avewright/proofwiki-math huggingface dataset:
Theorem: First Order ODE/(y^2 exp x y + cosine x) dx + (exp x y + x y exp x y) dy = 0

The first order ordinary differential equation:
:$(1): \quad \paren {y^2 e^{x y} + \cos x} \rd x + \paren {e^{x y} + x y e^{x y} } \rd y = 0$
is an exact differential equation with solution:
:$y e^{x y} + \sin x = C$

Proof:
Let:
:$\map M {x, y} = y^2 e^{x y} + \cos x$
:$\map N {x, y} = e^{x y} + x y e^{x y}$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial M} {\partial y}
      | r = 2 y e^{x y} + x y^2 e^{x y}
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = y e^{x y} + x y^2 e^{x y} + y e^{x y}
      | c = 
}}
{{eqn | r = 2 y e^{x y} + x y^2 e^{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^2 e^{x y} + \cos x} \rd x + \map g y
      | c = 
}}
{{eqn | r = y e^{x 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 {e^{x y} + x y e^{x y} } \rd y + \map h x
      | c = 
}}
{{eqn | r = \frac {e^{x y} } x + y e^{x y} - \frac {e^{x y} } x + \map h x
      | c = Primitive of $x e^{a x}$
}}
{{eqn | r = y e^{x y} + \map h x
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = y e^{x y} + \sin x$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$y e^{x y} + \sin x = C$
{{qed}}
Theorem: First Order ODE/y' ln (x - y) = 1 + ln (x - y)

The first order ordinary differential equation:
:$(1): \quad \dfrac {\d y} {\d x} \map \ln {x - y} \d x = 1 + \map \ln {x - y}$
is an exact differential equation with solution:
:$\paren {x - y} \map \ln {x - y} = C - y$

Proof:
Let $(1)$ be expressed as:
:$\paren {1 + \map \ln {x - y} } \rd x - \map \ln {x - y} \rd y = 0$
Let:
:$\map M {x, y} = 1 + \map \ln {x - y}$
:$\map N {x, y} = -\map \ln {x - y}$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial M} {\partial y}
      | r = -\frac 1 {x - y}
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = -\frac 1 {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 \left({1 + \map \ln {x - y} }\right) \rd x + \map g y
      | c = 
}}
{{eqn | r = x + \paren {x - y} \map \ln {x - y} - \paren {x - y} + \map g y
      | c = Primitive of $\ln x$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = f
      | r = \int \map N {x, y} \rd y + \map h x
      | c = 
}}
{{eqn | r = \int \left({-\map \ln {x - y} }\right) \rd y + \map h x
      | c = 
}}
{{eqn | r = \paren {x - y} \map \ln {x - y} - \paren {x - y} + \map h x
      | c = Primitive of $\ln x$
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \map f {x, y}
      | r = x + \paren {x - y} \map \ln {x - y} - \paren {x - y}
      | c = 
}}
{{eqn | r = \paren {x - y} \map \ln {x - y} + y
      | c = 
}}
{{end-eqn}}
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
{{begin-eqn}}
{{eqn | l = \paren {x - y} \map \ln {x - y} + y
      | r = C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x - y} \map \ln {x - y}
      | r = C - y
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Linear First Order ODE/y' + 2 x y = exp -x^2

The linear first order ODE:
:$\dfrac {\d y} {\d x} + 2 x y = \map \exp {-x^2}$
has the general solution:
:$y = \paren {x + C} \map \exp {-x^2}$

Proof:
$(1)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = 2 x$
:$\map Q x = \map \exp {-x^2}$
Thus:
{{begin-eqn}}
{{eqn | l = \int \map P x \rd x
      | r = \int 2 x \rd x
      | c = 
}}
{{eqn | r = x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{\int P \rd x}
      | r = \map \exp {x^2}
      | c = 
}}
{{end-eqn}}
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac {\d} {\d x} } {\map \exp {x^2} y} = \map \exp {-x^2} \map \exp {x^2} = 1$
Hence:
{{begin-eqn}}
{{eqn | l = \map \exp {x^2} y
      | r = \int 1 \rd x
      | c = 
}}
{{eqn | r = x + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \paren {x + C} \map \exp {-x^2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/(y^2 - 3 x y - 2 x^2) dx = (x^2 - x y) dy

The first order ordinary differential equation:
:$(1): \quad \paren {y^2 - 3 x y - 2 x^2} \rd x = \paren {x^2 - x y} \rd y$
is a homogeneous differential equation with solution:
:$y^2 x^2 - 2 y x^3 + x^4 = C$

