hcju/proofwikiqa huggingface dataset:
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = p \frac {\d p} {\d y}       | r = p^2 + 1       | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto       | l = \int \rd y       | r = \int \frac {p \rd p} {p^2 + 1}       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = y       | r = \frac 1 2 \, \map \ln {p^2 + 1} + k       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{eqn | ll= \leadsto       | l = p = \frac {\d y} {\d x}       | r = \sqrt {A_1^2 e^{2 y} - 1}       | c = where $A_1^2 = e^{2 k}$ }} {{eqn | n = 2       | ll= \leadsto       | l = \int \rd x       | r = \int \frac {\d y} {\sqrt {A_1^2 e^{2 y} - 1} }       | c = Separation of Variables }} {{end-eqn}} Making the subtitution $u = A_1 e^y$: :$\d u = A_1 e^y \rd y = u \rd y$ Thus $(2)$ transforms into: {{begin-eqn}} {{eqn | l = \int \rd x       | r = \int \frac {\d u} {u \sqrt {u^2 - 1} }       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \arcsec u + A_2       | c = Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$ }} {{eqn | r = \map \arcsec {A_1 e^y} + A_2       | c =  }} {{eqn | ll= \leadsto       | l = \map \sec {x - A_2}       | r = A_1 e^y       | c =  }} {{eqn | ll= \leadsto       | l = \map {\ln \sec} {x - A_2}       | r = \ln A_1 + \ln e^y       | c =  }} {{eqn | r = \ln A_1 + y       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \map {\ln \sec} {x - A_2} - \ln A_1       | c =  }} {{eqn | r = \map {\ln \sec} {x + C_1} + C_2       | c = setting $C_1 = -A_2$ and $C_2 = -\ln A_1$ }} {{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 = \dfrac {\d p} {\d x}       | r = p^2 + 1       | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto       | l = \int \rd x       | r = \int \frac {\d p} {p^2 + 1}       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = x       | r = \arctan p + A_1       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | ll= \leadsto       | l = p = \frac {\d y} {\d x}       | r = \map \tan {x - A_1}       | c =  }} {{eqn | ll= \leadsto       | l = \int \rd y       | r = \int \map \tan {x - A_1} \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = y       | r = \map {\ln \sec} {x - A_1} + A_2       | c = Primitive of $\tan a x$ }} {{eqn | r = \map {\ln \sec} {x + C_1} + C_2       | c = rearranging constants }} {{end-eqn}} {{qed}}
:500px By the Principle of Conservation of Energy, the speed of the bead at a particular height is determined by its loss in potential energy in getting there. Thus, at the point $\tuple {x, y}$, we have: :$(1): \quad v = \dfrac {\d s} {\d t} = \sqrt {2 g y}$ This can be written: {{begin-eqn}} {{eqn | l = \d t       | r = \frac {\d s} {\sqrt {2 g y} }       | c =  }} {{eqn | r = \frac {\sqrt {\d x^2 + \d y^2} } {\sqrt {2 g y} }       | c =  }} {{end-eqn}} Thus the time taken for the bead to slide down the wire is given by: :$\displaystyle T_1 = \int \sqrt {\dfrac {\d x^2 + \d y^2} {2 g y} }$ From Equation of Cycloid, we have: :$x = a \paren {\theta - \sin \theta}$ :$y = a \paren {1 - \cos \theta}$ Substituting these in the above integral: :$\displaystyle T_1 = \int_0^{\theta_1} \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {1 - \cos \theta} } } \rd \theta = \theta_1 \sqrt {\dfrac a g}$ This is the time needed for the bead to reach the bottom when released when $\theta_1 = \pi$, and so: :$T_1 = \pi \sqrt {\dfrac a g}$ Now suppose the bead is released at any intermediate point $\tuple {x_0, y_0}$. Take equation $(1)$ and replace it with: :$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$ Thus the total time to reach the bottom is: {{begin-eqn}} {{eqn | l = T       | r = \int_{\theta_0}^\pi \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {\cos \theta_0 - \cos \theta} } } \rd \theta       | c =  }} {{eqn | r = \sqrt {\dfrac a g} \int_{\theta_0}^\pi \sqrt {\dfrac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta       | c =  }} {{eqn | n = 2       | r = \sqrt {\dfrac a g} \int_{\theta_0}^\pi \dfrac {\sin \frac 1 2 \theta \rd \theta} {\sqrt {\cos^2 \frac 1 2 \theta_0 - \cos^2 \frac 1 2 \theta} } \rd \theta       | c = Double Angle Formula for Cosine }} {{end-eqn}} Setting: :$u = \dfrac {\cos \frac 1 2 \theta} {\cos \frac 1 2 \theta_0}$ and so: :$\d u = -\dfrac 1 2 \dfrac {\sin \frac 1 2 \theta \rd \theta} {\cos \frac 1 2 \theta_0}$ Then $(2)$ becomes: {{begin-eqn}} {{eqn | l = T       | r = -2 \sqrt {\dfrac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }       | c =  }} {{eqn | r = 2 \sqrt {\dfrac a g} \Big [{\arcsin u}\Big]_0^1       | c =  }} {{eqn | r = \pi \sqrt {\dfrac a g}       | c =  }} {{end-eqn}} That is, wherever the bead is released from, it takes that same time to reach the bottom. Hence the result. {{qed}}
By Brachistochrone is Cycloid, $C$ is a brachistochrone. {{ProofWanted}}
Let $\epsilon - \delta \ge \map d {x, y}$. Then $\epsilon \ge \map d {x, y} + \delta$. If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$. So: {{begin-eqn}} {{eqn | l = \map d {x, z}       | o = \le       | r = \map d {x, y} + \map d {y, z}       | c = metric space axioms }} {{eqn | o = <       | r = \map d {x, y} + \delta       | c = {{Defof|Open Ball}} }} {{eqn | o = \le       | r = \epsilon }} {{end-eqn}} Thus $z \in \map {B_\epsilon} x$. So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$. {{qed}}
The components of the velocity of $b$ are: :$\dfrac {\mathrm d x} {\mathrm d t} = - b \cos \theta$ :$\dfrac {\mathrm d y} {\mathrm d t} = - a + b \sin \theta$ Hence: {{begin-eqn}} {{eqn | l = \frac {\mathrm d y} {\mathrm d x}       | r = \frac {-a + b \sin \theta} {-b \cos \theta}       | c =  }} {{eqn | r = \frac {-a + b \left({-\dfrac y {\sqrt {x^2 + y^2} } }\right)} {-b \left({\dfrac x {\sqrt {x^2 + y^2} } }\right)}       | c =  }} {{eqn | n = 1       | r = \frac {-a \sqrt {x^2 + y^2} + b y} {b x}       | c =  }} {{end-eqn}} $(1)$ is a homogeneous differential equation. Let $z = \dfrac y x$. Then by Solution to Homogeneous Differential Equation: {{begin-eqn}} {{eqn | l = \ln x       | r = \int \frac {\mathrm d z} {\dfrac {-a \sqrt {1 + z^2} + b z} b - z} + C       | c =  }} {{eqn | r = \int \frac {b \, \mathrm d z} {-a \sqrt {1 + z^2} + b z - b z} + C       | c =  }} {{eqn | r = \int \frac {b \, \mathrm d z} {-a \sqrt {1 + z^2} } + C       | c =  }} {{eqn | r = -\frac b a \int \frac {\mathrm d z} {\sqrt {1 + z^2} } + C       | c =  }} {{eqn | r = -\frac b a \ln \left({z + \sqrt {1 + z^2} }\right)  + C       | c = Primitive of $\dfrac 1 {\sqrt{x^2 + a^2} }$ }} {{end-eqn}} Substituting back for $z$ and rearranging: :$C^k \left({y + \sqrt {x^2 + x^2} }\right) = x^{k + 1}$ {{qed}}
