hcju/proofwikiqa huggingface dataset:
We have: {{begin-eqn}} {{eqn | l = \paren {x + y} \rd x       | r = \paren {x - y} \rd y       | c =  }} {{eqn | ll= \leadsto       | l = x \rd y - y \rd x       | r = x \rd x + y \rd y       | c = rearranging }} {{eqn | ll= \leadsto       | l = \frac {x \rd y - y \rd x} {x^2 + y^2}       | r = \frac {x \rd x + y \rd y} {x^2 + y^2}       | c = dividing through by $x^2 + y^2$ }} {{eqn | ll= \leadsto       | l = \frac {x \rd y - y \rd x} {x^2 + y^2}       | r = \frac {\map \d {x^2 + y^2} } {2 \paren {x^2 + y^2} }       | c = Differential of Sum of Squares }} {{eqn | ll= \leadsto       | l = \map \d {\arctan \dfrac y x}       | r = \frac {\map \d {x^2 + y^2} } {2 \paren {x^2 + y^2} }       | c = Differential of Arctangent of Quotient }} {{eqn | ll= \leadsto       | l = \arctan \dfrac y x       | r = \frac {\map \ln {x^2 + y^2} } 2 + C       | c = integrating }} {{eqn | ll= \leadsto       | l = \arctan \dfrac y x       | r = \ln \sqrt {x^2 + y^2} + C       | c =  }} {{end-eqn}} {{qed}}
Divide both sides of $(1)$ by $x^2 \rd x$ to get: :$\dfrac 1 x \dfrac {\d y} {\d x} = \dfrac 1 x \paren {\dfrac y x } + 1 + 9 \paren {\dfrac y x}^2$ Now apply the substitution: :$y = u x$ This implies then that: :$\dfrac {\d y} {\d x} = u + x \dfrac {\d u} {\d x}$ Now substitute everything into $(1)$ to get: {{begin-eqn}} {{eqn | l = \dfrac 1 x \paren {u + x \dfrac {\d u} {\d x} }       | r = \dfrac 1 x u + 1 + 9 u^2       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d u} {\d x}       | r = 1 + 9 u^2       | c =  }} {{end-eqn}} Now it becomes Separation of Variables and we end up with: {{begin-eqn}} {{eqn | l = \frac {\d u} {9 u^2 + 1}       | r = \d x       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 3 \tan^{-1} \paren {3 u}       | r = x + K       | c =  }} {{eqn | ll= \leadsto       | l = \tan^{-1} \paren {3 u}       | r = 3 x + C       | c =  }} {{end-eqn}} Substitute back for $u$: :$\tan^{-1} \paren {\dfrac {3 y} x} = 3 x + C$ {{qed}}
Let $z = \map \arctan {3y / x}$. Then: :$\dfrac {\partial z} {\partial x} = \dfrac 1 {1 + \paren {3 y / x}^2} \dfrac {-3 y} {x^2} = \dfrac {-3 y} {x^2 + 9 y^2}$ :$\dfrac {\partial z} {\partial y} = \dfrac 1 {1 + \paren {3 y / x}^2} \dfrac 3 x = \dfrac 3 {x^2 + 9 y^2}$ So: :$\d z = \dfrac {3 x \rd y - 3 y \rd x} {x^2 + 9 y^2}$ Multiplying $(1)$ by $3$ and manipulating: :$\dfrac {3 x \rd y - 3 y \rd x} {x^2 + 9 y^2} = 3 \rd x$ From Differential of Arctangent of Quotient: :$\map \d {\map \arctan {\dfrac {3 y} x} } = 3 \rd x$ and so :$\map \arctan {\dfrac {3 y} x} = 3 x + C$ {{qed}}
This is a special case of: :Linear First Order ODE: $\dfrac {\d y} {\d x} + \dfrac y x = k x^n$ where $k = 3$ and $n = 1$, yielding: :$y = x^2 + \dfrac C x$ Multiplying through by $x$ reveals: :$x y = x^3 + C$ {{qed}}
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} + \paren {-\dfrac 3 x} y = x^3$ $(2)$ is a linear first order ODE in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = -\dfrac 3 x$ :$\map Q x = x^3$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int -\frac 3 x \rd x       | c =  }} {{eqn | r = -3 \ln x       | c =  }} {{eqn | r = \ln x^{-3}       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = e^{\ln x^{-3} }       | c =  }} {{eqn | r = x^{-3}       | c =  }} {{eqn | r = \frac 1 {x^3}       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\map {\dfrac \d {\d x} } {\dfrac y {x^3} } = 1$ and the general solution is: :$\dfrac y {x^3} = x + C$ or: :$y = x^4 + \dfrac C {x^3}$ {{qed}}