Proof:
$(1)$ can also be rendered:
:$\dfrac {\rd y} {\rd x} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$
Let:
:$\map M {x, y} = y^2 - 3 x y - 2 x^2$
:$\map N {x, y} = x^2 - x y$
Put $t x, t y$ for $x, y$:
{{begin-eqn}}
{{eqn | l = \map M {t x, t y}
      | r = \paren {t y}^2 - 3 t x t y - 2 \paren {t x}^2
      | c = 
}}
{{eqn | r = t^2 \paren {y^2 - 3 x y - 2 x^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 - t x t y
      | c = 
}}
{{eqn | r = t^2 \paren {x^2 - x y}
      | 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:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$
Hence:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {\frac {z^2 - 3 z - 2} {1 - z} - z} + C_1
      | c = 
}}
{{eqn | r = \int \frac {1 - z} {2 z^2 - 4 z - 2} \rd z + C_1
      | c = 
}}
{{eqn | r = -\frac 1 4 \int \frac {4 z - 4} {2 z^2 - 4 z - 2} \rd z + C_1
      | c = 
}}
{{eqn | r = -\frac 1 4 \map \ln {2 z^2 - 4 z - 2} + C_1
      | c = Primitive of Function under its Derivative
}}
{{end-eqn}}
Substituting back for $z$ and tidying up, the result is obtained:
:$y^2 x^2 - 2 y x^3 + x^4 = C$
{{qed}}
Theorem: Linear First Order ODE/(1 + x^2) y' + 2 x y = 4 x^3

The linear first order ODE:
:$\paren {1 + x^2} \dfrac {\d y} {\d x} + 2 x y = 4 x^3$
has the general solution:
:$y = \dfrac {x^4} {1 + x^2} + \dfrac C {1 + x^2}$

Proof:
It is noticed that:
:$\dfrac {\d} {\d x} \paren {1 + x^2} = 2 x$
and so $(1)$ can be rendered in the form:
{{begin-eqn}}
{{eqn | l = \map \d {\paren {1 + x^2} y}
      | r = 4 x^3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {1 + x^2} y
      | r = \int 4 x^3 \rd x
      | c = 
}}
{{eqn | r = x^4 + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \frac {x^4} {1 + x^2} + \frac C {1 + x^2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/exp x sine y dx + exp x cos y dy = y sine x y dx + x sine x y dy

The first order ordinary differential equation:
:$(1): \quad e^x \sin y \rd x + e^x \cos y \rd y = y \sin x y \rd x + x \sin x y \rd y$
is an exact differential equation with solution:
:$e^x \sin y + \cos x y = C$

Proof:
Let $(1)$ be expressed as:
:$\paren {e^x \sin y - y \sin x y} \rd x + \paren {e^x \cos y - x \sin x y} \rd y = 0$
Let:
:$\map M {x, y} = e^x \sin y - y \sin x y$
:$\map N {x, y} = e^x \cos y - x \sin x y$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial M} {\partial y}
      | r = e^x \cos y - x y \cos x y - y \sin x y
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = e^x \cos y - x y \cos x y - y \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 {e^x \sin y - y \sin x y} \rd x + \map g y
      | c = 
}}
{{eqn | r = e^x \sin y + \cos 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 {e^x \cos y - x \sin x y} \rd y + \map h x
      | c = 
}}
{{eqn | r = e^x \sin y + \cos x y + \map h x
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = e^x \sin y + \cos x y$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$e^x \sin y + \cos x y = C$
{{qed}}
Theorem: First Order ODE/(1 + x^2) y' + x y = 0

The first order ODE:
:$\paren {1 + x^2} y' + x y = 0$
has the general solution:
:$y = \dfrac C {\sqrt {1 + x^2} }$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {1 + x^2} \dfrac {\d y} {\d x}
      | r = -x y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d y} y
      | r = -\int \frac {x \rd x} {1 + x^2}
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = \ln y
      | r = -\frac 1 2 \, \map \ln {1 + x^2} + C
      | c = Primitive of $\dfrac x {x^2 + a^2}$
}}
{{eqn | ll= \leadsto
      | l = \ln y
      | r = \ln \frac {C_1} {\sqrt {1 + x^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \frac C {\sqrt {1 + x^2} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Examples of Linear First Order ODEs
Category:Examples of Solutions to Separable Differential Equation
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Theorem: Linear Second Order ODE/(1 + x^2) y'' + x y' = 0

The second order ODE:
:$\paren {1 + x^2} y' ' + x y' = 0$
has the general solution:
:$y = C_1 \map \ln {x + \sqrt {x^2 + 1} } + C_2$