Let: :$E_L$ be the drop in electromotive force across $L$ :$E_R$ be the drop in electromotive force across $R$ :$E_C$ be the drop in electromotive force across $C$. From Kirchhoff's Voltage Law: :$E - E_L - E_R - E_C = 0$ From Ohm's Law: :$E_R = R I$ From Drop in EMF caused by Inductance is proportional to Rate of Change of Current: :$E_L = L \dfrac {\mathrm d I} {\mathrm d t}$ From Drop in EMF caused by Capacitance is proportional to Accumulated Charge: :$E_C = \dfrac 1 C Q$ where $Q$ is the electric charge $Q$ that has accumulated on $C$. Thus: :$E - L \dfrac {\mathrm d I} {\mathrm d t} - R I - \dfrac 1 C Q = 0$ which can be rewritten: :$L \dfrac {\mathrm d I} {\mathrm d t} + R I + \dfrac 1 C Q = E$ Differentiating {{WRT|Differentiation}} $t$ gives: :$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} + R \dfrac {\mathrm d I} {\mathrm d t} + \dfrac 1 C I = \dfrac {\mathrm d E} {\mathrm d t}$ {{qed}}
Let: :$E_L$ be the drop in electromotive force across $L$ :$E_R$ be the drop in electromotive force across $R$ :$E_C$ be the drop in electromotive force across $C$. From Kirchhoff's Voltage Law: :$E - E_L - E_R - E_C = 0$ From Ohm's Law: :$E_R = R I$ From Drop in EMF caused by Inductance is proportional to Rate of Change of Current: :$E_L = L \dfrac {\mathrm d I} {\mathrm d t}$ From Drop in EMF caused by Capacitance is proportional to Accumulated Charge: :$E_C = \dfrac 1 C Q$ where $Q$ is the electric charge $Q$ that has accumulated on $C$. Thus: :$E - L \dfrac {\mathrm d I} {\mathrm d t} - R I - \dfrac 1 C Q = 0$ which can be rewritten: :$L \dfrac {\mathrm d I} {\mathrm d t} + R I + \dfrac 1 C Q = E$ By definition of electric current: :$I = \dfrac {\mathrm d Q} {\mathrm d t}$ and so: :$L \dfrac {\mathrm d^2 Q} {\mathrm d t^2} + R \dfrac {\mathrm d Q} {\mathrm d t} + \dfrac 1 C Q = E$ {{qed}}
Let: :$E_L$ be the drop in electromotive force across $L$ :$E_R$ be the drop in electromotive force across $R$ From Kirchhoff's Voltage Law: :$E - E_L - E_R = 0$ From Ohm's Law: :$E_R = R I$ From Drop in EMF caused by Inductance is proportional to Rate of Change of Current: :$E_L = L \dfrac {\d I} {\d t}$ Thus: :$E - L \dfrac {\d I} {\d t} - R I = 0$ which can be rewritten: :$L \dfrac {\d I} {\d t} + R I = E$ {{qed}}
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\d I} {\d t} + R I = E_0$ defines the behaviour of $I$. This can be written as: :$(1): \quad \dfrac {\d I} {\d t} = \dfrac R L \paren {\dfrac {E_0} R - I}$ $(1)$ is in the form: :$\dfrac {\d y} {\d x} = k \paren {y_a - y}$ where: :$k \in \R: k > 0$ :$y = y_0$ at $x = 0$ This is an example of the Decay Equation, where: :$k = \dfrac R L$ :$y_a = \dfrac {E_0} R$ :$y_0 = I_0$. whose particular solution is: :$y = y_a + \paren {y_0 - y_a} e^{-k x}$ Hence the particular solution to $(1)$ is: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$ {{qed}}
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$: :$I = \dfrac {E_0} R + \left({I_0 - \dfrac {E_0} R}\right) e^{-R t / L}$ We have that: :$\displaystyle \lim_{t \mathop \to \infty} e^{-R t / L} \to 0$ and so: :$\displaystyle \lim_{t \mathop \to \infty} I \to \dfrac {E_0} R$ Hence the result. {{qed}}