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where $\map P x = 1$. 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 =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\map {\dfrac \d {\d x} } {e^x y} = \dfrac {e^x} {1 + e^{2 x} }$ and the general solution becomes: :$\displaystyle y {e^x} = \int \frac {e^x} {1 + e^{2 x} } \rd x$ The integral on the {{RHS}} can be solved by substituting: :$u = e^x \implies \d u = e^x \rd x$ and so: {{begin-eqn}} {{eqn | l = e^x y       | r = \int \frac {1 \rd u} {1 + u^2}       | c =  }} {{eqn | r = \map \arctan u + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \map \arctan {e^x} + C       | c =  }} {{end-eqn}} or: :$y = e^{-x} \map \arctan {e^x} + C e^{-x}$ {{qed}}
$(1)$ can be written as: :$(2): \quad \paren {1 + x^2} \dfrac {\rd y} {\rd x} + 2 x y = \cot x$ We notice straight away that: :$\dfrac {\rd} {\rd x} \paren {1 + x^2} = 2 x$ and so: :$\dfrac {\rd} {\rd x} \paren {1 + x^2} y = \cot x$ Thus the general solution becomes: {{begin-eqn}} {{eqn | l = \paren {1 + x^2} y       | r = \int \cot x \rd x       | c =  }} {{eqn | r = \map \ln {\sin x} + C       | c =  }} {{end-eqn}} or: :$y = \dfrac {\map \ln {\sin x} } {1 + x^2} + \dfrac C {1 + x^2}$ {{qed}}
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = 1$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int 1 \rd x       | c =  }} {{eqn | r = x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = e^x       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\dfrac \d {\d x} e^x y = 2 x + x^2 e^x$ and the general solution is: :$e^x y = x^2 + e^x \paren {x^2 - 2 x + 2} + C$ or: :$y = x^2 e^{-x} + x^2 - 2 x + 2 + C e^{-x}$ {{qed}}
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = \cot x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int \cot x \rd x       | c =  }} {{eqn | r = \map \ln {\sin x}       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = \sin x       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\dfrac {\d} {\d x} y \sin x = 2 x$ and the general solution is: :$y \sin x = x^2 + C$ or: :$y = x^2 \csc x + C \csc x$ {{qed}}
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = - x^2$ $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where $\map P x = -\dfrac 2 x$. Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int -\dfrac 2 x \rd x       | c =  }} {{eqn | r = -2 \ln x       | c =  }} {{eqn | r = \map \ln {\frac 1 {x^2} }       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = \frac 1 {x^2}       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\map {\dfrac \d {\d x} } {\dfrac y {x^2} } = -1$ and the general solution is: :$\dfrac y {x^2} = -x + C$ or: :$y = -x^3 + C x^2$ {{qed}}
Make the substitution: :$z = y^{1 - n}$ in $(1)$. Then we have: {{begin-eqn}} {{eqn | l = \frac {\d z} {\d y}       | r = \paren {1 - n} y^{-n}       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d y} \frac {\d y} {\d x} + \map P x y \paren {1 - n} y^{-n}       | r = \map Q x y^n \paren {1 - n} y^{-n}       | c = }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x} + \paren {1 - n} \map P x y^{1 - n}       | r = \paren {1 - n} \map Q x       | c = Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x} + \paren {1 - n} \map P x z       | r = \paren {1 - n} \map Q x       | c = }} {{end-eqn}} This is now a linear first order ordinary differential equation in $z$. It has an integrating factor: {{begin-eqn}} {{eqn | l = \map \mu x       | r = e^{\int \paren {1 - n} \map P x \rd x} }} {{eqn | r = e^{\paren {1 - n} \int \map P x \rd x}       | c = }} {{end-eqn}} and this can be used to obtain: :$\displaystyle \map \mu x z = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$ Substituting $z = y^{1 - n} = \dfrac 1 {y^{n - 1} }$ finishes the proof. {{qed}}
