hcju/proofwikiqa huggingface dataset:
By Set is Subset of its Topological Closure: :$A^0 \subseteq \left({A^0}\right)^-$ To prove the equality by definition of set equality it suffices to show the inclusion: :$\left({A^0}\right)^- \subseteq A^0$ Let $x \in \left({A^0}\right)^-$. We will prove that :$(1): \quad \forall U \in \tau: x \in U \implies A \cap U$ is uncountable Let $U$ be an open subset of $S$ such that :$x \in U$ By Condition for Point being in Closure: :$A^0 \cap U \ne \varnothing$ By definition of empty set: :$\exists y: y \in A^0 \cap U$ By definition of intersection: :$y \in A^0 \land y \in U$ By definition of set of condensation points: :$y$ is condensation point of $A$ Then by definition of condensation point: :$A \cap U$ is uncountable We will prove that :$x$ is limit point of $A$ Let $U$ be an open subset of $S$ such that :$x \in U$ By $(1)$: :$A \cap U$ is uncountable :$A \cap \left({U \setminus \left\{{x}\right\}}\right)$ is uncountable Thus :$A \cap \left({U \setminus \left\{{x}\right\}}\right) \ne \varnothing$ Then by definition and $(1)$: :$x$ is condensation point of $A$ Thus by definition of set of condensation points: :$x \in A^0$ {{qed}}
$(1)$ can also be rendered: :$\dfrac {\mathrm d y} {\mathrm d x} = -\dfrac {x^2 - 2 y^2} {x y}$ Let: :$\map M {x, y} = x^2 - 2 y^2$ :$\map N {x, y} = x y$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y}       | r = \paren {t x}^2 - 2 \paren {t y}^2       | c =  }} {{eqn | r = t^2 \paren {x^2 - 2 y^2}       | c =  }} {{eqn | r = t^2 \, \map M {x, y}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map N {t x, t y}       | r = t x t y       | c =  }} {{eqn | r = t^2 \paren {x y}       | c =  }} {{eqn | r = t^2 \, N \left({x, y}\right)       | c =  }} {{end-eqn}} Thus both $M$ and $N$ are homogeneous functions of degree $2$. Thus, by definition, $(1)$ is a homogeneous differential equation. By Solution to Homogeneous Differential Equation, its solution is: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ where: :$\map f {x, y} = -\dfrac {x^2 - 2 y^2} {x y}$ Hence: {{begin-eqn}} {{eqn | l = \ln x       | r = \int \frac {\d z} {-\frac {1 - 2 z^2} {z} - z} + C_1       | c =  }} {{eqn | r = \int \frac {z \rd z} {z^2 - 1} + C_1       | c =  }} {{eqn | r = \frac 1 2 \, \map \ln {z^2 - 1} + C_1       | c = Primitive of $\dfrac x {x^2 - a^2}$ }} {{eqn | ll= \leadsto       | l = \map \ln {z^2 - 1}       | r = \map \ln {x^2} + \ln C       | c =  }} {{eqn | ll= \leadsto       | l = z^2 - 1       | r = C x^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {y^2} {x^2} - 1       | r = C x^2       | c =  }} {{eqn | ll= \leadsto       | l = y^2       | r = x^2 + C x^4       | c =  }} {{end-eqn}} {{qed}}
Let: :$\map M {x, y} = 3 x y + 2 y^2$ :$\map N {x, y} = x^2$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y}       | r = 3 t x t y + 2 \paren {t y}^2       | c =  }} {{eqn | r = t^2 \paren {3 x y + 2 y^2}       | c =  }} {{eqn | r = t^2 \, \map M {x, y}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map N {t x, t y}       | r = \paren {t x}^2       | c =  }} {{eqn | r = t^2 \paren {x^2}       | c =  }} {{eqn | r = t^2 \, \map N {x, y}       | c =  }} {{end-eqn}} Thus both $M$ and $N$ are homogeneous functions of degree $2$. Thus, by definition, $(1)$ is a homogeneous differential equation. By Solution to Homogeneous Differential Equation, its solution is: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ where: :$\map f {x, y} = \dfrac {3 x y + 2 y^2} {x^2}$ {{begin-eqn}} {{eqn | l = \ln x       | r = \int \frac {\d z} {\dfrac {3 z + 2 z^2} {1^2} - z} + C_1       | c =  }} {{eqn | r = \int \frac {\d z} {2 z \paren {1 + z} } + C_1       | c =  }} {{eqn | r = \frac 1 2 \ln \frac z {z + 1} + \ln C_2       | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$ }} {{eqn | ll= \leadsto       | l = \ln x^2       | r = \ln \frac z {z + 1} + \ln C_3       | c =  }} {{eqn | r = \ln \frac {C_3 z} {z + 1}       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | r = \frac {C_3 y / x} {y / x + 1}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = C x^2 \paren {x + y}       | c = where $C := \dfrac 1 {C_3}$ }} {{end-eqn}} {{qed}}
