hcju/proofwikiqa huggingface dataset:
Let $(1)$ be expressed as: :$(2): \quad \paren {y + x^2 y^3} \rd x + \paren {x - x^3 y^2} \rd y = 0$ We note that $(2)$ is in the form: :$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$ but is not exact. So, let: :$\map M {x, y} = y + x^2 y^3$ :$\map N {x, y} = x - x^3 y^2$ Let: :$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$ Thus: {{begin-eqn}} {{eqn | l = \map P {x, y}       | r = \paren {1 + 3 x^2 y^2} - \paren {1 - 3 x^2 y^2}       | c =  }} {{eqn | r = 6 x^2 y^2       | c =  }} {{end-eqn}} It can be observed that: {{begin-eqn}} {{eqn | l = \frac {\map P {x, y} } {\map N {x, y} y - \map M {x, y} x}       | r = \frac {6 x^2 y^2} {\paren {x y - x^3 y^3} - \paren {x y + x^3 y^3} }       | c =  }} {{eqn | r = -\frac {6 x^2 y^2} {2 x^3 y^3}       | c =  }} {{eqn | r = -\frac 3 {x y}       | c =  }} {{end-eqn}} Thus $\dfrac {\map P {x, y} } {\map M {x, y} }$ is a function of $x y$. So Integrating Factor for First Order ODE: Function of Product of Variables can be used. Let $z = x y$. Then: :$\map \mu {x y} = \map \mu z = e^{\int \map g z \rd z}$ Hence: {{begin-eqn}} {{eqn | l = \int \map g z \rd z       | r = \int -\frac 3 z \rd z       | c =  }} {{eqn | r = - 3 \ln z       | c =  }} {{eqn | r = \map \ln {z^{-3} }       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int \map g z \rd z}       | r = \frac 1 {z^3}       | c =  }} {{eqn | r = \frac 1 {x^3 y^3}       | c =  }} {{end-eqn}} Thus an integrating factor for $(1)$ has been found: :$\mu = \dfrac 1 {x^3 y^3}$ which yields, when multiplying it throughout $(2)$: :$\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$ which is now exact. By First Order ODE: $\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$, its general solution is: :$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$ {{qed}}
Using Solution of Second Order Differential Equation with Missing Independent Variable: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} + p^2 - 2 y p       | r = 0       | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto       | l = \frac {\d p} {\d y} + \frac p y       | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = p y       | r = y^2 + C       | c = Linear First Order ODE: $y' + \dfrac y x = k x^n$: $k = 2, n = 0$ }} {{eqn | ll= \leadsto       | l = \dfrac {\d y} {\d x}       | r = \frac {y^2 + C} y       | c =  }} {{eqn | ll= \leadsto       | l = \int \dfrac {y \rd y} {y^2 + C}       | r = \int \d x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \frac 1 2 \, \map \ln {y^2 + C}       | r = x + k       | c = Primitive of Function under its Derivative }} {{end-eqn}} After algebra, and reassigning constants: :$y^2 = C_2 e^{2 x} + C_1$ {{qed}}
Let $T = \struct {\R, \tau}$ be the Sorgenfrey line. Define: :$\BB := \set {\hointr x y: x, y \in \R}$ where $\hointr x y$ denotes the right half-open real interval between $x$ and $y$. By definition of Sorgenfrey line: :$\BB$ is basis of $T$. By definition of subset: :$\Q^- \subseteq \R$ where $\Q^-$ denotes the topological closure of $\Q$ in $T$. By definition of set equality to prove the equality: $\Q^- = \R$, it is necessary to show: :$\R \subseteq \Q^-$ Let $x \in \R$. By Characterization of Closure by Basis it suffices to prove that: :$\forall U \in \BB: x \in U \implies U \cap \Q \ne \O$ Let $U \in \BB$. By definition of $\BB$: :$\exists y, z \in \R: U = \hointr y z$ Assume: :$x \in U$ By definition of half-open real interval: :$y \le x < z$ By Between two Real Numbers exists Rational Number: :$\exists q \in \Q: y < q < z$ By definition of half-open real interval: :$q \in U$ Thus by definitions of intersection and empty set: :$U \cap \Q \ne \O$ {{qed}}