Let $(1)$ be rearranged as: :$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = x^3 y^3$ 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: :$\map P x = \dfrac 1 x$ :$\map Q x = x^3$ :$n = 3$ 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 - 3} \int \dfrac 1 x \rd x       | c =  }} {{eqn | r = -2 \ln x       | c =  }} {{eqn | r = \map \ln {\frac 1 {x^2} }       | c =  }} {{eqn | ll= \leadsto       | l = \map \mu x       | r = e^{\map \ln {\frac 1 {x^2} } }       | c =  }} {{eqn | r = \frac 1 {x^2}       | c =  }} {{end-eqn}} and so substituting into $(3)$: {{begin-eqn}} {{eqn | l = \frac 1 {x^2} \frac 1 {y^2}       | r = \paren {1 - 3} \int x^3 \frac 1 {x^2} \rd x       | c =  }} {{eqn | r = -2 \int x \rd x       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 {x^2 y^2}       | r = -2 \dfrac {x^2} 2 + C       | c =  }} {{end-eqn}} Hence the general solution to $(1)$ is: :$\dfrac 1 {y^2} = - x^4 + C x^2$ {{qed}}
Let $(1)$ be rearranged as: :$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = \dfrac {\cos x} {y^2}$ 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: :$\map P x = \dfrac 1 x$ :$\map Q x = \cos x$ :$n = -2$ 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 - \paren {-2} } \int \dfrac 1 x \rd x       | c =  }} {{eqn | r = 3 \ln x       | c =  }} {{eqn | r = \ln x^3       | c =  }} {{eqn | ll= \leadsto       | l = \map \mu x       | r = e^{\ln x^3}       | c =  }} {{eqn | r = x^3       | c =  }} {{end-eqn}} and so substituting into $(3)$: {{begin-eqn}} {{eqn | l = x^3 \frac 1 {y^{-3} }       | r = \paren {1 - \paren {-2} } \int x^3 \cos x \rd x       | c =  }} {{eqn | r = 3 \int x^3 \cos x \rd x       | c =  }} {{eqn | ll= \leadsto       | l = x^3 y^3       | r = 3 \paren {\paren {3 x^2 - 6} \cos x + \paren {x^3 - 6 x} \sin x} + C       | c = Primitive of $x^3 \cos a x$ }} {{end-eqn}} Hence the general solution to $(1)$ is: :$y^3 = 3 \sin x + \dfrac {9 \cos x} x - \dfrac {18 \sin x} {x^2} - \dfrac {18 \cos x} {x^3} + \dfrac C {x^3}$ {{qed}}
Let $(1)$ be rearranged as: :$\dfrac {\d y} {\d x} = \dfrac {y^2} {e^y - 2 x y}$ Hence: :$(2): \quad \dfrac {\d x} {\d y} + \dfrac 2 y x = \dfrac {e^y} {y^2}$ It can be seen that $(2)$ is a linear first order ODE in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y$ where: :$\map P y = \dfrac 2 y$ :$\map Q y = \dfrac {e^y} {y^2}$ Thus: {{begin-eqn}} {{eqn | l = \int \map P y \rd y       | r = \int \dfrac 2 y \rd y       | c =  }} {{eqn | r = 2 \ln y       | c =  }} {{eqn | r = \ln y^2       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd y}       | r = y^2       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(2)$ can be rewritten as: :$\map {\dfrac \d {\d y} } {x y^2} = e^y$ and the general solution is: :$x y^2 = e^y + C$ {{qed}}
Let $(1)$ be rearranged as: :$\dfrac {\d y} {\d x} = \dfrac y {y^2 e^y + x}$ Hence: :$(2): \quad \dfrac {\d x} {\d y} - \dfrac 1 y x = y e^y$ It can be seen that $(2)$ is a linear first order ODE in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y$ where: :$\map P y = -\dfrac 1 y$ :$\map Q y = y e^y$ Thus: {{begin-eqn}} {{eqn | l = \int \map P y \rd y       | r = \int -\dfrac 1 y \rd y       | c =  }} {{eqn | r = -\ln y       | c =  }} {{eqn | r = \ln y^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd y}       | r = \dfrac 1 y       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(2)$ can be rewritten as: :$\map {\dfrac {\d} {\d y} } {\dfrac x y} = e^y$ and the general solution is: :$\dfrac x y = e^y + C$ or: :$x = y e^y + C y$ {{qed}}
