hcju/proofwikiqa huggingface dataset:
Let: :$\map M {x, y} = \dfrac 1 {x^3 y^2} + \dfrac 1 x$ :$\map N {x, y} = \dfrac 1 {x^2 y^3} - \dfrac 1 y$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y}       | r = -\dfrac 2 {x^3 y^3}       | c =  }} {{eqn | l = \dfrac {\partial N} {\partial x}       | r = -\dfrac 2 {x^3 y^3}       | c =  }} {{end-eqn}} Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact. By Solution to Exact Differential Equation, the solution to $(1)$ is: :$\map f {x, y} = C$ where: {{begin-eqn}} {{eqn | l = \dfrac {\partial f} {\partial x}       | r = \map M {x, y}       | c =  }} {{eqn | l = \dfrac {\partial f} {\partial y}       | r = \map N {x, y}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = f       | r = \map M {x, y} \rd x + \map g y       | c =  }} {{eqn | r = \int \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \map g y       | c =  }} {{eqn | r = -\frac 1 {2 x^2 y^2} + \ln 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 {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y + \map h x       | c =  }} {{eqn | r = -\frac 1 {2 x^2 y^2} - \ln y + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = -\dfrac 1 {2 x^2 y^2} + \ln x - \ln y$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$-\dfrac 1 {2 x^2 y^2} + \ln x - \ln y = C$ or: :$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$ {{qed}} Category:Examples of Exact Differential Equation Category:Examples of First Order ODEs d7d8aehr5sm521jwtav2qh493ooxzmm
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = k x^n$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int \frac 1 x \rd x       | c =  }} {{eqn | r = \ln x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = e^{\ln x}       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\map {\dfrac \d {\d x} } {x y} = k x^{n + 1}$ When $n + 1 \ne -1$, Primitive of Power is used to obtain: :$x y = \dfrac {k x^{n + 2} } {n + 2} + C$ or: :$y = \dfrac {k x^{n + 1} } {n + 2} + \dfrac C x$ When $n + 1 = -1$, we have: :$\map {\dfrac \d {\d x} } {x y} = \dfrac k x$ and Primitive of Reciprocal is used, yielding: :$x y = k \ln x + C$ or: :$y = \dfrac {k \ln x} x + \dfrac C x$ {{qed}} Category:Examples of Linear First Order ODEs 081vts8gx9yr5g78be7x8dw3nbboh71
Let $(1)$ be rearranged as: :$(2): \quad \dfrac {\d x} {\d y} - \dfrac x y = -\dfrac {x^2} {y^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$ where: :$\map P y = -\dfrac 1 y$ :$\map Q y = -\dfrac 1 {y^2}$ :$n = 2$ and so is an example of Bernoulli's equation. By Solution to Bernoulli's Equation it has the general solution: :$(3): \quad \displaystyle \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \, \map \mu y \rd y + C$ where: :$\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$ Thus $\map \mu y$ is evaluated: {{begin-eqn}} {{eqn | l = \paren {1 - n} \int \map P y \rd y       | r = \paren {1 - 2} \int -\dfrac 1 y \rd y       | c =  }} {{eqn | r = -\paren {-\ln y}       | c =  }} {{eqn | r = \ln y       | c =  }} {{eqn | ll= \leadsto       | l = \map \mu y       | r = e^{\ln y}       | c =  }} {{eqn | r = y       | c =  }} {{end-eqn}} and so substituting into $(3)$: {{begin-eqn}} {{eqn | l = \frac y x       | r = \paren {1 - 2} \int -\dfrac 1 {y^2} y \rd y       | c =  }} {{eqn | r = \int \dfrac 1 y \rd y       | c =  }} {{eqn | r = \ln y + C       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = x \ln y + C x       | c =  }} {{end-eqn}} {{qed}}
Let $(1)$ be rearranged as: :$(2): \quad y^2 \rd x = \paren {x y - x^2} \rd y$ Let: :$\map M {x, y} = y^2$ :$\map N {x, y} = x y - x^2$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y}       | r = \paren {t y}^2       | c =  }} {{eqn | r = t^2 \paren {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 - \paren {t x}^2       | c =  }} {{eqn | r = t^2 \paren {x y - 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: :$\dfrac {\d x} {\d y} = \dfrac {x^2 - x y} {y^2}$ By Solution to Homogeneous Differential Equation, its solution is: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ where: :$\map f {y, x} = \dfrac {x y - x^2} {y^2}$ Hence: {{begin-eqn}} {{eqn | l = \ln y       | r = \int \frac {\d z} {\frac {z - z^2} 1 - z} + C       | c =  }} {{eqn | r = -\int \frac {\d z} {z^2} + C       | c =  }} {{eqn | r = \frac 1 z + C       | c = Primitive of Power }} {{eqn | ll= \leadsto       | l = \ln y       | r = \frac y x + C       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = x \ln y + C x       | c = multiplying through by $x$ }} {{end-eqn}} {{qed}}
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - \dfrac y x = x$ $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = -\dfrac 1 x$ :$\map Q x = x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int -\dfrac 1 x \rd x       | c =  }} {{eqn | r = -\ln x       | c =  }} {{eqn | r = \map \ln {\frac 1 x}       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = \frac 1 x       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\map {\dfrac \d {\d x} } {\dfrac y x} = 1$ and the general solution is: :$\dfrac y x = x + C$ or: :$y = x^2 + C x$ {{qed}}
