hcju/proofwikiqa huggingface dataset:
Let: :$\map M {x, y} = -\dfrac 1 y \sin \dfrac x y$ :$\map N {x, y} = \dfrac x {y^2} \sin \dfrac x y$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y}       | r = \frac 1 {y^2} \sin \frac x y + \paren {-\frac 1 y} \paren {-\frac 1 {y^2} } x \cos \frac x y       | c =  }} {{eqn | r = \frac 1 {y^2} \sin \frac x y + \frac x {y^3} \cos \frac x y       | c =  }} {{eqn | l = \frac {\partial N} {\partial x}       | r = \frac 1 {y^2} \sin \frac x y + \frac x {y^2} \frac 1 y \cos \frac x y       | c =  }} {{eqn | r = \frac 1 {y^2} \sin \frac x y + \frac x {y^3} \cos \frac x y       | 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 {-\dfrac 1 y \sin \dfrac x y} \rd x + \map g y       | c =  }} {{eqn | r = -\dfrac 1 y y \cos \dfrac x y + \map g y       | c =  }} {{eqn | r = -\cos \dfrac x y + \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 x {y^2} \sin \dfrac x y} \rd y + \map h x       | c =  }} {{end-eqn}} Set $z = \dfrac x y$: {{begin-eqn}} {{eqn | l = z       | r = \frac x y       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = -\frac x {y^2}       | c =  }} {{end-eqn}} Thus by Integration by Substitution: {{begin-eqn}} {{eqn | l = \int \frac x {y^2} \sin \frac x y \rd y       | r = \int \frac x {y^2} \frac {-y^2} x \sin z \rd z       | c =  }} {{eqn | r = -\int \sin z \rd z       | c =  }} {{eqn | r = \cos z       | c =  }} {{eqn | r = \cos \frac x y       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = -\cos \dfrac x y$ and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is: {{begin-eqn}} {{eqn | l = \cos \dfrac x y       | r = C_1       | c =  }} {{eqn | ll= \leadsto       | l = \map \arccos {\cos \dfrac x y}       | r = \arccos C_1       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac x y       | r = C       | c = where $C = \arccos C_1$ }} {{end-eqn}} {{qed}}
Let: :$\map M {x, y} = 2 x y^3 + y \cos x$ :$\map N {x, y} = 3 x^2 y^2 + \sin x$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y}       | r = 2 x \cdot 3 y^2 + \cos x       | c =  }} {{eqn | r = 6 x y^2 + \cos x       | c =  }} {{eqn | l = \frac {\partial N} {\partial x}       | r = 3 x^2 \cdot 2 y^2 + \cos x       | c =  }} {{eqn | r = 6 x y^2 + \cos x       | 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 {2 x y^3 + y \cos x} \rd x + \map g y       | c =  }} {{eqn | r = x^2 y^3 + y \sin 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 {3 x^2 y^2 + \sin x} \rd y + \map h x       | c =  }} {{eqn | r = x^2 y^3 + y \sin x + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = x^2 y^3 + y \sin x$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$x^2 y^3 + y \sin x = C$ {{qed}}
First express $(1)$ in the form: :$(2): \quad \paren {\dfrac y {1 - x^2 y^2} - 1} + \paren {\dfrac x {1 - x^2 y^2} } \dfrac {\d y} {\d x}$ Let: :$\map M {x, y} = \dfrac y {1 - x^2 y^2} - 1$ :$\map N {x, y} = \dfrac x {1 - x^2 y^2}$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y}       | r = \frac 1 {1 - x^2 y^2} + \frac {2 x^2 y^2} {\paren {1 - x^2 y^2}^2}       | c =  }} {{eqn | r = \frac {1 + x^2 y^2} {\paren {1 - x^2 y^2}^2}       | c =  }} {{eqn | l = \frac {\partial N} {\partial x}       | r = \frac {1 + x^2 y^2} {\paren {1 - x^2 y^2}^2}       | c = similarly }} {{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 {\dfrac y {1 - x^2 y^2} - 1} \rd x + \map g y       | c =  }} {{eqn | r = \int \dfrac {y \rd x} {1 - x^2 y^2} - \int \rd x + \map g y       | c =  }} {{end-eqn}} Substitute $z = x y$ to obtain: :$\dfrac {\d z} {\d x} = y$ This gives: {{begin-eqn}} {{eqn | l = f       | r = \map M {x, y} \rd x + \map g y       | c =  }} {{eqn | r = \frac 1 2 \map \ln {\frac {1 + z} {1 - z} } - \int \rd x + \map g y       | c =  }} {{eqn | ll= \leadsto       | l = \map M {x, y} \rd x + \map g y       | r = \frac 1 2 \map \ln {\frac {1 + x y} {1 - x 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 {\frac {1 + x^2 y^2} {\paren {1 - x^2 y^2}^2} } \rd y + \map h x       | c =  }} {{eqn | r = \frac 1 2 \map \ln {\frac {1 + x y} {1 - x y} } + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = \dfrac 1 2 \map \ln {\dfrac {1 + x y} {1 - x y} } - x$ and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is: :$\map \ln {\dfrac {1 + x y} {1 - x y} } - 2 x = C$ {{qed}}
Assume $x$ is limit point of $A$. By definition of limit point it suffices to prove :$\forall U \in \tau: x \in U \implies B \cap \paren {U \setminus \set x} \ne \O$ Let $U \in \tau$ such that :$x \in U$ By definition of limit point: :$A \cap \paren {U \setminus \set x} \ne \O$ By Set Intersection Preserves Subsets/Corollary: :$A \cap \paren {U \setminus \set x} \subseteq B \cap \paren {U \setminus \set x}$ Thus: :$B \cap \paren {U \setminus \set x} \ne \O$ {{qed}} Category:Limit Points 9wavooswmawzbhvl8fgyr4757e2ghdf
Let $\map F f$ be any function of $f$ which is an integrating factor of $(1)$. Then: :$\ds \mu \map F f \paren {\map M {x, y} \rd x + \map N {x, y} \rd y} = \map F f \rd f = \map \d {\int \map F f \rd f}$ so $\mu \map F f$ is also an integrating factor. {{qed}}
Let us for ease of manipulation express $(1)$ in the form of differentials: :$(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$ Let $\mu$ be an integrating factor for $(2)$. Then, by definition: :$\mu \, \map M {x, y} \rd x + \mu \, \map N {x, y} \rd y = 0$ is an exact differential equation. By Solution to Exact Differential Equation: :$\dfrac {\map \partial {\mu M} } {\partial y} = \dfrac {\map \partial {\mu N} } {\partial x}$ Evaluating this, using the Product Rule: :$\mu \dfrac {\partial M} {\partial y} + M \dfrac {\partial \mu} {\partial y} = \mu \dfrac {\partial N} {\partial x} + N \dfrac {\partial \mu} {\partial x}$ which leads us to: :$\dfrac 1 \mu \paren {N \dfrac {\partial \mu} {\partial x} - M  \dfrac {\partial \mu} {\partial y} } = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$ Let us use $\map P {x, y}$ for $\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$. Then: :$(3): \quad \dfrac 1 \mu = \dfrac {\map P {x, y} } {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y}}$
=== Preliminary Work === {{:Integrating Factor for First Order ODE/Preliminary Work}} === Proof for Function of $x$ or $y$ only === Suppose that $\mu$ is a function of $x$ only. Then: :$\dfrac {\partial \mu} {\partial x} = \dfrac {d \mu} {d x}, \dfrac {\partial \mu} {\partial y} = 0$ which, when substituting in $(3)$, leads us to: :$\dfrac 1 \mu \dfrac {\mathrm d \mu} {\mathrm d x} = \dfrac {P \left({x, y}\right)} {N \left({x, y}\right)} = g \left({x}\right)$ where $g \left({x}\right)$ is the function of $x$ that we posited. Similarly, if $\mu$ is a function of $y$ only, we find that :$\dfrac 1 \mu \dfrac {\mathrm d \mu} {\mathrm d y} = \dfrac {P \left({x, y}\right)} {-M \left({x, y}\right)} = h \left({y}\right)$ where $h \left({y}\right)$ is the function of $y$ that we posited. === Conclusion of Proof === {{:Integrating Factor for First Order ODE/Conclusion}}