Proof:
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$:
{{begin-eqn}}
{{eqn | l = \paren {1 + x^2} \dfrac {\d p} {\d x} + x p
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = p = \dfrac {\d y} {\d x}
      | r = \frac {C_1} {\sqrt {1 + x^2} }
      | c = First Order ODE: $\paren {1 + x^2} y' + x y = 0$
}}
{{eqn | ll= \leadsto
      | l = y
      | r = C_1 \int \frac 1 {\sqrt {1 + x^2} } \rd x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = C_1 \map \ln {x + \sqrt {x^2 + 1} } + C_2
      | c = Primitive of $\dfrac 1 {\sqrt{x^2 + a^2} }$
}}
{{end-eqn}}
{{qed}}
Theorem: First Order ODE/(x exp y + y - x^2) dy = (2 x y - exp y - x) dx

The first order ordinary differential equation:
:$(1): \quad \paren {x e^y + y - x^2} \rd y = \paren {2 x y - e^y - x} \rd x$
is an exact differential equation with solution:
:$2 x e^y + x^2 + y^2 - 2 x^2 y = C$

Proof:
Let $(1)$ be expressed in the form:
:$\paren {e^y + x - 2 x y} \rd x + \paren {x e^y + y - x^2} rd y = 0$
Let:
:$\map M {x, y} = e^y + x - 2 x y$
:$\map N {x, y} = x e^y + y - x^2$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial M} {\partial y}
      | r = e^y - 2 x
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = e^y - 2 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 {e^y + x - 2 x y} \rd x + \map g y
      | c = 
}}
{{eqn | r = x e^y + \frac {x^2} 2 - x^2 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 + y - x^2} \rd y + \map h x
      | c = 
}}
{{eqn | r = x e^y + \frac {y^2} 2 - x^2 y + \map h x
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = x e^y + \dfrac {x^2 + y^2} 2 - x^2 y$
and by Solution to Exact Differential Equation, the solution to $(1)$ is:
:$2 x e^y + x^2 + y^2 - 2 x^2 y = C$
after multiplying by $2$ to clear the fraction.
{{qed}}
Theorem: First Order ODE/exp x (1 + x) dx = (x exp x - y exp y) dy

The first order ordinary differential equation:
:$(1): \quad e^x \paren {1 + x} \rd x = \paren {x e^x - y e^y} \rd y$
has the solution:
:$2 x e^x e^{-y} + y^2 + C$

Proof:
Dividing $(1)$ through by $e^y$:
:$(2): \quad \paren {1 + x} e^x e^{-y} \rd x = \paren {x e^x e^{-y} - y} \rd y$
Let $z = x e^x e^{-y}$.
Then:
:$\dfrac {\d z} {\d x} = \paren {1 + x} e^x e^{-y} - x e^x e^{-y} \dfrac {\d y} {\d x}$
or:
:$\d z = \paren {1 + x} e^x e^{-y} \rd x - x e^x e^{-y} \rd y$
Substituting $z$ and $\d z$ in $(2)$:
:$\d z = \paren {z - y} \rd y - z \rd y = - y \rd y$
which directly leads to:
:$\ds \int \rd z = -\int y \rd y$
from which:
:$z = -\dfrac {y^2} 2 + k$
Substituting for $z$ and letting $C = 2 k$:
:$2 x e^x e^{-y} + y^2 + C$
{{qed}}
Theorem: General Solution of Riccati Equation from Particular Solution

Consider the Riccati equation:
:$(1): \quad y' = \map p x + \map q x y + \map r x y^2$
Let $\map {y_1} x$ be a particular solution to $(1)$.
Then the general solution to $(1)$ has the form:
:$\map y x = \map {y_1} x + \map z x$
where $\map z x$ is the general solution to the Bernoulli equation:
:$z' - \paren {q - 2 r y_1} z = r z^2$

Proof:
Let $\map y x = \map {y_1} x + \map z x$ be a particular solution to $(1)$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\d y} {\d x}
      | r = y_1' + z'
      | c = 
}}
{{eqn | r = \map p x + \map q x \paren {y_1 + z} + \map r x \paren {y_1 + z}^2
      | c = 
}}
{{eqn | r = \map p x + \map q x \paren {y_1 + z} + \map r x \paren {y_1^2 + 2 y_1 z + z^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z'
      | r = \map q x z + \map r x \paren {2 y_1 z + z^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z' - \paren {\map q x + 2 y_1 \map r x z}
      | r = \map r x z^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Riccati Equation/y' = (y over x) + x^3 y^2 - x^5

The Riccati equation:
:$(1): \quad y' = \dfrac y x + x^3 y^2 - x^5$
has the general solution:
:$C \exp \dfrac {2 x^5} 5 = \dfrac {y - x} {y + x}$

