hcju/proofwikiqa huggingface dataset:
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = \sin x       | c =  }} {{eqn | ll= \leadsto       | l = y'       | r = \cos x       | c = Derivative of Sine Function }} {{eqn | ll= \leadsto       | l = y''       | r = -\sin x       | c = Derivative of Cosine Function }} {{end-eqn}} and so: :$y_1 = x$ is a particular solution of $(1)$. $(1)$  is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = 0$ :$\map Q x = 1$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int 0 \rd x       | c =  }} {{eqn | r = k       | c = where $k$ is arbitrary }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-k}       | c =  }} {{eqn | r = C       | c =  where $C$ is also arbitrary }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \dfrac 1 {\sin^2 x} C \rd x       | c =  }} {{eqn | r = \int C \csc^2 x \rd x       | c = {{Defof|Cosecant}} }} {{eqn | r = -C \cot x       | c = Primitive of $\csc^2 x$ }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = \sin x \paren {-C \cot x}       | c =  }} {{eqn | r = -C \cos x       | c = {{Defof|Cotangent}}, and simplifying }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 \sin x + k \paren {-C \cos x}$ where $k$ is arbitrary. Setting $C_2 = - k C$ yields the result: :$y = C_1 \sin x + C_2 \cos x$ {{qed}}
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = e^x       | c =  }} {{eqn | ll= \leadsto       | l = y'       | r = e^x       | c = Derivative of Exponential Function }} {{eqn | ll= \leadsto       | l = y''       | r = e^x       | c = Derivative of Exponential Function }} {{end-eqn}} and so by inspection: :$y_1 = e^x$ is a particular solution of $(1)$. $(1)$  is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = 0$ :$\map Q x = -1$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int 0 \rd x       | c =  }} {{eqn | r = k       | c = where $k$ is arbitrary }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-k}       | c =  }} {{eqn | r = C       | c =  where $C$ is also arbitrary }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \dfrac 1 {e^{2 x} } C \rd x       | c =  }} {{eqn | r = \int C e^{-2 x} \rd x       | c =  }} {{eqn | r = -\frac {C e^{-2 x} } 2       | c = Primitive of $e^{a x}$ }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = e^x \paren {-\frac C 2 e^{-2 x} }       | c =  }} {{eqn | r = -\frac C 2 e^{-x}       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 \sin x + k \paren {-\dfrac C 2 e^{-x} }$ where $k$ is arbitrary. Setting $C_2 = - \dfrac {k C} 2$ yields the result: :$y = C_1 e^x + C_2 e^{-x}$ {{qed}}
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = y'       | r = 0       | c = Derivative of Constant }} {{eqn | ll= \leadsto       | l = y''       | r = 0       | c = Derivative of Constant }} {{end-eqn}} and so by inspection: :$y_1 = 1$ is a particular solution of $(1)$. $(1)$ can be expressed as: :$(2): \quad y'' + \dfrac 3 x y' = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = \dfrac 3 x$ :$\map Q x = 0$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int \dfrac 3 x \rd x       | c =  }} {{eqn | r = 3 \ln x       | c = Primitive of Reciprocal }} {{eqn | r = \ln x^3       | c =  }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-\ln x^3}       | c =  }} {{eqn | r = \frac 1 {x^3}       | c =   }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \frac 1 {x^3} \rd x       | c =  }} {{eqn | r = -\frac 1 {2 x^2}       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = -\frac 1 {2 x^2}       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 + k \paren {-\frac 1 {2 x^2} }$ where $k$ is arbitrary. Setting $C_2 = - \dfrac k 2$ yields the result: :$y = C_1 + \dfrac {C_2} {x^2}$ {{qed}} </onlyinclude>