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$ Setting $I_0 = 0$: {{begin-eqn}} {{eqn | l = I       | r = \dfrac {E_0} R + \paren {0 - \dfrac {E_0} R} e^{-R t / L}       | c =  }} {{eqn | r = \dfrac {E_0} R \paren {1 - e^{-R t / L} }       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$ Setting $E_0 = 0$: {{begin-eqn}} {{eqn | l = I       | r = \dfrac 0 R + \paren {I_0 - \dfrac 0 R} e^{-R t / L}       | c =  }} {{eqn | r = I_0 e^{-R t / L}       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E_0 e^{-k t}$ defines the behaviour of $I$. This can be written as: :$(1): \quad \dfrac {\mathrm d I} {\mathrm d t} + \dfrac R L I = \dfrac {E_0} L  e^{-k t}$ $(1)$ is a linear first order ODE in the form: :$\dfrac {\mathrm d I} {\mathrm d t} + P \left({t}\right) I = Q \left({t}\right)$ where: :$P \left({t}\right) = \dfrac R L$ :$Q \left({t}\right) = \dfrac {E_0} L  e^{-k t}$ Thus: {{begin-eqn}} {{eqn | l = \int P \left({t}\right) \, \mathrm d t       | r = \int \dfrac R L \, \mathrm d t       | c =  }} {{eqn | r = \dfrac {R t} L       | c =  }} {{eqn | ll= \implies       | l = e^{\int P \, \mathrm d t}       | r = e^{R t / L}       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\dfrac {\mathrm d} {\mathrm d t} \left({e^{R t / L} I}\right) = e^{R t / L} \dfrac {E_0} L e^{-k t}$ and the general solution becomes: :$\displaystyle e^{R t / L} I = \int e^{R t / L} \dfrac {E_0} L e^{-k t} \, \mathrm d t$ and so: {{begin-eqn}} {{eqn | l = e^{R t / L} I       | r = \frac {E_0} L  \int e^{R t / L} e^{-k t} \, \mathrm d t       | c =  }} {{eqn | r = \frac {E_0} L  \int e^{\left({R / L - k}\right) t} \, \mathrm d t       | c =  }} {{eqn | r = \frac {E_0} {L \left({\frac R L - k}\right)} e^{\left({R / L - k}\right) t} + C       | c =  }} {{eqn | r = \frac {E_0} {R - k L} e^{\left({R / L - k}\right) t} + C       | c =  }} {{end-eqn}} When $t = 0$, we have $I = I_0$. So: {{begin-eqn}} {{eqn | l = I_0       | r = \frac {E_0} {R - k L} + C       | c =  }} {{eqn | ll= \implies       | l = C       | r = I_0 - \frac {E_0} {R - k L}       | c =  }} {{end-eqn}} So: :$I e^{R t/ L} = \dfrac {E_0} {R - k L} e^{\left({R / L - k}\right) t} + I_0 - \dfrac {E_0} {R - k L}$ Multiplying by $e^{\frac {R t} L}$ and tidying up, we get: {{begin-eqn}} {{eqn | l = I       | r = \frac {E_0} {R - k L} e^{- R t / L + R t / L - k t} + I_0 e^{-R t / L} - \dfrac {E_0} {R - k L} e^{-R t / L}       | c =  }} {{eqn | r = \frac {E_0} {R - k L} e^{- k t} + I_0 e^{-R t / L} - \frac {E_0} {R - k L} e^{-R t / L}       | c =  }} {{eqn | r = \frac {E_0} {R - k L} e^{- k t} + \left({I_0 - \frac {E_0} {R - k L} }\right) e^{-R t / L}       | c =  }} {{end-eqn}} {{qed}}
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E_0 \sin \omega t$ defines the behaviour of $I$. This can be written as: :$(1): \quad \dfrac {\mathrm d I} {\mathrm d t} + \dfrac R L I = \dfrac {E_0} L \sin \omega t$ $(1)$ is a linear first order ODE in the form: :$\dfrac {\mathrm d I} {\mathrm d t} + P \left({t}\right) I = Q \left({t}\right)$ where: :$P \left({t}\right) = \dfrac R L$ :$Q \left({t}\right) = \dfrac {E_0} L  e^{-k t}$ Thus: {{begin-eqn}} {{eqn | l = \int P \left({t}\right) \, \mathrm