Let $(1)$ be expressed as: :$\paren {y e^x - 2 x y^3} \rd x + \paren {e^x - 3 x^2 y^2} \rd y = 0$ Let: :$\map M {x, y} = y e^x - 2 x y^3$ :$\map N {x, y} = e^x - 3 x^2 y^2$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y}       | r = e^x - 6 x y^2       | c =  }} {{eqn | l = \dfrac {\partial N} {\partial x}       | r = e^x - 6 x y^2       | c =  }} {{end-eqn}} Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact. By Solution to Exact Differential Equation, the solution to $(1)$ is: :$\map f {x, y} = C$ where: {{begin-eqn}} {{eqn | l = \dfrac {\partial f} {\partial x}       | r = \map M {x, y}       | c =  }} {{eqn | l = \dfrac {\partial f} {\partial y}       | r = \map N {x, y}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = f       | r = \map M {x, y} \rd x + \map g y       | c =  }} {{eqn | r = \int \paren {y e^x - 2 x y^3} \rd x + \map g y       | c =  }} {{eqn | r = y e^x - x^2 y^3 + \map g y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = f       | r = \int \map N {x, y} \rd y + \map h x       | c =  }} {{eqn | r = \int \paren {e^x - 3 x^2 y^2} \rd y + \map h x       | c =  }} {{eqn | r = y e^x + x^2 y^3 + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = y e^x + x^2 y^3$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$y e^x + x^2 y^3 = C$ {{qed}}
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$ in $(1)$ and rearranging: {{begin-eqn}} {{eqn | l = \dfrac {\d p} {\d x}       | r = -2 x p^2       | c =  }} {{eqn | ll= \leadsto       | l = \int \frac {\d p} {p^2}       | r = -2 \int x \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = -\frac 1 p       | r = -x^2 + k^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = x^2 + k^2       | c = substituting back for $p$ }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {x^2 + k^2}        | r = \int \rd y       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \frac 1 k \map \arctan {\frac x k}       | r = y + C_2       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | ll= \leadsto       | l = C_1 \map \arctan {C_1 x}       | r = y + C_2       | c = setting $C_1 = \dfrac 1 k$ }} {{end-eqn}} {{qed}}
Let $M$ and $N$ be defined as: :$\map M {x, y} = \dfrac y {x^2}$ :$\map N {x, y} = y - \dfrac 1 x$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y}       | r = \frac 1 {x^2}       | c =  }} {{eqn | l = \dfrac {\partial N} {\partial x}       | r = \dfrac 1 {x^2}       | c =  }} {{end-eqn}} Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact. By Solution to Exact Differential Equation, the solution to $(1)$ is: :$\map f {x, y} = C$ where: {{begin-eqn}} {{eqn | l = \dfrac {\partial f} {\partial x}       | r = \map M {x, y}       | c =  }} {{eqn | l = \dfrac {\partial f} {\partial y}       | r = \map N {x, y}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = f       | r = \map M {x, y} \rd x + \map g y       | c =  }} {{eqn | r = \int \dfrac y {x^2} \rd x + \map g y       | c =  }} {{eqn | r = - \frac y x + \map g y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = f       | r = \int \map N {x, y} \rd y + \map h x       | c =  }} {{eqn | r = \int \paren {y - \dfrac 1 x} \r d y + \map h x       | c =  }} {{eqn | r = \frac {y^2} 2 - \frac y x + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = \dfrac {y^2} 2 - \dfrac y x$ By Solution to Exact Differential Equation, the solution to $(1)$ is: :$\dfrac {y^2} 2 - \dfrac y x = C$ {{qed}} </onlyinclude> Category:Examples of Exact Differential Equation Category:Examples of First Order ODEs syuvq2v6cpkjnr7awwopekq6nio5baz
{{proofread}} Consider the following parabola whose axis of symmetry is the y-axis along with an incoming light ray from $B$ which is parallel to the axis of symmetry, reflects off the parabola at the point $P$ and intersects the y-axis at the point $F$. Points $B$, $P$, and $Q$ are co-linear and the point $Q$ also lies on the directrix of the parabola; the line $AT$ is the tangent of the parabola at the point $P$: 500px By the Law of Reflection the angle of incidence, $90 \degrees - \angle BPT$, is equal to the angle of reflection, $90 \degrees - \angle APF$. Hence :$\angle BPT = \angle APF$ Note: Incoming and outgoing light rays both follow the same Law of Reflection; hence, this argument is valid for both incoming and outgoing light rays. From Two Straight Lines make Equal Opposite Angles we also know that :$\angle BPT = \angle APQ$ We now consider the two triangles $APQ$ and $APF$. They have an equal side, $AP$, and an equal angle, $\angle APQ = \angle APF$, and by the Triangle Side-Angle-Side Equality triangles $APQ$ and $APF$ will be equal if and only if $FP$ and $PQ$. From the definition of a parabola it is the locus of the points that are equidistant from the focus and the directrix; hence the point $F$ is the focus.