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = {y_1}'       | r = 1       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_1}''       | r = 0       | c = Derivative of Constant }} {{end-eqn}} and so by inspection: :$y_1 = x$ is a particular solution of $(1)$. $(1)$ can be expressed as: :$(2): \quad y'' - \dfrac {2 x} {1 - x^2} y' + \dfrac 2 {1 - x^2} y = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = - \dfrac {2 x} {1 - x^2}$ :$\map Q x = \dfrac 2 {1 - x^2}$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int \paren {-\dfrac {2 x} {1 - x^2} } \rd x       | c =  }} {{eqn | r = \map \ln {1 - x^2}       | c = Primitive of Function under its Derivative }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-\map \ln {1 - x^2} }       | c =  }} {{eqn | r = \frac 1 {1 - x^2}       | c =   }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \dfrac 1 {x^2} \frac 1 {1 - x^2} \rd x       | c =  }} {{eqn | r = \int \dfrac 1 {x^2 \paren {1 - x^2} } \rd x       | c =  }} {{eqn | r = -\frac 1 x + \frac 1 2 \map \ln {\frac {1 + x} {1 - x} }       | c = Primitive of $\dfrac 1 {x^2 \paren {a^2 - x^2} }$ }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = \paren {-\frac 1 x + \frac 1 2 \map \ln {\frac {1 + x} {1 - x} } } x       | c =  }} {{eqn | r = -1 + \frac x 2 \map \ln {\frac {1 + x} {1 - x} }       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 x + C_2 \paren {\dfrac x 2 \, \map \ln {\dfrac {1 + x} {1 - x} } - 1}$ {{qed}}
=== Particular Solution === {{:Bessel's Equation/x^2 y'' + x y' + (x^2 - (1 over 4)) y = 0/Particular Solution}}{{qed|lemma}} $(1)$ can be expressed as: :$(2): \quad y'' + \dfrac 1 x y' + \paren {1 - \dfrac 1 {4 x^2} } y = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = 1 - \dfrac 1 {4 x^2}$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int \dfrac 1 x \rd x       | c =  }} {{eqn | r = \ln x       | c = Primitive of Reciprocal }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-\ln x}       | c =  }} {{eqn | r = \frac 1 x       | c =   }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \dfrac x {\sin^2 x} \frac 1 x \rd x       | c =  }} {{eqn | r = \int \csc^2 x \rd x       | c =  }} {{eqn | r = \cot x        | c = Primitive of $\csc^2 x$ }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = \paren {-\cot x} \dfrac {\sin x} {\sqrt x}       | c =  }} {{eqn | r = -\dfrac {\cos x} {\sqrt x}       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 \dfrac {\sin x} {\sqrt x} + C_2 \dfrac {\cos x} {\sqrt x}$ {{qed}}
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = \frac {\sin x} {\sqrt x}       | c =  }} {{eqn | r = x^{-1/2} \sin x       | c =  }} {{eqn | ll= \leadsto       | l = {y_1}'       | r = x^{-1/2} \cos x - \frac 1 2 x^{-3/2} \sin x        | c = Product Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_1}''       | r = -x^{-1/2} \sin x - \frac 1 2 x^{-3/2} \cos x - \frac 1 2 x^{-3/2} \cos x + \frac 3 4 x^{-5/2} \sin x        | c = Product Rule for Derivatives }} {{eqn | r = \left({\frac 3 4 x^{-5/2} - x^{-1/2} }\right) \sin x - x^{-3/2} \cos x        | c =  }} {{eqn | r = \left({\frac 3 {4 x^2} - 1}\right) x^{-1/2} \sin x - x^{-3/2} \cos x        | c =  }} {{end-eqn}} Inserting into $(1)$, and gradually simplifying: {{begin-eqn}} {{eqn | o =        | r = x^2 \left({\left({\frac 3 {4 x^2} - 1}\right) x^{-1/2} \sin x - x^{-3/2} \cos x}\right) + x \left({x^{-1/2} \cos x - \frac 1 2 x^{-3/2} \sin x}\right) + \left({x^2 - \dfrac 1 4}\right) x^{-1/2} \sin x       | c =  }} {{eqn | r = \left({\frac 3 4 - x^2}\right) x^{-1/2} \sin x - x^{1/2} \cos x + x^{1/2} \cos x - \frac 1 2 x^{-1/2} \sin x + x^{3/2} \sin x - \dfrac 1 4 x^{-1/2} \sin x       | c =  }} {{eqn | r = \frac 3 4 x^{-1/2} \sin x - x^{3/2} \sin x - \frac 1 2 x^{-1/2} \sin x + x^{3/2} \sin x - \dfrac 1 4 x^{-1/2} \sin x       | c =  }} {{eqn | r = \frac 3 4 x^{-1/2} \sin x - \frac 1 2 x^{-1/2} \sin x - \dfrac 1 4 x^{-1/2} \sin x       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} hence demonstrating that: :$y_1 = \dfrac {\sin x} {\sqrt x}$ is a particular solution of $(1)$.