=== Preliminary Work === {{:Integrating Factor for First Order ODE/Preliminary Work}} === Proof for Function of $x + y$ === Suppose that $\mu$ is a function of $z = x + y$. Then: :$\dfrac {\partial z} {\partial x} = 1 = \dfrac {\partial z} {\partial y}$ Thus: :$\dfrac {\partial \mu} {\partial x} = \dfrac {d \mu} {d z} \dfrac {\partial z} {\partial x} = \dfrac {\mathrm d \mu} {\mathrm d z} = \dfrac {\mathrm d \mu} {\mathrm d z} \dfrac {\partial z} {\partial y} = \dfrac {\partial \mu} {\partial y}$ which, when substituting in $(3)$, leads us to: :$\dfrac 1 \mu \dfrac {\mathrm d \mu} {\mathrm d z} = \dfrac {P \left({x, y}\right)} {N \left({x, y}\right) - M \left({x, y}\right)} = g \left({z}\right)$ where $g \left({z}\right)$ is the function of $z$ that we posited. === Conclusion of Proof === {{:Integrating Factor for First Order ODE/Conclusion}}
=== Preliminary Work === {{:Integrating Factor for First Order ODE/Preliminary Work}} === Proof for Function of x y === Suppose that $\mu$ is a function of $z = x y$. Then: :$\dfrac {\partial z} {\partial x} = y$ :$\dfrac {\partial z} {\partial y} = x$ Thus: :$\dfrac {\partial \mu} {\partial x} = \dfrac {\d \mu} {\d z} \dfrac {\partial z} {\partial x} = y \dfrac {\d \mu} {\d z}$ :$\dfrac {\partial \mu} {\partial y} = \dfrac {\d \mu} {\d z} \dfrac {\partial z} {\partial y} = x \dfrac {\d \mu} {\d z}$ which, when substituting in $(3)$, leads us to: :$\dfrac 1 \mu \dfrac {\d \mu} {\d z} = \dfrac {\map P {x, y} } {N y - M x} = \map g z$ where $\map g z$ is the function of $z$ that we posited. === Conclusion of Proof === {{:Integrating Factor for First Order ODE/Conclusion}}
=== Set of Condensation Points of Union Subset Union of Sets of Condensation Points === Let $x \in \left({A \cup B}\right)^0$. By definition of set of condensation points: :$x$ is condensation point of $A \cup B$ By Lemma: :$x$ is condensation point of $A$ or $x$ is condensation point of $B$ By definition of set of condensation points: :$x \in A^0$ or $x \in B^0$ Thus by definition of union: :$x \in A^0 \cup B^0$ {{qed|lemma}} === Union of Sets of Condensation Points Subset Set of Condensation Points of Union === By Set is Subset of Union: :$A \subseteq A \cup B \land B \subseteq A \cup B$ By Set of Condensation Points is Monotone: :$A^0 \subseteq \left({A \cup B}\right)^0 \land B^0 \subseteq \left({A \cup B}\right)^0$ Thus by Union is Smallest Superset: :$A^0 \cup B^0 \subseteq \left({A \cup B}\right)^0$ {{qed|lemma}} Thus by definition of set equality: :$\left({A \cup B}\right)^0 = A^0 \cup B^0$ {{qed}}
Assume $x$ is condensation point of $A \cup B$. Aiming for a contradiction suppose that :$x$ is not condensation point of $A$ and $x$ is not condensation point of $B$. By definition of condensation point: :$\exists U_1 \in \tau: x \in U_1 \land A \cap U_1$ is countable By definition of condensation point: :$\exists U_2 \in \tau: x \in U_2 \land B \cap U_2$ is countable By definition of intersection: :$x \in U_1 \cap U_2$ By definition of topological space: :$U_1 \cap U_2 \in \tau$ By Intersection is Subset: :$A \cap U_1 \cap U_2 \subseteq A \cap U_1 \land B \cap U_1 \cap U_2 \subseteq B \cap U_2$ By Set Union Preserves Subsets: :$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) \subseteq \left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$ By Countable Union of Countable Sets is Countable: :$\left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$ is countable Then by Subset of Countable Set is Countable: :$(1): \quad \left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right)$ is countable By Intersection Distributes over Union: :$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) = \left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$ By definition of condensation point: :$\left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$ is uncountable This contradicts $(1)$. Thus the result fallows by Proof by Contradiction. {{qed}}