Assume :$A \subseteq B$ Let $x \in A^0$. By definition of set of condensation points: :$x$ is condensation point of $A$ By definition of condensation point: :$x$ is limit point of $A$ such that $\forall U \in \tau: A \cap U$ is uncountable Thus by Limit Point of Subset is Limit Point of Set: :$x$ is limit point of $B$ Let $U \in \tau$. By definition of condensation point: :$A \cap U$ is uncountable By Set Intersection Preserves Subsets/Corollary: :$A \cap U \subseteq B \cap U$ Thus by Subset of Countable Set is Countable :$B \cap U$ is uncountable Then by definition: :$x$ is condensation point of $B$ Thus by definition of set of condensation points: :$x \in B^0$ {{qed}}
Let: :$\map M {x, y} = 3 \paren {x^2 + y^2} \arctan \dfrac y x + x y$ :$\map N {x, y} = x^2$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y}       | r = 3 \paren {\paren {t x}^2 + \paren {t y}^2} \arctan \dfrac {t y} {t x} + t x t y       | c =  }} {{eqn | r = t^2 \paren {3 \paren {x^2 + y^2} \arctan \dfrac y x + x y}       | c =  }} {{eqn | r = t^2 \, \map M {x, y}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map N {t x, t y}       | r = \paren {t x}^2       | c =  }} {{eqn | r = t^2 \paren {x^2}       | c =  }} {{eqn | r = t^2 \, \map N {x, y}       | c =  }} {{end-eqn}} Thus both $M$ and $N$ are homogeneous functions of degree $2$. Thus, by definition, $(1)$ is a homogeneous differential equation. By Solution to Homogeneous Differential Equation, its solution is: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ where: :$\map f {x, y} = \dfrac {3 \paren {x^2 + y^2} \arctan \dfrac y x + x y} {x^2}$ Thus: {{begin-eqn}} {{eqn | l = \ln x       | r = \int \frac {\d z} {\dfrac {3 \paren {1^2 + z^2} \arctan \frac z 1 + z} {1^2} - z} + C_1       | c =  }} {{eqn | r = \int \frac {\d z} {3 \paren {1 + z^2} \arctan z} + C_1       | c =  }} {{end-eqn}} Substituting $u = \arctan z$: {{begin-eqn}} {{eqn | l = \dfrac {\d u} {\d x}       | r = \frac 1 {z^2 + 1}       | c = Derivative of Arctangent Function }} {{eqn | ll= \leadsto       | l = \dfrac {\d z} {\d u}       | r = z^2 + 1       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \ln x       | r = \int \frac {\d z} {3 \paren {1 + z^2} \arctan z} + C_1       | c =  }} {{eqn | r = \int \frac {\d u} {3 u} + C_1       | c =  }} {{eqn | r = \frac 1 3 \ln u + C_1       | c = Primitive of Reciprocal }} {{eqn | ll= \leadsto       | l = \ln x^3 + \ln C       | r = \ln u       | c = putting $\ln C = - C_1$ }} {{eqn | ll= \leadsto       | l = C x^3       | r = u       | c =  }} {{eqn | ll= \leadsto       | l = C x^3       | r = \arctan \frac y x       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = x \tan C x^3       | c =  }} {{end-eqn}} {{qed}}
Let: :$\map M {x, y} = y \sin \dfrac y x + x$ :$\map N {x, y} = x \sin \dfrac y x$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y}       | r = t y \sin \dfrac t y t x + t x       | c =  }} {{eqn | r = t \paren {y \sin \dfrac y x + x}       | c =  }} {{eqn | r = t \, \map M {x, y}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map N {t x, t y}       | r = t x \sin \dfrac {t y} {t x}       | c =  }} {{eqn | r = t \paren {x \sin \dfrac y x}       | c =  }} {{eqn | r = t \, \map N {x, y}       | c =  }} {{end-eqn}} Thus both $M$ and $N$ are homogeneous functions of degree $1$. Thus, by definition, $(1)$ is a homogeneous differential equation. By Solution to Homogeneous