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = {y_1}'       | r = 1       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_1}''       | r = 0       | c = Derivative of Constant }} {{end-eqn}} and so by inspection: :$y_1 = x$ is a particular solution of $(1)$. $(1)$ can be expressed as: :$(2): \quad y'' - \dfrac x {x - 1} y' + \dfrac 1 {x - 1} y = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = -\dfrac x {x - 1}$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int \paren {-\dfrac x {x - 1} } \rd x       | c =  }} {{eqn | r = -x - \map \ln {x - 1}       | c = Primitive of $\dfrac x {a x + b}$ }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{x + \map \ln {x - 1} }       | c =  }} {{eqn | r = e^x \paren {x - 1}       | c =   }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \dfrac 1 {x^2} e^x \paren {x - 1} \rd x       | c =  }} {{eqn | r = \int \paren {\dfrac {e^x} x - \dfrac {e^x} {x^2} } \rd x       | c =  }} {{eqn | r = \int \dfrac {e^x} x \rd x - \int \dfrac {e^x} {x^2} \rd x       | c =  }} {{eqn | r = \int \dfrac {e^x} x \rd x + \frac {e^x} x - \int \frac {e^x} x \rd x       | c = Primitive of $\dfrac {e^{a x} } {x^n}$ }} {{eqn | r = \frac {e^x} x       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = \paren {\frac {e^x} x} x       | c =  }} {{eqn | r = e^x       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 x + C_2 e^x$ {{qed}} </onlyinclude>
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = {y_1}'       | r = 1       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_1}''       | r = 0       | c = Derivative of Constant }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = y'' - x \, \map f x y' + \map f x y       | r = 0 - x \, \map f x 1 + \map f x x       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} and so it has been demonstrated that: :$y_1 = x$ is a particular solution of $(1)$. $(1)$ is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = -x \, \map f x$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int \paren {-x \, \map f x} \rd x       | c =  }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-\int \paren {-x \, \map f x} \rd x}       | c =  }} {{eqn | r = e^{\int x \, \map f x \rd x}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \dfrac 1 {x^2} e^{\int x \, \map f x \rd x} \rd x       | c = as $y_1 = x$ }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = x \int x^{-2} e^{\int x \, \map f x \rd x} \rd x       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$\displaystyle y = C_1 x + C_2 x \int x^{-2} e^{\int x \, \map f x \rd x} \rd x$ {{qed}}
Note that: :$x - \paren {2 x + 1} + \paren {x + 1} = 0$ so if $y'' = y' = y$ we find that $(1)$ is satisfied. So: {{begin-eqn}} {{eqn | l = y_1       | r = e^x       | c =  }} {{eqn | ll= \leadsto       | l = {y_1}'       | r = e^x       | c = Derivative of Exponential Function }} {{eqn | ll= \leadsto       | l = {y_1}''       | r = e^x       | c = Derivative of Exponential Function }} {{end-eqn}} and so: :$y_1 = e^x$ is a particular solution of $(1)$. $(1)$ can be expressed as: :$(2): \quad y'' - \dfrac {2 x + 1} x y' + \dfrac {x + 1} x y = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = -\dfrac {2 x + 1} x$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int \paren {-\dfrac {2 x + 1} x} \rd x       | c =  }} {{eqn | r = \int \paren {-2 - \dfrac 1 x} \rd x       | c =  }} {{eqn | r = -2 x - \ln x       | c =  }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-\paren {-2 x - \ln x} }       | c =  }} {{eqn | r = e^{2 x + \ln x}       | c =  }} {{eqn | r = e^{2 x} e^{\ln x}       | c =  }} {{eqn | r = x e^{2 x}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \dfrac 1 {e^{2 x} } x e^{2 x} \rd x       | c = as $y_1 = e^x$ }} {{eqn | r = \int x \rd x       | c =  }} {{eqn | r = \frac {x^2} 2       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = \frac {x^2} 2 e^x       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 e^x + k \dfrac {x^2} 2 e^x$ and so setting $C_2 = \dfrac k 2$: :$y = C_1 e^x + C_2 x^2 e^x$ {{qed}}