Let $M$ and $N$ be defined as: :$\map M {x, y} = -\dfrac {2 x} {y^3}$ :$\map N {x, y} = \dfrac {3 x^2} {y^4} - \dfrac 1 {y^2}$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y}       | r = \frac {6 x} {2 y^4}       | c =  }} {{eqn | l = \dfrac {\partial N} {\partial x}       | r = \dfrac {6 x} {y^4}       | c =  }} {{end-eqn}} Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact. By Solution to Exact Differential Equation, the solution to $(1)$ is: :$\map f {x, y} = C$ where: {{begin-eqn}} {{eqn | l = \dfrac {\partial f} {\partial x}       | r = \map M {x, y}       | c =  }} {{eqn | l = \dfrac {\partial f} {\partial y}       | r = \map N {x, y}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = f       | r = \map M {x, y} \rd x + \map g y       | c =  }} {{eqn | r = \int - \dfrac {2 x} {y^3} \rd x + \map g y       | c =  }} {{eqn | r = -\dfrac {x^2} {y^3} + \map g y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = f       | r = \int \map N {x, y} \rd y + \map h x       | c =  }} {{eqn | r = \int \paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y + \map h x       | c =  }} {{eqn | r = -\dfrac {x^2} {y^3} + \dfrac 1 y + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = \dfrac 1 y - \dfrac {x^2} {y^3}$ and by Solution to Exact Differential Equation, the general solution to $(1)$ is: :$\dfrac 1 y - \dfrac {x^2} {y^3} = C$ {{qed}} Category:Examples of Exact Differential Equation Category:Examples of First Order ODEs hhsdvpmiuh51fgk6yttk6djrk9y3mdj
Let $M$ and $N$ be defined as: :$\map M {x, y} = y - \dfrac 1 x$ :$\map N {x, y} = x - y$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y}       | r = 1       | c =  }} {{eqn | l = \dfrac {\partial N} {\partial x}       | r = 1       | c =  }} {{end-eqn}} Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact. By Solution to Exact Differential Equation, the solution to $(1)$ is: :$\map f {x, y} = C$ where: {{begin-eqn}} {{eqn | l = \dfrac {\partial f} {\partial x}       | r = \map M {x, y}       | c =  }} {{eqn | l = \dfrac {\partial f} {\partial y}       | r = \map N {x, y}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = f       | r = \map M {x, y} \rd x + \map g y       | c =  }} {{eqn | r = \int \paren {y - \dfrac 1 x} \rd x + \map g y       | c =  }} {{eqn | r = x y - \ln x + \map g y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = f       | r = \int \map N {x, y} \rd y + \map h x       | c =  }} {{eqn | r = \int \paren {x - y} \rd y + \map h x       | c =  }} {{eqn | r = x y - \dfrac {y^2} 2 + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = x y - \ln x - \dfrac {y^2} 2$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$x y - \ln x - \dfrac {y^2} 2 + C$ {{qed}} Category:Examples of Exact Differential Equation Category:Examples of First Order ODEs pnrufusux3jtoaq69awzlj5ysojqhh3
