hcju/proofwikiqa huggingface dataset:
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = 4$ :$\map R x = \tan 2 x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' + 4 y = 0$ From Linear Second Order ODE: $y'' + 4 y = 0$, this has the general solution: :$y_g = C_1 \cos 2 x + C_2 \sin 2 x$ It remains to find a particular solution $y_p$ to $(1)$. Expressing $y_g$ in the form: :$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$ we have: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = \cos 2 x       | c =  }} {{eqn | l = \map {y_2} x       | r = \sin 2 x       | c =  }} {{eqn | ll= \leadsto       | l = \map { {y_1}'} x       | r = -2 \sin 2 x       | c = Derivative of Sine Function }} {{eqn | l = \map { {y_2}'} x       | r = 2 \cos 2 x       | c = Derivative of Cosine Function }} {{end-eqn}} By the Method of Variation of Parameters, we have that: :$y_p = v_1 y_1 + v_2 y_2$ where: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | l = v_2       | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{end-eqn}} where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$. We have that: {{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = y_1 {y_2}' - y_2 {y_1}'       | c = {{Defof|Wronskian}} }} {{eqn | r = \cos 2 x \paren {2 \cos 2 x} - \sin 2 x \paren {-2 \sin 2 x}       | c =  }} {{eqn | r = 2 \paren {\cos^2 2 x + \sin^2 2 x}       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int -\frac {\sin 2 x \tan 2 x} 2 \rd x       | c =  }} {{eqn | r = -\frac 1 2 \int \frac {\sin^2 2 x} {\cos 2 x} \rd x       | c =  }} {{eqn | r = -\frac 1 2 \paren {-\frac {\sin 2 x} 2 + \frac 1 2 \ln \map \tan {x + \frac \pi 4} }       | c = Primitive of $\dfrac {\sin^2 a x} {\cos a x}$ }} {{eqn | r = \frac 1 4 \paren {\sin 2 x - \map \ln {\sec 2 x + \tan 2 x} }       | c = Tangent of Half Angle plus $\dfrac \pi 4$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = v_2       | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int \frac {\cos 2 x \tan 2 x} 2 \rd x       | c =  }} {{eqn | r = \frac 1 2 \int \sin 2 x \rd x       | c =  }} {{eqn | r = \frac 1 2 \paren {-\frac {\cos 2 x} 2}       | c = Primitive of $\sin a x$ }} {{eqn | r = -\frac 1 4 \cos 2 x       | c =  }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = y_p       | r = \frac 1 4 \paren {\sin 2 x - \map \ln {\sec 2 x + \tan 2 x} } \cos 2 x - \frac 1 4 \cos 2 x \sin 2 x       | c =  }} {{eqn | r = -\frac 1 4 \cos 2 x \, \map \ln {\sec 2 x + \tan 2 x}       | c = simplifying }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \cos 2 x + C_2 \sin 2 x - \dfrac 1 4 \cos 2 x \, \map \ln {\sec 2 x + \tan 2 x}$ is the general solution to $(1)$. {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 2$ :$q = 1$ :$\map R x = e^{-x} \ln x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' + 2 y' + y = 0$ From Linear Second Order ODE: $y'' + 2 y' + y = 0$, this has the general solution: :$y_g = C_1 e^{-x} + C_2 x e^{-x}$ It remains to find a particular solution $y_p$ to $(1)$. Expressing $y_g$ in the form: :$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$ we have: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = e^{-x}       | c =  }} {{eqn | l = \map {y_2} x       | r = x e^{-x}       | c =  }} {{eqn | ll= \leadsto       | l = \map { {y_1}'} x       | r = -e^{-x}       | c = Derivative of Exponential Function }} {{eqn | l = \map { {y_2}'} x       | r = -x e^{-x} + e^{-x}       | c = Derivative of Exponential Function, Product Rule for Derivatives }} {{end-eqn}} By the Method of Variation of Parameters, we have that: :$y_p = v_1 y_1 + v_2 y_2$ where: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | l = v_2       | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{end-eqn}} where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$. We have that: {{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = y_1 {y_2}' - y_2 {y_1}'       | c = {{Defof|Wronskian}} }} {{eqn | r = e^{-x} \paren {-x e^{-x} + e^{-x} } - x e^{-x} \paren {-e^{-x} }       | c =  }} {{eqn | r = -x e^{-2 x} + e^{-2 x} + x e^{-2 x}       | c =  }} {{eqn | r = e^{-2 x}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int -\frac {x e^{-x} e^{-x} \ln x} {e^{-2 x} } \rd x       | c =  }} {{eqn | r = -\int x \ln x \rd x       | c =  }} {{eqn | r = -\frac {x^2} 2 \paren {\ln x - \frac 1 2}       | c = Primitive of $x \ln x$ }} {{eqn | r = \frac {x^2} 4 - \frac {x^2 \ln x} 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = v_2       | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int \frac {e^{-x} e^{-x} \ln x} {e^{-2 x} } \rd x       | c =  }} {{eqn | r = \int \ln x \rd x       | c =  }} {{eqn | r = x \ln x - x       | c = Primitive of $\ln x$ }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = y_p       | r = \paren {\frac {x^2} 4 - \frac {x^2 \ln x} 2} e^{-x} + \paren {x \ln x - x} x e^{-x}       | c =  }} {{eqn | r = \frac {x^2 e^{-x} } 4 - \frac {x^2 e^{-x} \ln x} 2 + x^2 e^{-x} \ln x - x^2 e^{-x}       | c = simplifying }} {{eqn | r = \frac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}       | c = simplifying }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^{-x} + C_2 x e^{-x} - \dfrac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}$ is the general solution to $(1)$. {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 - 2 m - 3 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = 3$ :$m_2 = -1$ These are real and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 e^{3 x} + C_2 e^{-x}$ {{qed}} Category:Examples of Constant Coefficient Homogeneous LSOODEs i6015diugly2x7vd35jz1fws8und39v
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form: :$y'' + p y' + q y = \map R x$ where: :$p = -2$ :$q = -3$ :$\map R x = 64 x e^{-x}$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - 2 y' - 3 y = 0$ From Linear Second Order ODE: $y'' - 2 y' - 3 y = 0$, this has the general solution: :$y_g = C_1 e^{3 x} + C_2 e^{-x}$ It remains to find a particular solution $y_p$ to $(1)$. Expressing $y_g$ in the form: :$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$ we have: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = e^{3 x}       | c =  }} {{eqn | l = \map {y_2} x       | r = e^{-x}       | c =  }} {{eqn | ll= \leadsto       | l = \map { {y_1}'} x       | r = 3 e^{3 x}       | c = Derivative of Exponential Function }} {{eqn | l = \map { {y_2}'} x       | r = -e^{-x}       | c = Derivative of Exponential Function }} {{end-eqn}} By the Method of Variation of Parameters, we have that: :$y_p = v_1 y_1 + v_2 y_2$ where: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | l = v_2       | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{end-eqn}} where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$. We have that: {{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = y_1 {y_2}' - y_2 {y_1}'       | c = {{Defof|Wronskian}} }} {{eqn | r = e^{3 x} \paren {-e^{-x} } - e^{-x} \paren {3 e^{3 x} }       | c =  }} {{eqn | r = -e^{2 x} - 3 e^{2 x}       | c =  }} {{eqn | r = -4 e^{2 x}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int -\frac {e^{-x} 64 x e^{-x} } {-4 e^{2 x} } \rd x       | c =  }} {{eqn | r = 16 \int x e^{-4 x} \rd x       | c =  }} {{eqn | r = 16 \frac {e^{-4 x} } {-4} \paren {x - \frac 1 {-4} }       | c = Primitive of $x e^{a x}$ }} {{eqn | r = -4 e^{-4 x} \paren {x + \frac 1 4}       | c =  }} {{eqn | r = -4 x e^{-4 x} - e^{-4 x}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = v_2       | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int \frac {e^{3 x} 64 x e^{-x} } {-4 e^{2 x} } \rd x       | c =  }} {{eqn | r = - 16 \int x \rd x       | c =  }} {{eqn | r = -8 x^2       | c = Primitive of Power }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = y_p       | r = \paren {-4 x e^{-4 x} - e^{-4 x} } e^{3 x} + \paren {-8 x^2} e^{-x}       | c =  }} {{eqn | r = -4 x e^{-x} - e^{-x} - 8 x^2 e^{-x}       | c = simplifying }} {{eqn | r = -e^{-x} \paren {8 x^2 + 4 x + 1}       | c = simplifying }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-x} - e^{-x} \paren {8 x^2 + 4 x + 1}$ is the general solution to $(1)$. {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad m^2 + 2 m + 5 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = -1 + 2 i$ :$m_2 = -1 - 2 i$ So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$ {{qed}} Category:Examples of Constant Coefficient Homogeneous LSOODEs tn73yr2w61wblyer24fduyyaxuq2kvc
Let $y, z \in \set x$. By definition of singleton: :$ y = x \land z = x$ By definition of reflexivity: :$y \precsim x \land z \precsim x$ Thus: :$\exists h \in \set x: y \precsim h \land z \precsim h$ Thus by definition: :$\set x$ is a directed subset of $S$. $\set x$ is a filtered subset of $S$ by mutatis mutandis. {{qed}}
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: :$(2): \quad y'' - \dfrac {2 x} {x^2 - 1} y' + \dfrac 2 {x^2 - 1} y = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = - \dfrac {2 x} {x^2 - 1}$ :$\map Q x = \dfrac 2 {x^2 - 1}$ 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 {2 x} {x^2 - 1} } \rd x       | c =  }} {{eqn | r = -\map \ln {x^2 - 1}       | c = Primitive of Function under its Derivative }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{\map \ln {x^2 - 1} }       | c =  }} {{eqn | r = x^2 - 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} \paren {x^2 - 1} \rd x       | c =  }} {{eqn | r = \int \paren {1 - \dfrac 1 {x^2} } \rd x       | c =  }} {{eqn | r = x + \frac 1 x       | c = Primitive of Power }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = \paren {x + \frac 1 x} x       | c =  }} {{eqn | r = x^2 + 1       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 x + C_2 \paren {x^2 + 1}$ {{qed}} Category:Examples of Homogeneous LSOODEs qg4jptdxydcikqwktj9ed8dkcpb9j3n
$(1)$ can be manipulated into the form: :$y'' - \dfrac {2 x} {x^2 - 1} y' + \dfrac 2 {x^2 - 1} y = x^2 - 1$ It can be seen that this is a nonhomogeneous linear second order ODE in the form: :$y'' + \map P x y' + \map Q x y = \map R x$ where: :$\map P x = -\dfrac {2 x} {x^2 - 1}$ :$\map Q x = \dfrac 2 {x^2 - 1}$ :$\map R x = x^2 - 1$ First we establish the solution of the corresponding homogeneous linear second order ODE: :$\paren {x^2 - 1} y'' - 2 x y' + 2 y = 0$ From Second Order ODE: $\paren {x^2 - 1} y'' - 2 x y' + 2 y = 0$, this has the general solution: :$y_g = C_1 x + C_2 \paren {x^2 + 1}$ It remains to find a particular solution $y_p$ to $(1)$. Expressing $y_g$ in the form: :$y_g = C_1 \map {y_1} x + C_2 \map {y_2} x$ we have: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = x       | c =  }} {{eqn | l = \map {y_2} x       | r = x^2 + 1       | c =  }} {{eqn | ll= \leadsto       | l = \map { {y_1}'} x       | r = 1       | c = Power Rule for Derivatives }} {{eqn | l = \map { {y_2}'} x       | r = 2 x       | c = Power Rule for Derivatives }} {{end-eqn}} By the Method of Variation of Parameters, we have that: :$y_p = v_1 y_1 + v_2 y_2$ where: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | l = v_2       | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{end-eqn}} where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$. We have that: {{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = y_1 {y_2}' - y_2 {y_1}'       | c = {{Defof|Wronskian}} }} {{eqn | r = x \paren {2 x} - \paren {x^2 + 1} \paren 1       | c =  }} {{eqn | r = 2 x^2 - x^2 - 1       | c =  }} {{eqn | r = x^2 - 1       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int -\frac {\paren {x^2 + 1} \paren {x^2 - 1} } {x^2 - 1} \rd x       | c =  }} {{eqn | r = -\int \paren {x^2 + 1} \rd x       | c =  }} {{eqn | r = -\frac {x^3} 3 - x       | c = Primitive of Power }} {{end-eqn}} {{begin-eqn}} {{eqn | l = v_2       | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int \frac {x \paren {x^2 - 1} } {x^2 - 1} \rd x       | c =  }} {{eqn | r = \int x \rd x       | c =  }} {{eqn | r = \frac {x^2} 2       | c = Primitive of Power }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = y_p       | r = \paren {-\frac {x^3} 3 - x} x + \paren {\frac {x^2} 2} \paren {x^2 + 1}       | c =  }} {{eqn | r = -\frac {x^4} 3 - x^2 + \frac {x^4} 2 + \frac {x^2} 2       | c = rearranging }} {{eqn | r = \frac {x^4} 6 - \frac {x^2} 2       | c = rearranging }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 x + C_2 \paren {x^2 + 1} + \dfrac {x^4} 6 - \dfrac {x^2} 2$ is the general solution to $(1)$. {{qed}}
=== Sufficient Condition === Let us assume that :$H$ is directed. Let $x, y \in H^\precsim$ By definition of lower closure: :$\exists x' \in H: x \precsim x'$ and :$\exists y' \in H: y \precsim y'$ By definition of directed subset: :$\exists z \in H: x' \precsim z \land y' \precsim z$ By definition of reflexivity; :$z \precsim z$ By definition of lower closure: :$z \in H^\precsim$ Thus by definition of transitivity :$\exists z \in H^\precsim: x \precsim z \land y \precsim z$ Thus by definition :$H^\precsim$ is directed. {{qed|lemma}} === Necessary Condition === Let us assume that :$H^\precsim$ is directed. Let $x, y \in H$. By definition of reflexivity; :$x \precsim x \land y \precsim y$ By definition of lower closure: :$x, y \in H^\precsim$ By definition of directed subset: :$\exists z \in H^\precsim: x \precsim z \land y \precsim z$ By definition of lower closure: :$\exists z' \in H: z \precsim z'$ Thus by definition of transitivity :$\exists z' \in H: x \precsim z' \land y \precsim z'$ Thus by definition :$H$ is directed. {{qed}}
This follows by mutatis mutandis of the proof of Directed iff Lower Closure Directed. {{qed}}
$(1)$ can be manipulated into the form: :$y'' + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = x + 1$ It can be seen that this is a nonhomogeneous linear second order ODE in the form: :$y'' + \map P x y' + \map Q x y = \map R x$ where: :$\map P x = \dfrac {2 - x^2} {x^2 + x}$ :$\map Q x = -\dfrac {2 + x} {x^2 + x}$ :$\map R x = x + 1$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$\paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = 0$ From Linear Second Order ODE: $\paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = 0$, this has the general solution: :$y_g = C_1 e^x + \dfrac {C_2} x$ It remains to find a particular solution $y_p$ to $(1)$. Expressing $y_g$ in the form: :$y_g = C_1 \map {y_1} x + C_2 \map {y_2} x$ we have: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = e^x       | c =  }} {{eqn | l =\map {y_2} x       | r = \dfrac 1 x       | c =  }} {{eqn | ll= \leadsto       | l = \map { {y_1}'} x       | r = e^x       | c = Derivative of Exponential Function }} {{eqn | l = \map { {y_2}'} x       | r = -\dfrac 1 {x^2}       | c = Power Rule for Derivatives }} {{end-eqn}} By the Method of Variation of Parameters, we have that: :$y_p = v_1 y_1 + v_2 y_2$ where: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | l = v_2       | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{end-eqn}} where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$. We have that: {{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = y_1 {y_2}' - y_2 {y_1}'       | c = {{Defof|Wronskian}} }} {{eqn | r = e^x \paren {-\dfrac 1 {x^2} } - \paren {\dfrac 1 x} \paren {e^x}       | c =  }} {{eqn | r = -\dfrac {e^x} {x^2} - \dfrac {e^x} x       | c =  }} {{eqn | r = -e^x \, \dfrac {x + 1} {x^2}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int -\frac {\paren {\dfrac 1 x} \paren {x + 1} } {-e^x \, \dfrac {x + 1} {x^2} } \rd x       | c =  }} {{eqn | r = \int x e^{-x} \rd x       | c =  }} {{eqn | r = -e^{-x} \paren {x + 1}       | c = Primitive of $x e^{a x}$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = v_2       | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int \frac {e^x \paren {x + 1} } {-e^x \, \dfrac {x + 1} {x^2} } \rd x       | c =  }} {{eqn | r = -\int x^2 \rd x       | c =  }} {{eqn | r = -\frac {x^3} 3       | c = Primitive of Power }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = y_p       | r = \paren {-e^{-x} \paren {x + 1} } e^x + \paren {-\frac {x^3} 3} \paren {\dfrac 1 x}       | c =  }} {{eqn | r = -x - 1 - \frac {x^2} 3       | c = rearranging }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^x + \dfrac {C_2} x - x - 1 - \dfrac {x^2} 3$ is the general solution to $(1)$. {{qed}}
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = \frac 1 x       | c =  }} {{eqn | ll= \leadsto       | l = {y_1}'       | r = -\frac 1 {x^2}       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_1}''       | r = \frac 2 {x^3}       | c = Power Rule for Derivatives }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | o =        | r = \paren {x^2 + x} {y_1}'' + \paren {2 - x^2} {y_1}' - \paren {2 + x} y       | c =  }} {{eqn | r = 2 \frac {x^2 + x} {x^3} - \frac {2 - x^2} {x^2} - \frac {2 + x} x       | c =  }} {{eqn | r = \frac 2 x + \frac 2 {x^2} - \frac 2 {x^2} + 1 - \frac 2 x - 1       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} Thus: :$y_1 = \dfrac 1 x$ is a particular solution of $(1)$. $(1)$ can be expressed as: :$(2): \quad y'' + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = \dfrac {2 - x^2} {x^2 + x}$ :$\map Q x = \dfrac {2 + x} {x^2 + x}$ 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^2} {x^2 + x} } \rd x       | c =  }} {{eqn | r = \int \dfrac 2 {x \paren {x + 1} } \rd x - \int \paren {\dfrac x {x + 1} } \rd x       | c =  }} {{eqn | r = 2 \, \map \ln {\frac x {x + 1} } - \int \paren {\dfrac x {x + 1} } \rd x       | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$ }} {{eqn | r = 2 \, \map \ln {\frac x {x + 1} } - x + \, \map \ln {x + 1}       | c = Primitive of $\dfrac x {a x + b}$ }} {{eqn | r = \map \ln {\frac {x^2} {x + 1} } - x       | c =  }} {{eqn | r = \map \ln {\frac {e^{-x} x^2} {x + 1} }       | c =  }} {{eqn | r = -\map \ln {\frac {e^x \paren {x + 1} } {x^2} }       | c =  }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{\map \ln {\frac {e^x \paren {x + 1} } {x^2} } }       | c =  }} {{eqn | r = \frac {e^x \paren {x + 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 x^2 \paren {\frac {e^x \paren {x + 1} } {x^2} } \rd x       | c =  }} {{eqn | r = \int e^x \paren {x + 1} \rd x       | c =  }} {{eqn | r = \int x e^x \rd x + \int e^x \rd x       | c =  }} {{eqn | r = e^x \paren {x - 1} + \int e^x \rd x       | c = Primitive of $x e^{a x}$ }} {{eqn | r = e^x \paren {x - 1} + e^x       | c = Primitive of $e^{a x}$ }} {{eqn | r = x e^x       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = \paren {x e^x} \frac 1 x       | c =  }} {{eqn | r = e^x       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 e^x + \dfrac {C_2} x$ {{qed}}