Note that: :$1 - \map f x + \paren {\map f x - 1} = 0$ so if $y'' = y' = y$ we find that $(1)$ is satisfied. So: {{begin-eqn}} {{eqn | l = y_1       | r = e^x       | c =  }} {{eqn | ll= \leadsto       | l = {y_1}'       | r = e^x       | c = Derivative of Exponential Function }} {{eqn | ll= \leadsto       | l = {y_1}''       | r = e^x       | c = Derivative of Exponential Function }} {{end-eqn}} and so: :$y_1 = e^x$ is a particular solution of $(1)$. $(1)$ is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = -\map f x$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int \paren {-\map f x} \rd x       | c =  }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-\int \paren {-\map f x} \rd x}       | c =  }} {{eqn | r = e^{\int \map f x \rd x}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \dfrac 1 {e^{2 x} } e^{\int \map f x \rd x} \rd x       | c = as $y_1 = e^x$ }} {{eqn | r = \int e^{-2 x + \int \map f x \rd x} \rd x       | c = as $y_1 = e^x$ }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = e^x \int e^{-2 x + \int \map f x \rd x} \rd x       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$\displaystyle y = C_1 e^x + C_2 e^x \int e^{-2 x + \int \map f x \rd x} \rd x$ {{qed}}
Consider the equation: : $y = e^{m_1 x}$ Differentiating {{WRT|Differentiation}} $x$: {{begin-eqn}} {{eqn | l = y       | r = e^{m_1 x}       | c =  }} {{eqn | l = y'       | r = m_1 e^{m_1 x}       | c =  }} {{eqn | l = y''       | r = {m_1}^2 e^{m_1 x}       | c =  }} {{end-eqn}} === Sufficient Condition === Let $y = e^{m_1 x}$ be a solution to $(1)$. Substituting for $y$ and its derivatives in $(1)$: {{begin-eqn}} {{eqn | l = y'' + p y' + q y       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = {m_1}^2 e^{m_1 x} + p m_1 e^{m_1 x} + q e^{m_1 x}        | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren { {m_1}^2 + p m_1 + q} e^{m_1 x}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = {m_1}^2 + p m_1 + q       | r = 0       | c = as $e^{m_1 x}$ is never zero }} {{end-eqn}} That is, by definition, $m_1$ is a root of $m^2 + p m + q = 0$. {{qed|lemma}} === Necessary Condition === Let $m_1$ be a root of $m^2 + p m + q = 0$. Thus: {{begin-eqn}} {{eqn | l = {m_1}^2 + p m_1 + q       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren { {m_1}^2 + p m_1 + q} e^{m_1 x}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = {m_1}^2 e^{m_1 x} + p m_1 e^{m_1 x} + q e^{m_1 x}        | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = y'' + p y' + q y       | r = 0       | c = substituting for $y$ and its derivatives }} {{end-eqn}} Thus $y = e^{m_1 x}$ satisfies $(1)$. {{qed}}
=== Real Roots of Auxiliary Equation === {{:Solution of Constant Coefficient Homogeneous LSOODE/Real Roots of Auxiliary Equation}}{{qed|lemma}} === Equal Real Roots of Auxiliary Equation === {{:Solution of Constant Coefficient Homogeneous LSOODE/Equal Real Roots of Auxiliary Equation}}{{qed|lemma}} === Complex Roots of Auxiliary Equation === {{:Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation}}{{qed}}
Consider the auxiliary equation of $(1)$: :$(2): \quad m^2 + p m + q$ Let $p^2 > 4 q$. From Solution to Quadratic Equation with Real Coefficients, $(2)$ has two real roots: {{begin-eqn}} {{eqn | l = m_1       | r = -\frac p 2 + \sqrt {\frac {p^2} 4 - q} }} {{eqn | l = m_2       | r = -\frac p 2 - \sqrt {\frac {p^2} 4 - q} }} {{end-eqn}} As $p^2 > 4 q$ we have that: :$\sqrt {\dfrac {p^2} 4 - q} \ne 0$ and so: :$m_1 \ne m_2$ From Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation: {{begin-eqn}} {{eqn | l = y_1       | r = e^{m_1 x} }} {{eqn | l = y_2       | r = e^{m_2 x} }} {{end-eqn}} are both particular solutions to $(1)$. We also have that: {{begin-eqn}} {{eqn | l = \frac {y_1} {y_2}       | r = \frac {e^{m_1 x} } {e^{m_2 x} } }} {{eqn | r = e^{\paren {m_1 - m_2} x} }} {{eqn | o = \ne       | r = 0       | c = as $m_1 \ne m_2$ }} {{end-eqn}} Thus $y_1$ and $y_2$ are linearly independent. It follows from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution that: :$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$ is the general solution to $(1)$.