Define: :$\BB := \ds \bigcup_{x \mathop \in S} \NN_x$ According to Topology Defined by Basis it should be proved that :$(\text B 1): \quad \forall A_1, A_2 \in \BB: \forall x \in A_1 \cap A_2: \exists A \in \BB: x \in A \subseteq A_1 \cap A_2$ :$(\text B 2): \quad \forall x \in S: \exists A \in \BB: x \in A$ Ad $(\text B 1)$: Let $A_1, A_2 \in \BB$. Let $x \in A_1 \cap A_2$. By definition of intersection: :$x \in A_1 \land x \in A_2$ By definition of union: :$\exists x_1 \in S: A_1 \in \NN_{x_1}$ and :$\exists x_2 \in S: A_2 \in \NN_{x_2}$ By $(\text N 2)$: :$\exists V_1 \in \NN_x: V_1 \subseteq A_1$ and :$\exists V_2 \in \NN_x: V_2 \subseteq A_2$ By $(\text N 3)$: :$\exists U \in \NN_x: U \subseteq V_1 \cap V_2$ By definition of union: :$U \in \BB$ By Set Intersection Preserves Subsets: :$V_1 \cap V_2 \subseteq A_1 \cap A_2$ Thus by Subset Relation is Transitive: :$U \subseteq A_1 \cap A_2$ Ad $(\text B 2)$: Let $x \in S$. By definition of empty set: :$\exists U: U \in \NN_x$ By $(\text N 1)$: :$x \in U$ By definition of union: :$U \in \BB$ Thus: :$\exists A \in \BB: x \in A$ By definition of $\tau$ and $\BB$: :$\tau = \set {\bigcup \GG: \GG \subseteq \BB}$ Thus by Topology Defined by Basis: :$T = \struct {S, \tau}$ is a topological space It remains to prove that :$\family {\NN_x}_{x \mathop \in S}$ is a neighborhood system Let $x \in S$. Let $U \in \tau$. By definition of $\tau$: :$\exists \GG \subseteq \BB: U = \bigcup \GG$ Let $x \in U$. By definition of union: :$\exists V \in \GG: x \in V$ By definition of subset: :$V \in \BB$ By definition of union: :$\exists y \in S: V \in \NN_y$ By $(\text N 2)$: :$\exists W \in \NN_x: W \subseteq V$ By Set is Subset of Union/Set of Sets: :$V \subseteq U$ Thus by Subset Relation is Transitive: :$\exists W \in \NN_x: W \subseteq U$ Thus by definition: :$\NN_x$ is a local basis. {{qed}}
We use the technique of formation of ordinary differential equation by elimination. Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives: :$\dfrac {\d y} {\d x} = 1 - C e^{-x}$ Eliminating $C$: {{begin-eqn}} {{eqn | l = C       | o = =       | r = \frac {y - x} {e^{-x} }       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = 1 - \frac {y - x} {e^{-x} } e^{-x}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x} + y       | r = x + 1       | c =  }} {{end-eqn}} Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by: :$-\dfrac {\d y} {\d x} = - x = 1 - y$ The integrating factor is $e^y$, giving: :$\displaystyle e^y x = \int y e^y - e^y \rd y$ Using Primitive of $x e^{a x}$: :$\displaystyle \int y e^y \rd y = y e^y - e^y$ Thus we get: :$e^y x = y e^y - e^y  - e^y + C$ which gives us: :$x = y - 2 + C e^{-y}$ {{qed}}
Differentiating $(1)$ {{WRT}} $x$ gives: :$(2): \quad \dfrac {\d y} {\d x} = C \map {f'} x + \map {g'} x$ Rearranging $(1)$, we have: :$C = \dfrac {y - \map g x} {\map f x}$ Substituting for $C$ in $(2)$: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x}       | r = \dfrac {y - \map g x} {\map f x} \map {f'} x + \map {g'} x       | c =  }} {{eqn | r = \dfrac {\map {f'} x} {\map f x} y - \dfrac {\map g x \map {f'} x} {\map f x} + \map g x       | c =  }} {{end-eqn}} which leaves: :$\dfrac {\d y} {\d x} - \dfrac {\map {f'} x} {\map f x} y = \map g x \paren {1 - \dfrac {\map {f'} x} {\map f x} }$ which is linear and of first order. {{qed}}
Consider the second order ordinary differential equation: :$(1): \quad \map f {x, y', y''} = 0$ Let a new dependent variable $p$ be introduced: :$y' = p$ :$y'' = \dfrac {\d p} {\d x}$ Then $(1)$ can be transformed into: :$(2): \quad \map f {x, p, \dfrac {\d p} {\d x} } = 0$ which is a first order ODE. If $(2)$ has a solution which can readily be found, it will be expressible in the form: :$(3): \quad \map g {x, p}$ which can then be expressed in the form: :$\map g {x, \dfrac {\d y} {\d x} } = 0$ which is likewise subject to the techniques of solution of a first order ODE. Hence such a second order ODE is reduced to the problem of solving two first order ODEs in succession. {{qed}}