Differential Equation, its solution is: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ where: :$\map f {x, y} = \dfrac {y \sin \dfrac y x + x} {x \sin \dfrac y x}$ Thus: {{begin-eqn}} {{eqn | l = \ln x       | r = \int \frac {\d z} {\dfrac {z \sin z + 1} {\sin z} - z} + C_1       | c =  }} {{eqn | r = \int \sin z \rd z + C_1       | c =  }} {{eqn | r = -\cos z + C_1       | c = Primitive of $\sin z$ }} {{eqn | ll= \leadsto       | l = \ln x + \ln C       | r = -\cos z       | c = putting $\ln C = - C_1$ }} {{eqn | ll= \leadsto       | l = \cos \frac y x + \ln C x = 0       | r = 0       | c =  }} {{end-eqn}} {{qed}}
Let: :$\map M {x, y} = y + 2 x e^{-y/x}$ :$\map N {x, y} = x$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y}       | r = t y + 2 t x e^{-t y / t x}       | c =  }} {{eqn | r = t \paren {y + 2 x e^{-y / x} }       | c =  }} {{eqn | r = t \, \map M {x, y}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map N {t x, t y}       | r = t x       | c =  }} {{eqn | r = t \, \map N {x, y}       | c =  }} {{end-eqn}} Thus both $M$ and $N$ are homogeneous functions of degree $1$. Thus, by definition, $(1)$ is a homogeneous differential equation. By Solution to Homogeneous Differential Equation, its solution is: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ where: :$\map f {x, y} = \dfrac {y + 2 x e^{-y / x} } x$ Thus: {{begin-eqn}} {{eqn | l = \ln x       | r = \int \frac {\d z} {\dfrac {z + 2 e^{-z} } 1 - z} + C_1       | c =  }} {{eqn | r = \int \frac {e^z} 2 \rd z + C_1       | c =  }} {{eqn | r = \frac {e^z} 2 + C_1       | c = Primitive of Exponential Function }} {{eqn | ll= \leadsto       | l = 2 \ln x + C       | r = e^z       | c = putting $C = -\dfrac {C_1} 2$ }} {{eqn | ll= \leadsto       | l = e^{y / x}       | r = \ln x^2 + C       | c =  }} {{end-eqn}} {{qed}}
We have: :$\dfrac {\mathrm d y} {\mathrm d x} = f \left({a x + b y + c}\right)$ Put: : $z := a x + b y + c$ Then: {{begin-eqn}} {{eqn | l = z       | r = a x + b y + c       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = a + b \dfrac {\mathrm d y} {\mathrm d x}       | c =  }} {{eqn | ll= \implies       | l = \dfrac {\mathrm d y} {\mathrm d x}       | r = \frac 1 b \left({\frac {\mathrm d z} {\mathrm d x} - a}\right)       | c =  }} {{eqn | ll= \implies       | l = \frac 1 b \left({\frac {\mathrm d z} {\mathrm d x} - a}\right)       | r = f \left({z}\right)       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = b f \left({z}\right) + a       | c =  }} {{end-eqn}} This can be solved by Separation of Variables: :$\displaystyle x = \int \frac {\mathrm d z} {b f \left({z}\right) + a}$ {{qed}}
Make the substitution: :$z = x + y$ Then from First Order ODE in form $y' = f (a x + b y + c)$ with $a = b = 1$: {{begin-eqn}} {{eqn | l = x       | r = \int \frac {\d z} {z^2 + 1}       | c =  }} {{eqn | r = \arctan z + C_1       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | ll= \leadsto       | l = y - C_1       | r = \map \arctan {x + y}       | c =  }} {{eqn | ll= \leadsto       | l = x + y       | r = \map \tan {x + C}       | c = letting $C = - C_1$ }} {{end-eqn}} {{qed}}
Make the substitution: :$z = x - y + 1$ Then from First Order ODE in form $y' = f (a x + b y + c)$ with $a = 1, b = - 1$: {{begin-eqn}} {{eqn | l = x       | r = \int \frac {\d z} {- \sin^2 z + 1}       | c =  }} {{eqn | r = \int \frac {\d z} {\cos^2 z}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \int \sec^2 z \rd z       | c = Secant is Reciprocal of Cosine }} {{eqn | r = \tan z + C_1       | c = Primitive of $\sec^2 z$ }} {{eqn | ll= \leadsto       | l = \map \tan {x - y + 1}       | r = x + C       | c = letting $C = - C_1$ }} {{end-eqn}} {{qed}}
