avewright/proofwiki-math huggingface dataset:
Theorem: First Order ODE/y dx + x dy + 3 x^3 y^4 dy

The first order ODE:
:$(1): \quad y \rd x + x \rd y + 3 x^3 y^4 \rd y = 0$
has the general solution:
:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$

Proof:
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}}
Theorem: Curved Mirror producing Parallel Rays is Paraboloid

Let $M$ be a curved mirror embedded in a real cartesian $3$- space.
Let there be a source of light at the origin.
Let $M$ reflect the light in a beam parallel to the $x$-axis.
Then $M$ is the solid of revolution produced by rotating about the $x$-axis the parabola whose equation is:
:$y^2 = 2 c x + c^2$

Proof:
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 $P = \tuple {x, y}$ be an arbitrary point on $APB$.
:500px
From the Law of Reflection:
:$\alpha = \beta$
From Parallelism implies Equal Corresponding Angles:
:$\phi = \beta$
From External Angle of Triangle equals Sum of other Internal Angles
:$\theta = \alpha + \phi = 2 \beta$
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= \leadsto
      | l = \frac y x
      | r = \dfrac {2 \paren {\dfrac {\d y} {\d x} } } {1 - \paren {\dfrac {\d y} {\d x} }^2}
      | c = $\dfrac {\d y} {\d x} = \tan \beta$ and $\tan \theta = \dfrac y x$
}}
{{eqn | ll= \leadsto
      | l = 0
      | r = y \paren {\dfrac {\d y} {\d x} }^2 + 2 x \paren {\frac {\d y} {\d x} } - y
}}
{{end-eqn}}
Using the Quadratic Formula:
:$\dfrac {\d y} {\d x} = \dfrac {-x \pm \sqrt {x^2 + y^2} } y$
which can be expressed as:
:$x \rd x + y \rd y = \pm \sqrt {x^2 + y^2} \rd x$
Using Differential of Sum of Squares:
:$\pm \dfrac {\map \d {x^2 + y^2} } {2 \sqrt {x^2 + y^2} } = \d x$
The {{LHS}} can be integrated by defining $u = x^2 + y^2$:
{{begin-eqn}}
{{eqn | l = \pm \dfrac {\map \d {x^2 + y^2} } {2 \sqrt {x^2 + y^2} }
      | r = \pm \dfrac {\d u} {2 \sqrt u }
}}
{{eqn | ll= \leadsto
      | l = \pm \int \dfrac {\d u} {2 \sqrt u }
      | r = \pm \sqrt u + c_1
      | c = Primitive of Power
}}
{{eqn | r = \pm \sqrt {x^2 + y^2} + c_1
}}
{{end-eqn}}
The {{RHS}} integrates to:
:$\ds \int \d x = x + c_2$
Hence:
{{begin-eqn}}
{{eqn | l = \pm \sqrt {x^2 + y^2} + c_1
      | r = x + c_2
}}
{{eqn | l = \pm \sqrt {x^2 + y^2}
      | r = x + c
      | c = $c = c_2 - c_1$
}}
{{eqn | ll= \leadsto
      | l = x^2 + y^2
      | r = x^2 + 2 c x + c^2
      | c = squaring both sides
}}
{{eqn | ll= \leadsto
      | l = y^2
      | r = 2 c x + c^2
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/x dy - y dx = (1 + y^2) dy

The first order ODE:
:$(1): \quad x \rd y - y \rd x = \paren {1 + y^2} \rd y$
has the general solution:
:$\dfrac x y = \dfrac 1 y - y + C$

Proof:
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}}
Theorem: Differential of Sum of Squares

:$\map \d {x^2 + y^2} = 2 \paren {x \map \rd x + y \map \rd y}$

Proof:
{{begin-eqn}}
{{eqn | l = \map \rd {x^2 + y^2; \vec h}
      | r = 2 x \vec h_1 + 2 y \vec h_2
      | c = {{Defof|Differential of Mapping|subdef = Real-Valued Function}}
}}
{{eqn | r = 2 x \map \rd {x; \vec h} + 2 y \map \rd {y; \vec h}
}}
{{eqn | r = 2 \paren {x \map \rd {x; \vec h} + y \map \rd {y; \vec h} }
}}
{{end-eqn}}
{{explain|what are $h$, $h_1$ and $h_2$? The notation used does not match that of the statement of the result}}
Theorem: First Order ODE/y dx + (x^2 y - x) dy = 0/Proof 1

{{:First Order ODE/y dx + (x^2 y - x) dy = 0}}
This can also be presented in the form:
:$\dfrac {\d y} {\d x} + \dfrac y {x^2 y - x}$

Proof:
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}}
Theorem: First Order ODE/y dx - x dy = x y^3 dy

The first order ODE:
:$(1): \quad y \rd x - x \rd y = x y^3 \rd y$
has the general solution:
:$\ln \dfrac x y = \dfrac {y^3} 3 + C$

Proof:
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}}
Theorem: First Order ODE/x dy = (x^5 + x^3 y^2 + y) dx