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$ in $(1)$: :$x \dfrac {\d p} {\d x} - p = 3 x^2$ and divide through by $x$: :$\dfrac {\d p} {\d x} - \dfrac p x = 3 x$ From: :Linear First Order ODE: $y' - \dfrac y x = 3 x$ its solution is: :$p = 3 x^2 + C_1 x$ Substituting back for $p$: :$\dfrac {\d y} {\d x} = 3 x^2 + C_1 x$ which is separable, leading to: :$y = x^3 + \dfrac {C_1 x^2} 2 + C^2$ {{qed}}
Consider the second order ordinary differential equation: :$(1): \quad \map g {y, \dfrac {\d y} {\d x}, \dfrac {\d^2 y} {\d x^2} } = 0$ Let a new dependent variable $p$ be introduced: :$y' = p$ Hence: :$y'' = \dfrac {\d p} {\d x} = \dfrac {\d p} {\d y} \dfrac {\d y} {\d x} = p \dfrac {\d p} {\d y}$ Then $(1)$ can be transformed into: :$(2): \quad \map g {y, p, p \dfrac {\d p} {\d y} = 0}$ which is a first order ODE. If $(2)$ has a solution which can readily be found, it will be expressible in the form: :$(3): \quad \map g {x, p}$ which can then be expressed in the form: :$\map g {x, \dfrac {\d y} {\d x} }$ which is likewise subject to the techniques of solution of a first order ODE. Hence such a second order ODE is reduced to the problem of solving two first order ODEs in succession. {{qed}}
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} + p^2       | r = 0       | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto       | l = y \rd p + p \rd y       | r = 0       | c = multiplying by $\dfrac {\d y} p$ }} {{eqn | ll= \leadsto       | l = p y       | r = C       | c = First Order ODE: $y \rd x + x \rd y = 0$ }} {{eqn | ll= \leadsto       | l = y \dfrac {\d y} {\d x}       | r = C       | c = substituting $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto       | l = y^2       | r = 2 C x + C_2       | c = First Order ODE: $y \dfrac {\d y} {\d x} = k$ }} {{eqn | r = C_1 x + C_2       | c = reassigning constant }} {{end-eqn}} {{qed}}
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$ in $(1)$: {{begin-eqn}} {{eqn | l = x \dfrac {\d p} {\d x}       | r = p + p^3       | c =  }} {{eqn | ll= \leadsto       | l = p = \frac {\d y} {\d x}       | r = \frac x {\sqrt {C_1^2 - x^2} }       | c = First Order ODE: $x \rd y = \paren {y + y^3} \rd x$ }} {{eqn | ll= \leadsto       | l = \int \rd y       | r = \int \frac x {\sqrt {C_1^2 - x^2} }       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = y       | r = -\sqrt {C_1^2 - x^2} + C_2       | c = Primitive of $\dfrac x {\sqrt{a^2 - x^2} }$ }} {{eqn | ll= \leadsto       | l = x^2 + \paren {y - C_2}^2       | r = C_1^2       | c = rearranging }} {{end-eqn}} {{qed}}
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$: {{begin-eqn}} {{eqn | l = x^2 \dfrac {\d p} {\d x}        | r = 2 x p + p^2       | c =  }} {{eqn | ll= \leadsto       | l = p = \frac {\d y}{\d x}       | r = -\frac {x^2} {x - C_1}       | c = Bernoulli's Equation: $x^2 \rd y = \paren {2 x y + y^2} \rd x$ }} {{eqn | ll= \leadsto       | l = \int \rd y       | r = -\int \frac {x^2} {x - C_1} \rd x       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = -\frac {x^2} 2 - C_1 x - {C_1}^2 \, \map \ln {x - C_1} + C_2       | c = Primitive of $\dfrac {x^2} {a x + b}$ }} {{end-eqn}} {{qed}}