We have: :$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$ Make the substitutions: :$x := z - h$ :$y := w - k$ We have: :$\dfrac {\d x} {\d z} = 1$ :$\dfrac {\d y} {\d w} = 1$ Thus: {{begin-eqn}} {{eqn | l = \frac {\d w} {\d z}       | r = \map F {\frac {a \paren {z - h} + b \paren {w - k} + c} {d \paren {z - h} + e \paren {w - k} + f} }       | c =  }} {{eqn | r = \map F {\frac {a z + b w - a h - b k + c} {d z + e w - d h - e k + f} }       | c =  }} {{end-eqn}} In order to simplify this appropriately, we wish to reduce it to the form: :$\dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$ by finding values of $h$ and $k$ such that: :$a h + b k = c$ :$d h + e k = f$ So: {{begin-eqn}} {{eqn | l = a h + b k       | r = c       | c =  }} {{eqn | l = d h + e k       | r = f       | c =  }} {{eqn | ll= \leadsto       | l = a d h + b d k       | r = c d       | c =  }} {{eqn | l = a d h + a e k       | r = a f       | c =  }} {{eqn | ll= \leadsto       | l = k \paren {a e - b d}       | r = a f - c d       | c =  }} {{eqn | ll= \leadsto       | l = k       | r = \frac {a f - c d} {a e - b d}       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = a h + b k       | r = c       | c =  }} {{eqn | l = d h + e k       | r = f       | c =  }} {{eqn | ll= \leadsto       | l = a e h + b e k       | r = c e       | c =  }} {{eqn | l = b d h + b e k       | r = b f       | c =  }} {{eqn | ll= \leadsto       | l = h \paren {a e - b d}       | r = c e - b f       | c =  }} {{eqn | ll= \leadsto       | l = h       | r = \frac {c e - b f} {a e - b d}       | c =  }} {{end-eqn}} We note that the above works {{iff}}: :$a e - b d \ne 0 \implies a e \ne b d$ Thus: :$(1): \quad \dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$ Letting: :$\map M {z, w} = a z + b w$ :$\map N {z, w} = d z + e w$ we see: :$\map M {t z, t w} = a t z + b t w = t \paren {a z + b w} = t \, \map M {z, w}$ :$\map N {t z, t w} = d t z + e t w = t \paren {d z + e w} = t \, \map N {z, w}$ Thus, by definition, $(1)$ is homogeneous. {{qed}}
We note that $(1)$ is in the form: :$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$ where: :$a e = -1 \ne b d = 1$ Hence we can use First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$. Let: :$x = z - h$ where $h = 2 / -2 = -1$ :$y = w - k$ where $k = -10 / -2 = 5$. Then: :$\dfrac {\d w} {\d z} = \dfrac {z + w} {z - w}$ From the solution to $\dfrac {\d y} {\d x} = \dfrac {x + y} {x - y}$: :$\arctan \dfrac w z = \ln \sqrt {z^2 + w^2} + C$ We have: :$x = z - \paren {-1}$ :$y = w - 5$ and so: :$z = x - 1$ :$w = y + 5$ This gives: :$\map \arctan {\dfrac {y + 5} {x - 1} } = \ln \sqrt {\paren {x - 1}^2 + \paren {y + 5}^2} + C$ {{qed}}
We note that $(1)$ is in the form: :$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$ where: :$a e = b d = 1$ Hence we can use First Order ODE in form $y' = \map F {\dfrac{a x + b y + c} {d x + e y + f} }$ where $a e = b d$. Let: :$z = x + y$ to obtain: :$\dfrac {\d z} {\d x} = b \, \map F {\dfrac {a z + a c} {d z + f} } + a$ where: :$a = 1$ :$c = 4$ :$d = 1$ :$f = -6$ which gives: {{begin-eqn}} {{eqn | l = \dfrac {\d z} {\d x}       | r = \dfrac {z + 4} {z - 6} + 1       | c =  }} {{eqn | r = \frac {z + 4 + z - 6} {z - 6}       | c =  }} {{eqn | r = \frac {2 \paren {z - 1} } {z - 6}       | c =  }} {{eqn | ll= \leadsto       | l = \int \rd x       | r = \int \frac {z - 6} {2 \paren {z - 1} } \rd z       | c =  }} {{end-eqn}} Substitute $v = z - 1$ which gives: :$\dfrac {\d z} {\d v} = 1$ and thence: :$v = x + y - 1$: {{begin-eqn}} {{eqn | l = \int \rd x       | r = \frac 1 2 \int \frac {v - 5} v \rd v       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \frac v 2 - \frac 5 2 \ln v + C_1       | c =  }} {{eqn | ll= \leadsto       | l = 2 x       | r = v - 5 \ln v + C_2       | c =  }} {{end-eqn}} Substituting $x + y - 1$ for $v$: {{begin-eqn}} {{eqn | l = 2 x       | r = x + y - 1 - 5 \, \map \ln {x + y - 1} + C_2       | c =  }} {{eqn | ll= \leadsto       | l = x - y       | r = -5 \, \map \ln {x + y - 1} + C_3       | c =  }} {{eqn | ll= \leadsto       | l = y - x       | r = 5 \, \map \ln {x + y - 1} + C       | c =  }} {{end-eqn}} {{qed}}