Let $M$ and $N$ be defined as: :$\map M {x, y} = \dfrac 1 {x^3 y^2}$ :$\map N {x, y} = \dfrac 1 {x^2 y^3} + 3 y$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y}       | r = -\frac 1 {x^3 y^2}       | c =  }} {{eqn | l = \dfrac {\partial N} {\partial x}       | r = -\frac 1 {x^3 y^2}       | c =  }} {{end-eqn}} By Solution to Exact Differential Equation, the solution to $(1)$ is: :$\map f {x, y} = C$ where: {{begin-eqn}} {{eqn | l = \dfrac {\partial f} {\partial x}       | r = \map M {x, y}       | c =  }} {{eqn | l = \dfrac {\partial f} {\partial y}       | r = \map N {x, y}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = f       | r = \int \map M {x, y} \rd x + \map g y       | c =  }} {{eqn | r = \int \frac 1 {x^3 y^2} \rd x + \map g y       | c =  }} {{eqn | r = -\frac 1 {2 x^2 y^2} + \map g y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = f       | r = \int \map N {x, y} \rd y + \map h x       | c =  }} {{eqn | r = \int \paren {\frac 1 {x^2 y^3} + 3 y} \rd y + \map h x       | c =  }} {{eqn | r = -\frac 1 {2 x^2 y^2} + 3 \dfrac {y^2} 2 + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = -\dfrac 2 {x^2 y^2} + \dfrac {3 y^2} 2$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$ {{qed}} Category:Examples of Exact Differential Equation Category:Examples of First Order ODEs h02f2apyn104iuz6ibkxxmdj318amx8
This is an instance of Linear First Order ODE: $y' - \dfrac y x = k x$. Its solution is: :$\dfrac y x = k x + C$ or: :$y = k x^2 + C x$ from which the solution to $(1)$ is found by substituting $3$ for $k$. {{qed}} Category:Examples of Linear First Order ODEs 5zl5xx9teqim55grzediaigsppbf6nk
=== Sufficient Condition === Let $x \in A^-$. Let $U \in \mathcal B$. By definition of basis, $U$ is an open set of $T$. Thus from Condition for Point being in Closure: :if $x \in U$ then $A \cap U \ne \varnothing$. {{qed|lemma}} === Necessary Condition === Let $x$ be such that for every $U \in \mathcal B$: :if $x \in U$ :then $A \cap U \ne \varnothing$. By Condition for Point being in Closure, to prove that $x \in \operatorname{Fr} A$ it is enough to prove that: :for every open set $U$ of $T$: ::if $x \in U$ then $A \cap U \ne \varnothing$. Let $U$ be an open set of $T$. Let $x \in U$. By definition of (analytic) basis, there exists $V \in \mathcal B$ such that: :$x \in V \subseteq U$ By assumption: :$A \cap V \ne \varnothing$ From the corollary to Set Intersection Preserves Subsets: :$A \cap V \subseteq A \cap U$ So: :$A \cap U \ne \varnothing$ and hence the result. {{qed}}
By Rationals are Everywhere Dense in Sorgenfrey Line: :$\Q$ is dense in the Sorgenfrey line. By Rational Numbers are Countably Infinite: :$\Q$ is countable. Thus by definition: :The Sorgenfrey line is separable. {{qed}}
$(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = x \rd y       | r = -y \rd x       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = C x^{-1}       | c = First Order ODE: $x \rd y = k y \rd x$ }} {{eqn | ll= \leadsto       | l = x y       | r = C       | c = multiplying through by $x$ }} {{end-eqn}} {{qed}} Category:Examples of First Order ODEs Category:Examples of Separation of Variables 6ezpoyg7g40z9d75edixh9xyydjql0n
{{begin-eqn}} {{eqn | l = y \, \dfrac {\d y} {\d x}       | r = k       | c =  }} {{eqn | ll= \leadsto       | l = \int y \rd y       | r = \int k \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \frac {y^2} 2       | r = k x + \frac C 2       | c =  }} {{eqn | ll= \leadsto       | l = y^2       | r = 2 k x + C       | c =  }} {{end-eqn}} {{qed}} Category:Examples of First Order ODEs Category:Examples of Separation of Variables gkokkgsocrjtw5p1968f3m1dbe3iagj