Proof:
It can be seen by inspection that:
:$\map {y_1} x = x$
is a particular solution to $(1)$.
Thus from General Solution of Riccati Equation from Particular Solution:
:$y = x + \map z x$
where:
:$z' - \paren {\dfrac 1 x + 2 x^4} z = x^3 z^2$
From Bernoulli's Equation: $y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$:
:$z = \dfrac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}$
Thus:
{{begin-eqn}}
{{eqn | l = y
      | r = x + \frac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {y - x} \paren {C \map \exp {-\dfrac {2 x^5} 5} - 1}
      | r = 2 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y C \map \exp {-\dfrac {2 x^5} 5} - y - x C \map \exp {-\dfrac {2 x^5} 5} + x
      | r = 2 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y C \map \exp {-\dfrac {2 x^5} 5} - y - x C \map \exp {-\dfrac {2 x^5} 5} - x
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {y - x} C \map \exp {-\dfrac {2 x^5} 5}
      | r = y + x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {y - x} {y + x}
      | r = C \map \exp {\dfrac {2 x^5} 5}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Bernoulli's Equation/y' - (1 over x + 2 x^4) y = x^3 y^2

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

Proof:
It can be seen that $(1)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
where:
:$\map P x = -\paren {\dfrac 1 x + 2 x^4}$
:$\map Q x = x^3$
:$n = 2$
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 - 2} \int -\paren {\dfrac 1 x + 2 x^4} \rd x
      | c = 
}}
{{eqn | r = \ln x + \dfrac {2 x^5} 5
      | c = 
}}
{{eqn | r = \map \ln {x \exp \dfrac {2 x^5} 5}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \mu x
      | r = e^{\map \ln {x \exp \frac {2 x^5} 5} }
      | c = 
}}
{{eqn | r = x \exp \dfrac {2 x^5} 5
      | c = 
}}
{{end-eqn}}
and so substituting into $(3)$:
{{begin-eqn}}
{{eqn | l = \frac x y \exp \dfrac {2 x^5} 5
      | r = \paren {1 - 2} \int x^3 x \exp \dfrac {2 x^5} 5 \rd x
      | c = 
}}
{{eqn | r = -\int x^4 \exp \dfrac {2 x^5} 5 \rd x
      | c = 
}}
{{eqn | r = -\frac {\exp \dfrac {2 x^5} 5} 2 + \frac C 2
      | c = by using the substitution $w = x^5$
}}
{{eqn | ll= \leadsto
      | l = \frac x y
      | r = \frac C 2 \map \exp {-\dfrac {2 x^5} 5} - \frac 1 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac 1 y
      | r = \frac C {2 x} \map \exp {-\dfrac {2 x^5} 5} - \frac 1 {2 x}
      | c = 
}}
{{eqn | r = \frac {C \map \exp {-\dfrac {2 x^5} 5} - 1} {2 x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \frac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}
      | c = 
}}
{{end-eqn}}
Hence the general solution to $(1)$ is:
:$y = \dfrac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}$
{{qed}}
Category:Examples of Bernoulli's Equation
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Theorem: Motion of Body with Variable Mass

Let $B$ be a body undergoing a force $\mathbf F$.
Let $B$ be travelling at a velocity $\mathbf v$ at time $t$.
Let mass travelling at a velocity $\mathbf v + \mathbf w$ be added to $B$ at a rate of $\dfrac {\d m} {\d t}$.
Let $m$ be the mass of $B$ at time $t$.
Then the equation of motion of $B$ is given by:
:$\mathbf w \dfrac {\d m} {\d t} + \mathbf F = m \dfrac {\d \mathbf v} {\d t}$

Proof:
From Newton's Second Law of Motion:
:$\mathbf F = \map {\dfrac \d {\d t} } {m \mathbf v}$
Then the added momentum being added to $B$ by the mass being added to it is given by:
:$\paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t}$
Hence:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t} + \mathbf F
      | r = \map {\dfrac \d {\d t} } {m \mathbf v}
      | c = 
}}
{{eqn | r = m \dfrac {\d \mathbf v} {\d t} + \mathbf v \dfrac {\d m} {\d t}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \mathbf w \dfrac {\d m} {\d t} + \mathbf F
      | r = m \dfrac {\d \mathbf v} {\d t}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Burnout Height of Upward Rocket under Constant Gravity