d t       | r = \int \dfrac R L \, \mathrm d t       | c =  }} {{eqn | r = \dfrac {R t} L       | c =  }} {{eqn | ll= \implies       | l = e^{\int P \, \mathrm d t}       | r = e^{R t / L}       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: :$\dfrac {\mathrm d} {\mathrm d t} \left({e^{R t / L} I}\right) = e^{R t / L} \dfrac {E_0} L \sin \omega t$ and so the general solution becomes: {{begin-eqn}} {{eqn | l = e^{R t / L} I       | r = \frac {E_0} L \int e^{R t / L} \sin \omega t \, \mathrm d t       | c =  }} {{eqn | r = \frac {E_0} L \frac {e^{R t / L} \left({\frac R L \sin \omega t - \omega \cos \omega t}\right)} {\left({\frac R L}\right)^2 + \omega^2} + C       | c = Primitive of $e^{a x} \sin b x$ }} {{eqn | r = \frac {E_0 \left({R \sin \omega t - L \omega \cos \omega t}\right)} {R^2 + \omega^2 L^2} e^{R t / L} + C       | c =  }} {{end-eqn}} When $t = 0, I = I_0$. {{begin-eqn}} {{eqn | l = I_0       | r = E_0 \, {\frac {-L \omega} {R^2 + \omega^2 L^2} } + C       | c =  }} {{eqn | ll= \implies       | l = C       | r = I_0 + \frac {E_0 L \omega} {R^2 + \omega^2 L^2}       | c =  }} {{eqn | ll= \implies       | l = I       | r = \frac {E_0 \left({R \sin \omega t - L \omega \cos \omega t}\right)} {R^2 + \omega^2 L^2} + \left({I_0 + \frac {E_0 L \omega} {R^2 + \omega^2 L^2} }\right) e^{-R t / L}       | c =  }} {{end-eqn}} The result follows by application of Multiple of Sine plus Multiple of Cosine: Sine Form. {{qed}}
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E$ defines the behaviour of $I$. Let $I$ be at a maximum or a minimum. Then from Derivative at Maximum or Minimum: :$\dfrac {\mathrm d I} {\mathrm d t} = 0$ and so: {{begin-eqn}} {{eqn | l = E       | r = 0 + R I       | c =  }} {{eqn | r = R I       | c =  }} {{end-eqn}} which is Ohm's Law. {{qed}}
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E$ defines the behaviour of $I$. Taking the derivative: :$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} + R \dfrac {\mathrm d I} {\mathrm d t} = \dfrac {\mathrm d E} {\mathrm d t}$ From Second Derivative of Real Function at Minimum: :$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} \ge 0$ while from Derivative at Maximum or Minimum: :$R \dfrac {\mathrm d I} {\mathrm d t} = 0$ and so: :$\dfrac {\mathrm d E} {\mathrm d t} \ge 0$ The result follows from Increasing Function has Positive Derivative. {{qed}}
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E$ defines the behaviour of $I$. Taking the derivative: :$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} + R \dfrac {\mathrm d I} {\mathrm d t} = \dfrac {\mathrm d E} {\mathrm d t}$ From Second Derivative of Real Function at Minimum: :$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} \le 0$ while from Derivative at Maximum or Minimum: :$R \dfrac {\mathrm d I} {\mathrm d t} = 0$ and so: :$\dfrac {\mathrm d E} {\mathrm d t} \le 0$ The result follows from Decreasing Function has Negative Derivative. {{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 = p^2       | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto       | l = y \frac {\d p} {\d y}       | r = p       | c =  }} {{eqn | ll= \leadsto       | l = p = \dfrac {\d y} {\d x}       | r = C_1 y       | c = First Order ODE: $x \rd y = k y \rd x$ }} {{eqn | ll= \leadsto       | l = y       | r = C_2 e^{C_1 x}       | c = First Order ODE: $\d y = k y \rd x$ }} {{end-eqn}} {{qed}}