We have an equation of the form: :$\dfrac 1 \mu \dfrac {\d \mu} {\d w} = \map f w$ which is what you get when you apply the Chain Rule for Derivatives and Derivative of Logarithm Function to: :$\dfrac {\map \d {\ln \mu} } {\d w} = \map f w$ Thus: :$\displaystyle \ln \mu = \int \map f w \rd w$ and so: :$\mu = e^{\int \map f w \rd w}$ Hence the results as stated. {{qed}}
Assume :$A$ is countable. Aiming for a contradiction suppose that :$A^0 \ne \varnothing$ By definition of empty set: :$\exists x: x \in A^0$ Then by definition of set of condensation points: :$x$ is a condensation point of $A$. This contradicts Lemma. Thus the result follows by Proof by Contradiction. {{qed}}
Assume :$A$ is countable. Aiming for a contradiction suppose that there exists a point $x$ of $S$ such that : $x$ is a condensation point of $A$ By definition of topological space: :$S \in \tau$ Then by definition of condensation point: :$A \cap S$ is uncountable By Intersection with Subset is Subset: :$A \cap S = A$ $A$ is countable contradicts $A$ is uncountable. Thus the result follows by Proof by Contradiction. {{qed}}
We note that $(1)$ is in the form: :$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$ but that $(1)$ is not exact. So, let: :$\map M {x, y} = -2 x y$ :$\map N {x, y} = 3 x^2 - y^2$ 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 = -2 x - 6 x       | c =  }} {{eqn | r = -8 x       | c =  }} {{end-eqn}} It can be observed that: {{begin-eqn}} {{eqn | l = \frac {\map P {x, y} } {\map M {x, y} }       | r = \frac {-8 x} {-2 x y}       | c =  }} {{eqn | r = \frac 4 y       | c =  }} {{end-eqn}} Thus $\dfrac {\map P {x, y}} {\map M {x, y} }$ is a function of $y$ only. So Integrating Factor for First Order ODE: Function of One Variable can be used: :$\map \mu y = e^{\int -\map h y \rd y}$ Hence: {{begin-eqn}} {{eqn | l = \int -\map h y \rd y       | r = \int -\paren {4 / y} \rd y       | c =  }} {{eqn | r = - 4 \ln y       | c =  }} {{eqn | r = \map \ln {y^{-4} }       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int -\map h y \rd y}       | r = \frac 1 {y^4}       | c =  }} {{end-eqn}} Thus an integrating factor for $(1)$ has been found: :$\mu = \dfrac 1 {y^4}$ which yields, when multiplying it throughout $(1)$: :$\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$ which is now exact. By First Order ODE: $\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$, its solution is: :$\dfrac 1 y - \dfrac {x^2} {y^3} = C$ {{qed}}
We note that $(1)$ is in the form: :$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$ but that $(1)$ is not exact. So, let: :$\map M {x, y} = x y - 1$ :$\map N {x, y} = x^2 - x y = x \paren {x - y}$ 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 = x - \paren {2 x - y}       | c =  }} {{eqn | r = -\paren {x - y}       | c =  }} {{end-eqn}} It can be observed that: {{begin-eqn}} {{eqn | l = \frac {\map P {x, y} } {\map N {x, y} }       | r = \frac {-\paren {x - y} } {x \paren {x - y} }       | c =  }} {{eqn | r = -\frac 1 x       | c =  }} {{end-eqn}} Thus $\dfrac {\map P {x, y} } {\map N {x, y}}$ is a function of $x$ only. So Integrating Factor for First Order ODE: Function