The first order ODE:
:$(1): \quad x \rd y = \paren {x^5 + x^3 y^2 + y} \rd x$
has the general solution:
:$\arctan \dfrac x y = -\dfrac {x^4} 4 + C$

Proof:
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}}
Theorem: First Order ODE/(x + y) dx = (x - y) dy/Proof 1

The first order ordinary differential equation:
:$(1): \quad \paren {x + y} \rd x = \paren {x - y} \rd y$
is a homogeneous differential equation with solution:
:$\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$

Proof:
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:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
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}}
Theorem: First Order ODE/x dy = (y + x^2 + 9 y^2) dx

The first order ODE:
:$(1): \quad x \rd y = \paren {y + x^2 + 9 y^2} \rd x$
has the general solution:
:$\map \arctan {\dfrac {3 y} x} = 3 x + C$

Proof:
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 Solution to Separable Differential Equation 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}}
Theorem: First Order ODE/x dy = (y + x^2 + 9 y^2) dx

The first order ODE:
:$(1): \quad x \rd y = \paren {y + x^2 + 9 y^2} \rd x$
has the general solution:
:$\map \arctan {\dfrac {3 y} x} = 3 x + C$

Proof:
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}}
Theorem: Linear First Order ODE/y' + (y over x) = 3 x

The linear first order ODE:
:$\dfrac {\d y} {\d x} + \dfrac y x = 3 x$
has the general solution:
:$x y = x^3 + C$
or:
:$y = x^2 + \dfrac C x$

Proof:
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}}
Theorem: Linear First Order ODE/x y' - 3 y = x^4

The linear first order ODE:
:$(1): \quad x \dfrac {\d y} {\d x} - 3y = x^4$
has the general solution:
:$y = x^4 + \dfrac C {x^3}$

Proof:
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}}
Theorem: Linear First Order ODE/y' + y = 1 over (1 + exp 2 x)

The linear first order ODE:
:$(1): \quad y' + y = \dfrac 1 {1 + e^{2 x} }$
has the general solution:
:$y = e^{-x} \map \arctan {e^x} + C e^{-x}$

Proof:
$(1)$ 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:
:$\ds y {e^x} = \int \frac {e^x} {1 + e^{2 x} } \rd x$
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}}
Theorem: Linear First Order ODE/(1 + x^2) dy + 2 x y dx = cotangent x dx

The linear first order ODE:
:$(1): \quad \paren {1 + x^2} \rd y + 2 x y \rd x = \cot x \rd x$
has the general solution:
:$y = \dfrac {\map \ln {\sin x} } {1 + x^2} + \dfrac C {1 + x^2}$

Proof:
$(1)$ 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}}
Theorem: Linear First Order ODE/y' + y = 2 x exp -x + x^2

The linear first order ODE:
:$(1): \quad y' + y = 2 x e^{-x} + x^2$
has the general solution:
:$y = x^2 e^{-x} + x^2 - 2 x + 2 + C e^{-x}$

Proof:
$(1)$ 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}}
Theorem: Linear First Order ODE/y' + y cot x = 2 x cosec x

The linear first order ODE:
:$(1): \quad y' + y \cot x = 2 x \csc x$
has the general solution:
:$y = x^2 \csc x + C \csc x$

Proof:
$(1)$ 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}}
Theorem: Linear First Order ODE/(2 y - x^3) dx = x dy

The linear first order ODE:
:$(1): \quad \paren {2 y - x^3} \rd x = x \rd y$
has the general solution:
:$y = -x^3 + C x^2$

Proof:
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}}
Theorem: Solution to Bernoulli's Equation

'''Bernoulli's equation''':
:$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
where:
:$n \ne 0, n \ne 1$
has the general solution:
:$\ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
where:
:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$

Proof:
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:
:$\ds \map \mu x z = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
Substituting $z = y^{1 - n} = \dfrac 1 {y^{n - 1} }$ finishes the proof.
{{qed}}
Theorem: Bernoulli's Equation/x y' + y = x^4 y^3

The first order ODE:
:$(1): \quad x y' + y = x^4 y^3$
has the general solution:
:$\dfrac 1 {y^2} = - x^4 + C x^2$

Proof:
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:
:$(3): \quad \ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
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}}
Theorem: Bernoulli's Equation/x y^2 y' + y^3 = x cosine x

The first order ODE:
:$(1): \quad x y^2 y' + y^3 = x \cos x$
has the general solution:
:$y^3 = 3 \sin x + \dfrac {9 \cos x} x - \dfrac {18 \sin x} {x^2} - \dfrac {18 \cos x} {x^3} + \dfrac C {x^3}$

Proof:
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:
:$\ds (3): \quad \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \, \map \mu x \rd x + C$
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}}
Theorem: First Order ODE/(exp y - 2 x y) y' = y^2

The first order ODE:
:$(1): \quad \paren {e^y - 2 x y} y' = y^2$
has the general solution:
:$x y^2 = e^y + C$

Proof:
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}}
Theorem: First Order ODE/y - x y' = y' y^2 exp y

The first order ODE:
:$(1): \quad y - x y' = y' y^2 e^y$
has the general solution:
:$x y^2 = e^y + C$

Proof:
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}}