Let $(1)$ be rearranged as: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d y}       | r = \frac {1 - x y} {y^2}       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = \frac {\d x} {\d y} + \frac x y       | r = \frac 1 {y^2}       | c =  }} {{end-eqn}} 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: :$\map P y = \dfrac 1 y$ :$\map Q y = \dfrac 1 {y^2}$ 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 | ll= \leadsto       | l = e^{\int P \rd y}       | r = y       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(2)$ can be rewritten as: {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d y} } {x y}       | r = \dfrac 1 y       | c =  }} {{eqn | ll= \leadsto       | l = x y       | r = \int \frac {\d y} y       | c =  }} {{eqn | r = \ln y + C       | c =  }} {{end-eqn}} {{qed}}
Rewriting $(1)$ as: :$\dfrac {\d y} {\d x} = -\dfrac {2 x + 3 y + 1} {-3 x + 2 y + 5}$ we note that it is in the form: :$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$ where: :$a e = -4 \ne b d = 9$ Hence we can use: :First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$ which can be solved by substituting: :$x := z - h$ :$y := w - k$ where: :$h = \dfrac {c e - b f} {a e - b d}$ :$k = \dfrac {a f - c d} {a e - b d}$ So let: :$x = z - h$ where $h = \dfrac {\paren {1 \times 2} - \paren {3 \times 5} } {\paren {2 \times 2} - \paren {3 \times -3} } = \dfrac {-13} {13} = -1$ :$y = w - k$ where $k = \dfrac {\paren {2 \times 5} - \paren {1 \times -3} } {\paren {2 \times 2} - \paren {3 \times -3} } = \dfrac {13} {13} = 1$ Then $(1)$ is transformed into: :$(2): \quad \dfrac {\d w} {\d z} = -\dfrac {2 z + 3 w} {3 z - 2 w}$ Let: :$\map M {w, z} =  2 z + 3 w$ :$\map N {w, z} =  3 z - 2 w$ We have that: :$\map M {t w, t z} =  2 t z + 3 t w = t \paren {2 z + 3 w} = t \, \map M {x, y}$ :$\map N {t w, t z} =  3 t z - 2 t w = t \paren {3 z - 2 w} = t \, \map N {x, y}$ Thus both $M$ and $N$ are homogeneous functions of degree $1$. Thus by definition $(2)$ is a homogeneous differential equation. So: {{begin-eqn}} {{eqn | l = \frac {\d w} {\d z}       | r = \frac {2 z + 3 w} {3 z - 2 w}       | c =  }} {{eqn | r = \frac {2 + 3 w / z} {3 - 2 w / z}       | c = dividing top and bottom by $x$ }} {{eqn | r = \frac {2 + 3 u} {3 - 2 u}       | c = substituting $u$ for $w / z$ }} {{end-eqn}} By Solution to Homogeneous Differential Equation: :$\displaystyle \ln z = \int \frac {\d u} {\map f {1, u} - u} + C$ where: :$\map f {1, u} = \dfrac {2 + 3 u} {3 - 2 u}$ Hence: {{begin-eqn}} {{eqn | l = \ln z       | r = \int \frac {\d u} {\dfrac {2 + 3 u} {3 - 2 u} - u} + C       | c =  }} {{eqn | r = \int \frac {\paren {3 - 2 u} \rd z} {2 + 3 u - u \paren {3 - 2 u} } + C       | c =  }} {{eqn | r = \int \frac {\paren {3 - 2 u} \rd v} {2 \paren {1 + u^2} } + C       | c =  }} {{eqn | r = \dfrac 3 2 \int \dfrac {\d u} {1 + u^2} - \int \dfrac {u \rd u} {1 + u^2} + C       | c =  }} {{eqn | r = \dfrac 3 2 \arctan u - \dfrac 1 2 \, \map \ln {1 + u^2} + C       | c =  }} {{end-eqn}} Replacing all the substitutions: :$\map \ln {x - 1} = \dfrac 3 2 \, \map \arctan {\dfrac {y + 1} {x - 1} } - \dfrac 1 2 \, \map \ln {1 + \paren {\dfrac {y + 1} {x - 1} } } + C$ Tidying up and reassigning constants appropriately: :$3 \, \map \arctan {\dfrac {y + 1} {x - 1} } = \map \ln {\paren {y + 1}^2 + \paren {x - 1}^2} + C$ {{qed}}