of One Variable can be used: :$\map \mu y = e^{\int \map g x \rd x}$ Hence: {{begin-eqn}} {{eqn | l = \int \map g x \rd x       | r = \int -\paren {1 / x} \rd x       | c =  }} {{eqn | r = -\ln x       | c =  }} {{eqn | r = \map \ln {x^{-1} }       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int \map g x \rd x}       | r = \frac 1 x       | c =  }} {{end-eqn}} Thus an integrating factor for $(1)$ has been found: :$\mu = \dfrac 1 x$ which yields, when multiplying it throughout $(1)$: :$\paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$ which is now exact. By First Order ODE: $\paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$, its solution is: :$x y - \ln x - \dfrac {y^2} 2 + C$ {{qed}}
We note that $(1)$ is in the form: :$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$ but that $(1)$ is not exact. So, let: :$\map M {x, y} = y$ :$\map N {x, y} = 3 x^3 y^4 + x = x \paren {3 x^2 y^4 + 1}$ 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 = 1 - \paren {9 x^2 y^4 + 1}       | c =  }} {{eqn | r = -9 x^2 y^4       | 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 {-9 x^2 y^4} {\paren {3 x^3 y^4 + x} y - y x}       | c =  }} {{eqn | r = -\frac {-9 x^2 y^4} {3 x^3 y^5}       | 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 $(1)$: :$\dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$ which is now exact. By First Order ODE: $\dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$, its solution is: :$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$ {{qed}}
The mirror will have the shape of a surface of revolution generated by revolving a curve $APB$ in the cartesian plane around the $x$-axis. Let $P = \left({x, y}\right)$ be an arbitrary point on $APB$. :500px From the Law of Reflection: :$\alpha = \beta$ By the geometry of the situation: :$\phi = \beta$ :$\theta = \alpha + \phi = 2 \beta$ By definition of tangent: :$\tan \theta = \dfrac y x$ and so: {{begin-eqn}} {{eqn | l = \tan \theta       | r = \tan 2 \beta       | c =  }} {{eqn | r = \dfrac {2 \tan \beta} {1 - \tan^2 \beta}       | c = Double Angle Formula for Tangent }} {{eqn | ll= \implies       | l = \frac y x       | r = \dfrac {2 \dfrac {\mathrm d y} {\mathrm d x} } {1 - \left({\dfrac {\mathrm d y} {\mathrm d x} }\right)^2}       | c = Double Angle Formula for Tangent }} {{end-eqn}} Using the Quadratic Formula: :$\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {-x \pm \sqrt {x^2 + y^2} \, \mathrm d x} y$ which can be expressed as: :$x \, \mathrm d x + y  \, \mathrm d y = \pm \sqrt {x^2 + y^2} \, \mathrm d x$ Using Differential of Sum of Squares: :$\pm \dfrac {\mathrm d \left({x^2 + y^2}\right)} {2 \sqrt {x^2 + y^2} } = \mathrm d x$ and so: :$\pm \sqrt {x^2 + y^2} = x + c$ Hence the result. {{qed}}
Rearranging, we have: :$\dfrac {y \rd x - x \rd y} {y^2} = -\paren {\dfrac 1 {y^2} + 1} \rd y$ From the Quotient Rule for Derivatives: :$\map \d {\dfrac x y} = \dfrac {y \rd x - x \rd y} {y^2}$ from which: :$\map \d {\dfrac x y} = -\paren {\dfrac 1 {y^2} + 1} \rd y$ Hence the result: :$\dfrac x y = \dfrac 1 y - y + C$ {{qed}}