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x}       | r = y + y^3       | c =  }} {{eqn | ll= \leadsto       | l = \int \frac {\d y} {y \paren {1 + y^2} }       | r = \int \frac {\d x} x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \frac 1 2 \, \map \ln {\frac {y^2} {y^2 + 1} }       | r = \ln x + \ln k       | c = Primitive of $\dfrac 1 {x \paren {x^2 + a^2} }$ }} {{eqn | r = \map \ln {k x}       | c =  }} {{eqn | ll= \leadsto       | l = \map \ln {\frac {y^2} {y^2 + 1} }       | r = \map \ln {k^2 x^2}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {y^2} {y^2 + 1}        | r = k^2 x^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {y^2 + 1} {y^2}       | r = \frac 1 {k^2 x^2}       | c =  }} {{eqn | ll= \leadsto       | l = 1 + \frac 1 {y^2}       | r = \frac 1 {k^2 x^2}       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 {y^2}       | r = \frac 1 {k^2 x^2} - 1       | c =  }} {{eqn | r = \frac {1 - k^2 x^2} {k^2 x^2}       | c =  }} {{eqn | ll= \leadsto       | l = y^2       | r = \frac {k^2 x^2} {1 - k^2 x^2}       | c =  }} {{eqn | ll= \leadsto       | l = y^2       | r = \frac {x^2} {\frac 1 {k^2} - x^2}       | c =  }} {{eqn | ll= \leadsto       | l = y^2       | r = \frac {x^2} {C^2 - x^2}       | c = letting $C = \dfrac 1 k$ }} {{eqn | ll= \leadsto       | l = y       | r = \frac x {\sqrt {C^2 - x^2} }       | c =  }} {{end-eqn}} {{qed}} Category:Examples of First Order ODEs Category:Examples of Separation of Variables gp4lv30s4p8oxy2ivldhtbtf5kgzxi2
{{begin-eqn}} {{eqn | l = y \rd y       | r = k x \rd x       | c =  }} {{eqn | ll= \leadsto       | l = \int y \rd y       | r = \int k x \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \frac {y^2} 2       | r = \frac {k x^2} 2 + C_1       | c =  }} {{eqn | ll= \leadsto       | l = y^2       | r = k x^2 + C       | c = reassigning the arbitrary constant. }} {{end-eqn}} {{qed}} Category:Examples of First Order ODEs Category:Examples of Separation of Variables q7mp87fthztpa56s99ft03048efrdar
There are three cases: :$0 \le k$ :$-x^2 < k < 0$ :$k < -x^2$ === $(1): \quad 0 \le k$ === If $0 \le k$ then $k = a^2$ for some $a \in \R$. Then Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ applies: :$\displaystyle \int \frac {\d x} {\sqrt {x^2 + a^2} } = \map \ln {x + \sqrt {x^2 + a^2} } + C$ === $(2): \quad -x^2 < k < 0$ === If $-x^2 < k < 0$ then: :$k = -a^2$ for some $a \in \R$ and: :$x^2 - a^2 > 0$ Then Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ applies: :$\displaystyle \int \frac {\d x} {\sqrt {x^2 - a^2} } = \map \ln {x + \sqrt {x^2 - a^2} } + C$ === $(3): \quad k < -x^2$ === If $k < -x^2$ then: :$k = -a^2$ for some $a \in \R$ and: :$a^2 - x^2 > 0$ Then Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ applies: :$\displaystyle \int \frac {\d x} {\sqrt {a^2 - x^2} } = \map \ln {x + \sqrt {a^2 - x^2} } + C$ The result holds for all three cases. {{qed}} Category:Primitives involving Root of x squared plus a squared Category:Primitives involving Root of x squared minus a squared Category:Primitives involving Root of a squared minus x squared Category:Primitives involving Reciprocals d9jhgj45z2hif4ntthtibtbnusj6dwu