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$ in $(1)$: {{begin-eqn}} {{eqn | l = \paren {x^2 + 2 p} \dfrac {\d p} {\d x} + 2 x p       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = 2 x p \rd x + \paren {x^2 + 2 p} \rd p        | r = 0       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = p \paren {x^2 + p}       | r = C_1        | c = Bernoulli's Equation: $2 x y \rd x + \paren {x^2 + 2 y} \rd y = 0$ }} {{end-eqn}} Consider the initial condition: :$y' = p = 0$ when $x = 0$ Hence putting $p = x = 0$ in $(2)$ we get: :$0 \cdot 0^2 + 0^2 = C_1$ :$C_1 = 0$ and so $(2)$ becomes: {{begin-eqn}} {{eqn | l = p x^2       | r = -p^2       | c =  }} {{eqn | ll= \leadsto       | l = p \paren {x^2 - p}       | r = 0       | c =  }} {{end-eqn}} There are two possibilities here: {{begin-eqn}} {{eqn | l = p       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d y} {\d x}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = C_2       | c =  }} {{end-eqn}} From our initial condition: :$y = 1$ when $x = 0$ gives us: :$C_2 = 1$ and so the solution is obtained: :$y = 1$ {{qed|lemma}} The other option is: {{begin-eqn}} {{eqn | l = p = \dfrac {\d y} {\d x}       | r = -x^2        | c =  }} {{eqn | ll= \leadsto       | l = y       | r = -\int x^2 \rd x       | c =  }} {{eqn | n = 3       | r = -\frac {x^3} 3 + C_2       | c =  }} {{end-eqn}} From our initial condition: :$y = 1$ when $x = 0$ Hence putting $x = 0$ and $y = 1$ in $(3)$ we get: :$1 = - \dfrac {0^3} 3 = C_2$ and so $C_2 = 1$. Thus we have: :$y + \dfrac {x^3} 3 = 1$ or: :$3 y + x^3 = 3$ Hence the result. {{qed}}
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y}       | r = y^2 p + p^2       | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto       | l = \frac {\d p} {\d y} - \frac p y       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = p = \dfrac {\d y} {\d x}       | r = y \paren {y + C_1}       | c = Linear First Order ODE: $y' - \dfrac y x = k x$ }} {{eqn | ll= \leadsto       | l = \int \frac {\d y} {y \paren {y + C_1} }       | r = \int \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \frac 1 {C_1} \map \ln {\frac y {y + C_1} }       | r = x + C_2       | c = Primitive of Reciprocal of $\dfrac x {\paren {a x + b} }$ }} {{end-eqn}} Now to consider the initial conditions: :$y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$ After algebra: {{begin-eqn}} {{eqn | l = \map \ln {\frac y {y + C_1} }       | r = C_1 x + C_2       | c = reassigning $C_2$ }} {{eqn | ll= \leadsto       | l = \frac y {y + C_1}       | r = e^{C_1 x + C_2}       | c =  }} {{eqn | r = e^{C_2} e^{C_1 x}       | c =  }} {{end-eqn}} When $x = 0$ we have $y = -1/2$: {{begin-eqn}} {{eqn | l = \frac {-1/2} {-1/2 + C_1}       | r = \frac 1 {1 - 2 C_1}       | c =  }} {{eqn | r = e^{C_2}       | c =  }} {{eqn | ll= \leadsto       | l = \frac y {y + C_1}       | r = \frac {e^{C_1 x} } {1 - 2 C_1}       | c =  }} {{eqn | ll= \leadsto       | l = y \paren {1 - 2 C_1}       | r = \paren {y + C_1} e^{C_1 x}       | c =  }} {{end-eqn}} Differentiating to get $y'$: :$y' \paren {1 - 2 C_1} = \paren {y + C_1} C_1 e^{C_1 x} + e^{C_1 x} y'$ Putting $y' = 1$ when $x = 0$ we get: {{begin-eqn}} {{eqn | l = 1 - 2 C_1       | r = \paren {-\frac 1 2 + C_1} C_1 + 1       | c =  }} {{eqn | ll= \leadsto       | l = C_1       | r = -\frac 3 2       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = y \paren {1 - 2 C_1}       | r = \paren {y + C_1} e^{C_1 x}       | c =  }} {{eqn | ll= \leadsto       | l = y \paren {1 - 2 \frac {-3} 2}       | r = \paren {y - \frac 3 2} e^{\frac {-3 x} 2}       | c =  }} {{eqn | ll= \leadsto       | l = 8 y       | r = \paren {2 y - 3} e^{\frac {-3 x} 2}       | c =  }} {{eqn | ll= \leadsto       | l = 2 y - 3       | r = 8 y \, \map \exp {\dfrac {3 x} 2}       | c =  }} {{end-eqn}} {{qed}}