :$p \dfrac {\d p} {\d y} = -y$ where $p = \dfrac {\d y} {\d x}$. From: :First Order ODE: $y \rd y = k x \rd x$ with $k = 1$, this has the solution: :$p^2 = -y^2 + C$ or:  :$p^2 + y^2 = C$ As the {{LHS}} is the sum of squares, $C$ has to be positive for this to have any solutions. Thus, let $C = \alpha^2$. Then: {{begin-eqn}} {{eqn | l = p^2 + y^2       | r = \alpha^2       | c =  }} {{eqn | ll= \leadsto       | l = p = \dfrac {\d y} {\d x}       | r = \pm \sqrt {\alpha^2 - y^2}       | c =  }} {{eqn | ll= \leadsto       | l = \int \dfrac {\d y} {\sqrt {\alpha^2 - y^2} }       | r = \int \pm 1 \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \arcsin \dfrac y \alpha       | r = \pm x + \beta       | c = Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ }} {{eqn | ll= \leadsto       | l = y       | r = \alpha \map \sin {\pm x + \beta}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = A \map \sin {x + B}       | c =  }} {{end-eqn}} From Multiple of Sine plus Multiple of Cosine: Sine Form, this can be expressed as: :$y = C_1 \sin x + C_2 \cos x$ {{qed}}
$(1)$ can be seen to be a special case of: :$(2): \quad$ Linear Second Order ODE: $y'' + k^2 y = 0$ with $k = 1$. $(2)$ has the solution: :$y = C_1 \sin k x + C_2 \cos k x$ Hence setting $k = 1$: :$y = C_1 \sin x + C_2 \cos x$ {{qed}}
We have that: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \, \sin x       | r = \cos x       | c = Derivative of Sine Function }} {{eqn | l = \frac \d {\d x} \, \cos x       | r = -\sin x       | c = Derivative of Cosine Function }} {{eqn | ll= \leadsto       | l = \frac {\d^2} {\d x^2} \, \cos x       | r = -\cos x       | c =  }} {{eqn | l = \frac {\d^2} {\d x^2} \, \sin x       | r = -\sin x       | c =   }} {{end-eqn}} Hence it can be seen by inspection that: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = \sin x }} {{eqn | l = \map {y_2} x       | r = \cos x }} {{end-eqn}} are particular solutions to $(1)$. It is noted that: :$\dfrac {y_1} {y_2} = \tan x$ which is not a constant function on any closed interval $\closedint a b$. Calculating the Wronskian of $y_1$ and $y_2$: {{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = \begin{vmatrix} \sin x & \cos x \\ \cos x & -\sin x \end{vmatrix}       | c =  }} {{eqn | r = -\sin^2 x - \cos^2 x       | c =  }} {{eqn | r = -1       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} So the Wronskian of $y_1$ and $y_2$ is never zero. We have that $(1)$ is a homogeneous linear second order ODE in the form: :$y'' + \map P x y' + \map Q x y = 0$ where $\map P x = 0$ and $\map Q x$ are constant functions. So by Constant Real Function is Continuous: $P$ and $Q$ are continuous on $\closedint a b$ Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$(1)$ has the general solution: ::$y = C_1 \sin x + C_2 \cos x$ :on $\closedint a b$. Since $\closedint a b$ can be extended indefinitely without introducing points where $P$ or $Q$ are discontinuous, the general solution is valid for all $x$. {{qed}}
From Linear Second Order ODE: $y'' + y = 0$, the general solution of $(1)$ is: :$y = C_1 \sin x + C_2 \cos x$ Differentiating {{WRT|Differentiation}} $x$: :$y' = C_1 \cos x - C_2 \sin x$ Thus for the initial conditions: {{begin-eqn}} {{eqn | l = \map y 0       | r = C_1 \sin 0 + C_2 \cos 0       | c =  }} {{eqn | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = C_1 \times 0 + C_2 \times 1       | r = 2       | c = Sine of Zero is Zero, Cosine of Zero is One }} {{eqn | ll= \leadsto       | l = C_2       | r = 2       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map {y'} 0       | r = C_1 \cos 0 - C_2 \sin 0       | c =  }} {{eqn | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = C_1 \times 1 - C_2 \times 0       | r = 3       | c = Sine of Zero is Zero, Cosine of Zero is One }} {{eqn | ll= \leadsto       | l = C_1       | r = 3       | c =  }} {{end-eqn}} Hence the result. {{qed}}
We have that: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} \, e^x       | r = e^x       | c = Derivative of Exponential Function }} {{eqn | l = \frac {\d} {\d x} \, e^{-x}       | r = -e^{-x}       | c = Derivative of Exponential Function }} {{eqn | ll= \leadsto       | l = \frac {\d^2} {\d x^2} \, e^x       | r = e^x       | c =  }} {{eqn | l = \frac {\d^2} {\d x^2} \, e^{-x}       | r = e^{-x}       | c =   }} {{end-eqn}} Hence it can be seen by inspection that: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = e^x }} {{eqn | l = \map {y_2} x       | r = e^{-x} }} {{end-eqn}} are solutions to $(1)$. Calculating the Wronskian of $y_1$ and $y_2$: {{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = \begin{vmatrix} e^x & e^{-x} \\ e^x & -e^{-x} \end{vmatrix}       | c =  }} {{eqn | r = -e^x e^{-x} - e^x e^{-x}       | c =  }} {{eqn | r = -2       | c =  }} {{end-eqn}} So the Wronskian of $y_1$ and $y_2$ is never zero. Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent: :$y_1$ and $y_2$ are linearly independent. {{qed}}