Rearranging: :$\dfrac {\d y} {\d x} + \dfrac y x = x \cos x$ This is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = x \cos x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int \dfrac 1 x \rd x       | c =  }} {{eqn | r = \ln x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = x       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor: :$\map {\dfrac {\d} {\d x} } {x y} = x^2 \cos x$ {{begin-eqn}} {{eqn | l = \map {\dfrac {\d} {\d x} } {x y}       | r = x^2 \cos x       | c =  }} {{eqn | l = x y       | r = \int x^2 \cos x \rd x       | c =  }} {{eqn | r = 2 x \cos x + \paren {x^2 - 2} \sin x + C       | c = Primitive of $x^2 \cos a x$ }} {{eqn | ll= \leadsto       | l = y       | r = 2 \cos x + x \sin x - \frac 2 x \sin x + \frac C x       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y       | r = x^2 \cos x       | c =  }} {{eqn | ll= \leadsto       | l = x \rd y + y \rd x       | r = x^2 \cos x \rd x       | c =  }} {{eqn | ll= \leadsto       | l = \map \d {x y}       | r = x^2 \cos x \rd x       | c = Product Rule for Differentials }} {{eqn | ll= \leadsto       | l = x y       | r = \int x^2 \cos x \rd x       | c =  }} {{eqn | r = 2 x \cos x + \paren {x^2 - 2} \sin x + C       | c = Primitive of $x^2 \cos a x$ }} {{eqn | ll= \leadsto       | l = y       | r = 2 \cos x + x \sin x - \frac 2 x \sin x + \frac C x       | c =  }} {{end-eqn}} {{qed}}
Express it in the form: :$\paren {x \map \sin {x + y} - \map \cos {x + y} } \rd x + x \map \sin {x + y} \rd y$ Let: :$\map M {x, y} = x \map \sin {x + y} - \map \cos {x + y}$ :$\map N {x, y} = x \map \sin {x + y}$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y}       | r = x \map \cos {x + y} + \map \sin {x + y}       | c =  }} {{eqn | l = \frac {\partial N} {\partial x}       | r = x \map \cos {x + y} + \map \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 {x \map \sin {x + y} - \map \cos {x + y} } \rd x + \map g y       | c =  }} {{eqn | r = \map \sin {x + y} - x \map \cos {x + y} - \map \sin {x + y}       | c =  }} {{eqn | r = -x \map \cos {x + y}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = f       | r = \map N {x, y} \rd y + \map h x       | c =  }} {{eqn | r = \int x \map \sin {x + y} \rd y + \map h x       | c =  }} {{eqn | r = -x \map \cos {x + y} + \map h x       | c =  }} {{end-eqn}} Thus: :$\map f {x, y} = -x \map \cos {x + y}$ and by Solution to Exact Differential Equation, the solution to $(1)$ is: :$-x \map \cos {x + y} = C_1$ or setting $C = -C_1$: :$x \map \cos {x + y} = C$ {{qed}}
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y       | r = \map f x       | c =  }} {{eqn | ll= \leadsto       | l = x \rd y + y \rd x       | r = \map f x \rd x       | c =  }} {{eqn | ll= \leadsto       | l = \map \d {x y}       | r = \map f x \rd x       | c = Product Rule for Differentials }} {{eqn | ll= \leadsto       | l = x y       | r = \int \map f x \rd x + C       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y       | r = x^2 \cos x       | c =  }} {{eqn | ll= \leadsto       | l = x y       | r = \int x^2 \cos x \d x + C       | c = Linear First Order ODE: $x y' + y = \map f x$ }} {{eqn | r = 2 x \cos x + \paren {x^2 - 2} \sin x + C       | c = Primitive of $x^2 \cos a x$ }} {{eqn | ll= \leadsto       | l = y       | r = 2 \cos x + x \sin x - \frac 2 x \sin x + \frac C x       | c =  }} {{end-eqn}} {{qed}}
Rearranging: :$\dfrac {d y} {\d x} + \dfrac y x = \cos x$ This is a linear first order ODE in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x y = \dfrac 1 x$ :$\map Q x = \cos x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = \int \frac 1 x \rd x       | c =  }} {{eqn | r = \ln x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = e^{\ln x}       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor: {{begin-eqn}} {{eqn | l = \dfrac {\d} {\d x} \paren {x y}       | r = x \cos x       | c =  }} {{eqn | ll= \leadsto       | l = x y       | r = \int x \cos x \rd x       | c =  }} {{eqn | r = x \sin x + \cos x + C       | c = Primitive of $x \cos a x$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y       | r = x \cos x       | c =  }} {{eqn | ll= \leadsto       | l = x y       | r = \int x \cos x \rd x + C       | c = Linear First Order ODE: $x y' + y = \map f x$ }} {{eqn | r = x \sin x + \cos x + C       | c = Primitive of $x \cos a x$ }} {{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} + x p       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = x \dfrac {\d p} {\d x} + p       | r = \frac 1 x       | c =  }} {{eqn | ll= \leadsto       | l = x p       | r = \int \frac {\d x} x       | c = Linear First Order ODE: $x y' + y = \map f x$ }} {{eqn | r = \ln x + C_1       | c = Primitive of Reciprocal }} {{eqn | ll= \leadsto       | l = \dfrac {\d y} {\d x}       | r = \frac {\ln x} x + \frac {C_1} x       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \int \paren {\frac {\ln x} x + \frac {C_1} x} \rd x       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \frac {\paren {\ln x}^2} 2 + C_1 \ln x + C_2       | c = Primitive of $\dfrac {\ln x} x$ }} {{end-eqn}} {{qed}}
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}}
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}}
$(1)$ 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}}
$(1)$ 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: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ 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 \ln \map \{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}}
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}}
{{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 = Separation of Variables }} {{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 Separation of Variables q1ev8n414r0vr7rxjt0ng5cia3utj6i
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}}
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}}
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}}
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}}