Let: :$\map M {x, y} = e^y$ :$\map N {x, y} = x e^y + 2 y$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y}       | r = e^y       | c =  }} {{eqn | l = \dfrac {\partial N} {\partial x}       | r = e^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 = \int \map M {x, y} \rd x + \map g y       | c =  }} {{eqn | r = \int e^y \rd x + \map g y       | c =  }} {{eqn | r = x e^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 {x e^y + 2 y} \rd y + \map h x       | c =  }} {{eqn | r = x e^y + y^2 + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = x e^y + y^2$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$x e^y + y^2 = C$ {{qed}}
Let: :$\map M {x, y} = y$ :$\map N {x, y} = x + \dfrac 2 y$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\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 y \rd x + \map g y       | c =  }} {{eqn | r = x y + \map g y       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = f       | r = \map N {x, y} \rd y + \map h x       | c =  }} {{eqn | r = \int \paren {x + \dfrac 2 y} \rd y + \map h x       | c =  }} {{eqn | r = x y + 2 \ln y + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = x y + 2 \ln y$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$x y + 2 \ln y = C$ {{qed}}
Let: :$\map M {x, y} = y - x^3$ :$\map N {x, y} = x + y^3$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\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 - x^3} \rd x + \map g y       | c =  }} {{eqn | r = x y - \dfrac {x^4} 4 + \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 \left({x + y^3}\right) \rd y + \map h x       | c =  }} {{eqn | r = x y + \dfrac {y^4} 4 + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = x y - \dfrac {x^4} 4 + \dfrac {y^4} 4$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$x y - \dfrac {x^4} 4 + \dfrac {y^4} 4 = C_1$ which can be simplified by multiplying through by $4$ and setting $C = 4 C_1$: :$4 x y - x^4 + y^4 = C$ {{qed}}
Let: :$\map M {x, y} = y + y \cos x y$ :$\map N {x, y} = x + x \cos x y$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y}       | r = 1 + \cos x y  - x \sin x y       | c =  }} {{eqn | l = \dfrac {\partial N} {\partial x}       | r = 1 + \cos x y  - x \sin 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 {y + y \cos x y} \rd x + \map g y       | c =  }} {{eqn | r = x y + \sin 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 {x + x \cos x y} \rd y + \map h x       | c =  }} {{eqn | r = x y + \sin x y + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = x y + \sin x y$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$x y + \sin x y = C$ {{qed}}
First express $(1)$ in the form: :$(2): \quad -\paren {e^y + \cos x \cos y} + \paren {\sin x \sin y - x e^y} \dfrac {\d y} {\d x}$ Let: :$\map M {x, y} = -\paren {e^y + \cos x \cos y}$ :$\map N {x, y} = \sin x \sin y - x e^y$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y}       | r = -e^y + \cos x \sin y       | c =  }} {{eqn | l = \dfrac {\partial N} {\partial x}       | r = -e^y + \cos x \sin 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 {e^y + \cos x \cos y} \rd x + \map g y       | c =  }} {{eqn | r = -x e^y - \sin x \cos 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 {\sin x \sin y - x e^y} \rd y + \map h x       | c =  }} {{eqn | r = -\sin x \cos y - x e^y + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = -\sin x \cos y - x e^y$ and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is: :$\sin x \cos y + x e^y = C$ {{qed}}