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = \dfrac {y^2} {x^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = -\dfrac 2 x$ :$\map Q x = \dfrac 1 {x^2}$ :$n = 2$ 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 -\dfrac 2 x \rd x       | c =  }} {{eqn | r = 2 \ln x       | c =  }} {{eqn | r = \ln x^2       | c =  }} {{eqn | ll= \leadsto       | l = \map \mu x       | r = e^{\ln x^2}       | c =  }} {{eqn | r = x^2       | c =  }} {{end-eqn}} and so substituting into $(3)$: {{begin-eqn}} {{eqn | l = x^2 \frac 1 y       | r = \paren {1 - 2} \int \dfrac 1 {x^2} x^2 \rd x       | c =  }} {{eqn | r = -\int \rd x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {x^2} y       | r = -x + C       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = -\frac {x^2} {x + C}       | c =  }} {{end-eqn}} {{qed}} Category:Examples of Bernoulli's Equation pkb326kfdtfk18zkhwk69sp4ow2nx18
{{begin-eqn}} {{eqn | l = 2 x y \rd x + \paren {x^2 + 2 y} \rd y       | r = 0       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = \dfrac {\d x} {\d y} + \frac 1 {2 y} x       | r = -\frac 1 x       | c =  }} {{end-eqn}} It can be seen that $(2)$ is in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$ where: :$\map P y = \dfrac 1 {2 y}$ :$\map Q y = -1$ :$n = -1$ and so is an example of Bernoulli's equation. By Solution to Bernoulli's Equation it has the general solution: :$(3): \quad \displaystyle \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \, \map \mu y \rd y + C$ where: :$\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$ Thus $\map \mu x$ is evaluated: {{begin-eqn}} {{eqn | l = \paren {1 - n} \int \map P y \rd y       | r = \paren {1 - \paren {-1} } \int \dfrac 1 {2 y} \rd y       | c =  }} {{eqn | r = 2 \frac 1 2 \ln y       | c =  }} {{eqn | r = \ln y       | c =  }} {{eqn | ll= \leadsto       | l = \map \mu y       | r = e^{\ln y}       | c =  }} {{eqn | r = y       | c =  }} {{end-eqn}} and so substituting into $(3)$: {{begin-eqn}} {{eqn | l = \frac y {x^{\paren {-1} - 1} }       | r = \paren {1 - \paren {-1} } \int \paren {-1} y \rd y       | c =  }} {{eqn | r = -2 \int y \rd y       | c =  }} {{eqn | ll= \leadsto       | l = y x^2       | r = -2 \frac {y^2} 2 + C       | c =  }} {{eqn | ll= \leadsto       | l = y \paren {x^2 + y}       | r = C       | c =  }} {{end-eqn}} {{qed}} Category:Examples of Bernoulli's Equation 0jlmrbe1mo38cnp9og957tw0cmiv2wx
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where $\map P x = -\dfrac 1 x$. Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int -\frac 1 x \rd x       | c =  }} {{eqn | r = -\ln x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = e^{-\ln x}       | c =  }} {{eqn | r = \frac 1 x       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\map {\dfrac {\d} {\d x} } {\dfrac y x} = \dfrac 1 x k x = k$ and the general solution is: :$\dfrac y x = k x + C$ or: :$y = k x^2 + C x$ {{qed}} Category:Examples of Linear First Order ODEs 2c5s81vvm5t4d1tgvkb8zr7atz6asrc
$(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = x \rd y       | r = k y \rd x       | c =  }} {{eqn | ll= \leadsto       | l = \int \dfrac {\d y} y       | r = k \int \dfrac {\d x} x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \ln y       | r = k \ln x + \ln C       | c =  }} {{eqn | r = \map \ln {C x^k}       | c =  }} {{eqn | l = y       | r = C x^k       | c =  }} {{end-eqn}} {{qed}} Category:Examples of First Order ODEs Category:Examples of Separation of Variables Category:First Order ODE/x dy = k y dx i7raos39f81so6kkk5pupb6v1ws4dc6