We note that $(1)$ is in the form: :$\map M {x, y} \d x + \map N {x, y} \d y = 0$ but that $(1)$ is not exact. So, let: :$\map M {x, y} = y$ :$\map N {x, y} = x^2 y - x$ 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 = 1 - \paren {2 x y - 1}       | c =  }} {{eqn | r = 2 \paren {1 - x y}       | c =  }} {{end-eqn}} It can be observed that: {{begin-eqn}} {{eqn | l = \frac {\map P {x, y} } {\map N {x, y} }       | r = \frac {2 \paren {1 - x y} } {x^2 y - x}       | c =  }} {{eqn | r = \frac {-2 \paren {x y - 1} } {x \paren {x y - 1} }       | c =  }} {{eqn | r = -\frac 2 x       | c =  }} {{end-eqn}} Thus $\dfrac {\map P {x, y} } {\map N {x, y} }$ is a function of $x$ only. So Integrating Factor for First Order ODE: Function of One Variable can be used: :$\map \mu x = e^{\int \map g x \rd x}$ Hence: {{begin-eqn}} {{eqn | l = \int \map g x \rd x       | r = \int - \paren {2 / x} \rd x       | c =  }} {{eqn | r = - 2 \ln x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int \map g x \rd x}       | r = \frac 1 {x^2}       | c =  }} {{end-eqn}} Thus an integrating factor for $(1)$ has been found: :$\mu = \dfrac 1 {x^2}$ which yields, when multiplying it throughout $(1)$: :$\dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$ which is now exact. By First Order ODE: $\dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$, its solution is: :$\dfrac {y^2} 2 - \dfrac y x = C$ {{qed}}
Rearranging, we have: :$x^2 y \rd y - \paren {x \rd y - y \rd x} = 0$ Aiming to use Quotient Rule for Differentials, divide by $x^2$: :$y \rd y = \dfrac {x \rd y - y \rd x} {x^2}$ So from Quotient Rule for Differentials: Formulation 1 :$y \rd y = \map \d {\dfrac y x}$ from which the solution immediately drops: :$\dfrac {y^2} 2 - \dfrac y x = C$ {{qed}}
Rearranging, we have: :$\dfrac {y \rd x - x \rd y} {x y} = y^2 \rd y$ From Differential of Logarithm of Quotient: :$\map \d {\ln \dfrac y x} = \dfrac {y \rd x - x \rd y} {x y}$ from which: :$\map \d {\ln \dfrac x y} = y^2 \rd y$ Hence the result: :$\ln \dfrac x y = \dfrac {y^3} 3 + C$ {{qed}}
Rearranging, we have: :$y \rd x - x \rd y = -\paren {x^2 + y^2} x^3 \rd x$ from which: :$\dfrac {y \rd x - x \rd y} {x^2 + y^2} = - x^3 \rd x$ From Differential of Arctangent of Quotient: :$\map \d {\arctan \dfrac x y} = \dfrac {y \rd x - x \rd y} {x^2 + y^2}$ from which $(1)$ evolves into: :$\map \d {\arctan \dfrac x y} = -x^3 \rd x$ Hence the result: :$\arctan \dfrac x y = -\dfrac {x^4} 4 + C$ {{qed}}
Let: :$\map M {x, y} =  x + y$ :$\map N {x, y} =  x - y$ We have that: :$\map M {t x, t y} =  t x + t y = t \paren {x + y} = t \map M {x, y}$ :$\map N {t x, t y} =  t x - t y = t \paren {x - y} = t \map N {x, y}$ Thus both $M$ and $N$ are homogeneous functions of degree $1$. Thus by definition $(1)$ is a homogeneous differential equation. {{begin-eqn}} {{eqn | l = \frac {\d y} {\d x}       | r = \frac {x + y} {x - y}       | c =  }} {{eqn | r = \frac {1 + y / x} {1 - y / x}       | c = dividing top and bottom by $x$ }} {{eqn | r = \frac {1 + z} {1 - z}       | c = substituting $z$ for $y / x$ }} {{end-eqn}} By Solution to Homogeneous Differential Equation: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ where: :$\map f {1, z} = \dfrac {1 + z} {1 - z}$ Hence: {{begin-eqn}} {{eqn | l = \ln x       | r = \int \frac {\d z} {\dfrac {1 + z} {1 - z} - z}       | c =  }} {{eqn | r = \int \frac {\paren {1 - z} \rd z} {1 + z - z \paren {1 - z} }       | c =  }} {{eqn | r = \int \frac {\paren {1 - z} \rd z} {1 + z^2}       | c =  }} {{eqn | r = \int \frac {\d z} {1 + z^2} - \int \frac {z \rd z} {1 + z^2}       | c =  }} {{eqn | r = \arctan z - \int \frac {z \rd z} {1 + z^2} + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \arctan z - \frac 1 2 \map \ln {1 + z^2} + C       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{end-eqn}} Substituting $y / x$ for $z$ reveals the solution: :$\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$ {{qed}}