Let $R$ be a rocket whose structural mass is $m_1$.
Let $R$ contain fuel of initial mass $m_2$.
Let $R$ be fired straight up from the surface of a planet whose gravitational field exerts an Acceleration Due to Gravity of $g$, assumed constant.
Let $R$ burn fuel at a constant rate $a$, with a constant exhaust velocity $b$ relative to $R$.
Let all forces on $R$ except that due to  the gravitational field be neglected.
Then the burnout height of $R$ is given by:
:$h_b = -\dfrac {g m_2^2} {2 a^2} + \dfrac {b m_2} a + \dfrac {b m_1} a \ln \dfrac {m_1} {m_1 + m_2}$

Proof:
{{Proofread}}
{{MissingLinks}}
The total mass of the rocket at time $t$ is given by
:$m = m_1 + m_2 - at$
The time $t_b$ at which the rocket runs out of fuel is given by:
:$t_b = \dfrac {m_2} a$
Now we can calculate the total force on the rocket to be:
:$F = b a - mg$
where:
:$b a$ is the rocket's thrust as calculated from Newton's Second Law of Motion
:$-mg$ is the force of gravity on the rocket
{{explain|Not obvious why thrust is $b a$}}
We can now calculate the acceleration of the rocket as a function of time using our calculated total force in Newton's First Law of Motion to obtain
{{begin-eqn}}
{{eqn | l = \ddot h
      | r = \dfrac 1 m \paren {b a - m g}
      | c = 
}}
{{eqn | r = \dfrac {b a} {m_1 + m_2 - a t} - g
      | c = 
}}
{{end-eqn}}
where $\ddot h = \dfrac {\d^2 h} {\d t^2}$ is the second time derivative of the height of the rocket.
Integrating the acceleration function twice with respect to time will yield the height of the rocket as a function of time.
Performing the first integration:
{{begin-eqn}}
{{eqn | l = \dot h
      | r = \int_0^t \paren {\dfrac {b a} {m_1 + m_2 - at'} - g} \rd t'
      | c = 
}}
{{eqn | r = -b \ln \paren {m_1 + m_2 - a t} + b \map \ln {m_1 + m_2} - g t
      | c = 
}}
{{end-eqn}}
The second integration:
{{begin-eqn}}
{{eqn | l = \map h t
      | r = \int_0^t \paren {b \map \ln {m_1 + m_2} - b \map \ln {m_1 + m_2 - a t'} - g t'} \rd t'
      | c = 
}}
{{eqn | r = -\dfrac 1 2 g t^2 + b t \map \ln {m_1 + m_2} - b \ds \int_0^t \paren {\map \ln {m_1 + m_2 - at'} } \rd t
      | c = 
}}
{{eqn | r = -\dfrac 1 2 g t^2 + b t \map \ln {m_1 + m_2} + \dfrac b a \paren {\paren {m_1 + m_2 - a t} \map \ln {m_1 + m_2 - a t} + a t - \paren {m_1 + m_2} \map \ln {m_1 + m_2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map h {t_b}
      | r = -\dfrac {g m_2^2} {2 a^2} + \dfrac {b m_2} a \map \ln {m_1 + m_2} + \dfrac b a \paren {m_1 \ln m_1 + m_2 - \paren {m_1 + m_2} \map \ln {m_1 + m_2} }
      | c = substituting $t_b = \dfrac {m_2} a$ for $t$
}}
{{eqn | r = -\dfrac {g m_2^2} {2 a^2} + \dfrac {b m_2} a + {b m_1} a \map \ln {\dfrac {m_1} {m_1 + m_2} }
      | c = simplifying
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Linear First Order ODE/y' = x + y

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

Proof:
Rearranging $(1)$:
:$(2): \quad \dfrac {\d y} {\d x} - y = x$
$(2)$ is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = -1$
:$\map Q x = x$
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 = 
}}
{{eqn | r = \frac 1 {e^x}
      | c = 
}}
{{end-eqn}}
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
{{begin-eqn}}
{{eqn | l = \map {\dfrac {\d} {\d x} } {\dfrac y {e^x} }
      | r = \frac x {e^x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac y {e^x}
      | r = \int \frac x {e^x} \rd x
      | c = 
}}
{{eqn | r = -\frac 1 {e^x} \paren {x + 1} + C
      | c = Primitive of $x e^{a x}$
}}
{{eqn | ll= \leadsto
      | l = y
      | r = C e^x - x - 1
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Linear First Order ODE/y' = x + y/y(0) = 1

The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = x + y$
with initial condition:
:$\map y 0 = 1$
has the particular solution:
:$y = 2 e^x - x - 1$

Proof:
From Linear First Order ODE: $y' = x + y$, the general solution of $(1)$ is:
:$y = C e^x - x - 1$
Setting $y = 1$ when $x = 0$ gives:
:$1 = C + 1$
from which $C = 2$.
Hence the result.
{{qed}}