We divide through by $x$ to show that $(1)$ is homogeneous: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d y}       | r = \frac {\sqrt {x^2 + y^2} } x       | c =  }} {{eqn | r = \sqrt {\frac {x^2 + y^2} {x^2} }       | c =  }} {{eqn | r = \sqrt {1 + \paren {\frac y x}^2}       | c =  }} {{end-eqn}} By Solution to Homogeneous Differential Equation, its solution is: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ where: :$\map f {x, y} = \dfrac {\sqrt {x^2 + y^2} } x$ Thus: {{begin-eqn}} {{eqn | l = \ln x       | r = \int \frac {\d z} {\sqrt {1 + z^2} - z}       | c =  }} {{eqn | r = \int \paren {\sqrt{1 + z^2} + z} \rd z       | c =  }} {{eqn | r = \frac {z \sqrt {1 + z^2} } 2 + \frac {\map \ln {\sqrt{1 + z^2} + z} } 2 + \frac {z^2} 2 + k       | c = Primitive of $\sqrt {x^2 + a^2}$ }} {{eqn | ll= \leadsto       | l = 2 \ln x       | r = z \sqrt {1 + z^2} + \map \ln {\sqrt{1 + z^2} + z} + z^2 + C       | c = where $C = 2 k$ }} {{eqn | r = \frac y x \sqrt {1 + \paren {\frac y x}^2} + \map \ln {\sqrt {1 + \paren {\frac y x}^2} + \frac y x} + \paren {\frac y x}^2 + C       | c =  }} {{eqn | r = \frac y x \sqrt {\frac {x^2 + y^2} {x^2} } + \map \ln {\sqrt {\frac {x^2 + y^2} {x^2} } + \frac y x} + \paren {\frac y x}^2 + C       | c =  }} {{eqn | r = \frac y x \frac {\sqrt {x^2 + y^2} } x + \map \ln {\frac {\sqrt {x^2 + y^2} } x + \frac y x} + \paren {\frac y x}^2 + C       | c =  }} {{eqn | ll= \leadsto       | l = 2 x^2 \ln x       | r = y \sqrt {x^2 + y^2} + x^2 \, \map \ln {y + \sqrt {x^2 + y^2} } - x^2 \ln x + y^2 + C x^2       | c =  }} {{eqn | ll= \leadsto       | l = 3 x^2 \ln x       | r = y \sqrt {x^2 + y^2} + x^2 \, \map \ln {y + \sqrt {x^2 + y^2} } + y^2 + C x^2       | c =  }} {{end-eqn}} {{qed}}
Dividing $(1)$ by $y^2$ and rearranging: :$(2): \quad \dfrac {\d x} {\d y} + \dfrac x y = \dfrac {x^3} {y^2}$ 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: :$\map P y = \dfrac 1 y$ :$\map Q y = \dfrac 1 {y^2}$ :$n = 3$ and so is an example of Bernoulli's equation. By Solution to Bernoulli's Equation it has the general solution: :$(3): \quad \displaystyle \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \, \map \mu y \rd y + C$ 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 - 3} \int \dfrac 1 y \rd y       | c =  }} {{eqn | r = -2 \ln y       | c =  }} {{eqn | r = \map \ln {y^{-2} }       | c =  }} {{eqn | ll= \leadsto       | l = \map \mu y       | r = e^{\map \ln {y^{-2} } }       | c =  }} {{eqn | r = \dfrac 1 {y^2}       | c =  }} {{end-eqn}} and so substituting into $(3)$: {{begin-eqn}} {{eqn | l = \frac 1 {y^2} \frac 1 {x^2}       | r = \paren {1 - 3} \int \frac 1 {y^2} \frac 1 {y^2} \rd y       | c =  }} {{eqn | r = -2 \int \dfrac {\d y} {y^4}       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 {y^2 x^2}       | r = -\dfrac 2 {-3} \dfrac 1 {y^3} + C       | c =  }} {{eqn | ll= \leadsto       | l = 3 y       | r = 2 x^2 + C x^2 y^2       | c =  }} {{end-eqn}} {{qed}}