From Linear Second Order ODE: $x^2 y'' - 2 x y' + 2 y = 0$, the general solution of $(1)$ is: :$y = C_1 x + C_2 x^2$ Differentiating {{WRT|Differentiation}} $x$: :$y' = C_1 + 2 C_2 x$ Thus for the initial conditions: {{begin-eqn}} {{eqn | l = \map y 1       | r = C_1 \times 1 + C_2 \times 1^2       | c =  }} {{eqn | r = 3       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = C_1 + C_2       | r = 3       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map {y'} 1       | r = C_1 + 2 C_2 \times 1       | c =  }} {{eqn | r = 5       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = C_1 + 2 C_2       | r = 5       | c =  }} {{eqn | ll= \leadsto       | l = C_2       | r = 2       | c = $(3) - (2)$ }} {{eqn | ll= \leadsto       | l = C_1 + 2       | r = 3       | c = substituting for $C_2$ in $(2)$ }} {{eqn | ll= \leadsto       | l = C_1       | r = 1       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Linear Second Order ODE: $y'' - 3 y' + 2 y = 0$, the general solution of $(1)$ is: :$y = C_1 e^x + C_2 e^{2 x}$ Differentiating {{WRT|Differentiation}} $x$: :$y' = C_1 e^x + 2 C_2 e^{2 x}$ Thus for the initial conditions: {{begin-eqn}} {{eqn | l = \map y 0       | r = C_1 e^0 + C_2 e^{2 \times 0}       | c =  }} {{eqn | r = -1       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = C_1 + C_2       | r = -1       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map {y'} 1       | r = C_1 e^0 + 2 C_2 e^{2 \times 0}       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = C_1 + 2 C_2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = C_2       | r = 2       | c = $(3) - (2)$ }} {{eqn | ll= \leadsto       | l = C_1 + 2       | r = -1       | c = substituting for $C_2$ in $(2)$ }} {{eqn | ll= \leadsto       | l = C_1       | r = -3       | c =  }} {{end-eqn}} Hence the result. {{qed}}
Let $\map {y_1} x$ be a non-trivial particular solution to $(1)$. Thus, for all $C \in \R$, $C y_1$ is also a non-trivial particular solution to $(1)$. Let $\map v x$ be a function of $x$ such that: :$(2): \quad \map {y_2} x = \map v x \, \map {y_1} x$ is a particular solution to $(1)$ such that $y_1$ and $y_2$ are linearly independent. Thus: :$(3): \quad {y_2}'' + \map P x {y_2}' + \map Q x y = 0$ Differentiating $(2)$ twice {{WRT|Differentiation}} $x$: :${y_2}' = \map {v'} x \, \map {y_1} x + \map v x \, \map { {y_1}'} x$ and: :${y_2}'' = \map {v''} x \, \map {y_1} x + 2 \, \map {v'} x \, \map { {y_1}'} x + \map v x \, \map { {y_1}''} x$ Substituting these into $(3)$ and rearranging: :$v paren { {y_1}'' + P {y_1}' + Q y_1} + v'' y_1 + v' \paren {2 {y_1}'' + P y_1} = 0$ Since $y_1$ is a particular solution of $(1)$, this reduces to: :$v'' y_1 + v' \paren {2 {y_1}'' + P y_1} = 0$ which can be expressed as: :$\dfrac {v''} {v'} = -2 \dfrac { {y_1}'} {y_1} - P$ Integration gives: :$\displaystyle \ln v' = - 2 \ln y_1 - \int P \rd x$ This leads to: :$v' = \dfrac 1 { {y_1}^2} e^{-\int P \rd x}$ and so: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ From Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution, $y_1$ and $v y_1$ are linearly independent. {{qed}}
This will be demonstrated by calculating the Wronskian of $y_1$ and $y_2$ and demonstrating that it is non-zero everywhere. First we take the derivative of $v$ {{WRT}} $x$: :$v' = \dfrac 1 { {y_1}^2} e^{- \int P \rd x}$ {{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = y_1 {y_2}' - y_2 {y_1}'       | c = {{Defof|Wronskian}} }} {{eqn | r = y_1 \paren {v y_1}' - v y_1 {y_1}'       | c = Definition of $y_2$ }} {{eqn | r = y_1 \paren {v' y_1 + v {y_1}'} - v y_1 {y_1}'       | c = Product Rule }} {{eqn | r = {y_1}^2 v'        | c = simplifying }} {{eqn | r = {y_1}^2 \dfrac 1 { {y_1}^2} e^{-\int P \rd x}       | c = substituting for $v'$ }} {{eqn | r = e^{-\int P \rd x}       | c = simplifying }} {{end-eqn}} As $\displaystyle -\int P \rd x$ is a real function, $e^{-\int P \rd x}$ is non-zero wherever it is defined. Hence from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent, $y_1$ and $y_2$ are linearly independent. {{qed}}