Consider the auxiliary equation of $(1)$: :$(2): \quad m^2 + p m + q$ Let $p^2 < 4 q$. From Solution to Quadratic Equation with Real Coefficients, $(2)$ has two complex roots: {{begin-eqn}} {{eqn | l = m_1       | r = -\frac p 2 + i \sqrt {q - \frac {p^2} 4} }} {{eqn | l = m_2       | r = -\frac p 2 - i \sqrt {q - \frac {p^2} 4} }} {{end-eqn}} As $p^2 < 4 q$ we have that: :$\sqrt {q - \dfrac {p^2} 4} \ne 0$ and so: :$m_1 \ne m_2$ Let: {{begin-eqn}} {{eqn | l = m_1       | r = a + i b }} {{eqn | l = m_2       | r = a - i b }} {{end-eqn}} where $a = -\dfrac p 2$ and $b = \sqrt {q - \dfrac {p^2} 4}$. From Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation: {{begin-eqn}} {{eqn | l = y_a       | r = e^{m_1 x} }} {{eqn | l = y_b       | r = e^{m_2 x} }} {{end-eqn}} are both particular solutions to $(1)$. We can manipulate $y_a$ and $y_b$ into the following forms: {{begin-eqn}} {{eqn | l = y_a       | r = e^{m_1 x}       | c =  }} {{eqn | r = e^{\paren {a + i b} x}       | c =  }} {{eqn | r = e^{a x} e^{i b x}       | c =  }} {{eqn | n = 3       | r = e^{a x} \paren {\cos b x + i \sin b x}       | c = Euler's Formula }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y_b       | r = e^{m_2 x}       | c =  }} {{eqn | r = e^{\paren {a - i b} x}       | c =  }} {{eqn | r = e^{a x} e^{-i b x}       | c =  }} {{eqn | n = 4       | r = e^{a x} \paren {\cos b x - i \sin b x}       | c = Euler's Formula: Corollary }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = y_a + y_b       | r = e^{a x} \paren {\cos b x + i \sin b x} + e^{a x} \paren {\cos b x - i \sin b x}       | c = adding $(3)$ and $(4)$ }} {{eqn | r = 2 e^{a x} \cos b x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {y_a + y_b} 2       | r = e^{a x} \cos b x       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = y_b - y_a       | r = e^{a x} \paren {\cos b x - i \sin b x} - e^{a x} \paren {\cos b x + i \sin b x}       | c = subtracting $(4)$ from $(3)$ }} {{eqn | r = 2 e^{a x} \sin b x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {y_b - y_a} 2       | r = e^{a x} \sin b x       | c =  }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = y_1       | r = \frac {y_a + y_b} 2 }} {{eqn | r = e^{a x} \cos b x       | c =  }} {{eqn | l = y_2       | r = \frac {y_b - y_a} 2 }} {{eqn | r = e^{a x} \sin b x       | c =  }} {{end-eqn}} We have that: {{begin-eqn}} {{eqn | l = \frac {y_1} {y_2}       | r = \frac {e^{a x} \cos b x} {e^{a x} \sin b x} }} {{eqn | r = \cot b x }} {{end-eqn}} As $\cot b x$ is not zero for all $x$, $y_1$ and $y_2$ are linearly independent. From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE: :$y_1 = \dfrac {y_a + y_b} 2$ :$y_2 = \dfrac {y_b - y_b} 2$ are both particular solutions to $(1)$. It follows from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution that: :$y = C_1 e^{a x} \cos b x + C_2 e^{a x} \sin b x$ or: :$y = e^{a x} \paren {C_1 \cos b x + C_2 \sin b x}$ is the general solution to $(1)$.