hcju/proofwikiqa huggingface dataset:
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = 1$ :$\map R x = 2 \cos x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' + y = 0$ From Linear Second Order ODE: $y'' + y = 0$, this has the general solution: :$y_g = C_1 \sin x + C_2 \cos x$ We have that: :$\map R x = 2 \cos x$ and it is noted that $2 \cos x$ is a particular solution of $(2)$. So from the Method of Undetermined Coefficients for Sine and Cosine: :$y_p = A x \sin x + B x \cos x$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A x \sin x + B x \cos x       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = A x \cos x - B x \sin x + A \sin x + B \cos x       | c = Derivative of Sine Function, Derivative of Cosine Function }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = -A x \sin x - B x \cos x + A \cos x - B \sin x + A \cos x - B \sin x       | c = Power Rule for Derivatives }} {{eqn | r = -A x \sin x - B x \cos x + 2 A \cos x - 2 B \sin x       | c = Power Rule for Derivatives }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = -A x \sin x - B x \cos x + 2 A \cos x - 2 B \sin x + A x \sin x + B x \cos x       | r = 2 \cos x       | c =  }} {{eqn | ll= \leadsto       | l = -2 B \sin x       | r = 0       | c = equating coefficients }} {{eqn | l = 2 A \cos x       | r = 2 \cos x       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = 0       | c =  }} {{eqn | l = A       | r = 1       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \sin x + C_2 \cos x + x \sin x$ {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 - 2 m = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = 0$ :$m_2 = 2$ These are real and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 e^{0 x} + C_2 e^{2 x} = C_1 + C_2 e^{2 x}$ {{qed}} Category:Examples of Constant Coefficient Homogeneous LSOODEs inhed7gc83phylrmx05mo5zksgxz5cc
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 2$ :$q = 0$ :$\map R x = 12 x - 10$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - 2 y' = 0$ From Linear Second Order ODE: $y'' - 2 y' = 0$, this has the general solution: :$y_g = C_1 + C_2 e^{2 x}$ We have that: :$\map R x = 12 x - 10$ and it is noted that $12 x - 10$ is not itself a particular solution of $(2)$. So from the Method of Undetermined Coefficients for Polynomials: :$y_p = A_0 + A_1 x + A_2 x^2$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A_0 + A_1 x + A_2 x^2       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = A_1 + 2 A_2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = 2 A_2       | c = Power Rule for Derivatives }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = 2 A_2 - 2 \paren {A_1 + 2 A_2 x}       | r = 12 x - 10       | c =  }} {{eqn | ll= \leadsto       | l = -4 A_2 x       | r = 12 x       | c = equating coefficients }} {{eqn | l = 2 A_2 - 2 A_1       | r = -10       | c =  }} {{eqn | ll= \leadsto       | l = A_2       | r = -3       | c =  }} {{eqn | ll= \leadsto       | l = -6 - 2 A_1       | r = -10       | c =  }} {{eqn | ll= \leadsto       | l = -2 A_1       | r = -4       | c =  }} {{eqn | ll= \leadsto       | l = A_1       | r = 2       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 + C_2 e^{2 x} + 2 x - 3 x^2$ {{qed}}
{{begin-eqn}} {{eqn | n = 4       | l = {y_1}'' + \map P x {y_1}' + \map Q x y_1       | r = \map {R_1} x       | c = as $y_1$ is a particular solution to $(1)$ }} {{eqn | n = 5       | l = {y_2}'' + \map P x {y_2}' + \map Q x y_2       | r = \map {R_2} x       | c = as $y_2$ is a particular solution to $(2)$ }} {{eqn | ll= \leadsto       | l = \paren { {y_1}'' + {y_2}''} + \map P x \paren { {y_1}' + {y_2}'} + \map Q x \paren {y_1 + y_2}       | r = \map {R_1} x + \map {R_2} x       | c = adding $(4)$ and $(5)$ }} {{eqn | ll= \leadsto       | l = \paren {y_1 + y_2}'' + \map P x \paren {y_1 + y_2}' + \map Q x \paren {y_1 + y_2}       | r = \map {R_1} x + \map {R_2} x       | c = Linear Combination of Derivatives }} {{end-eqn}} Hence the result. {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = 4$ :$\map R x = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' + 4 y = 0$ From Linear Second Order ODE: $y'' + 4 y = 0$, this has the general solution: :$y_g = C_1 \sin 2 x + C_2 \cos 2 x$ We have that: :$\map R x = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$ Let: :$\map R x = \map {R_1} x + \map {R_2} x + \map {R_3} x$ where: :$\map {R_1} x = 4 \cos 2 x$ :$\map {R_2} x = 6 \cos x$ :$\map {R_3} x = 8 x^2 - 4 x$ Consider in turn the solutions to: :$y'' + p y' + q y = \map {R_1} x$ From Linear Second Order ODE: $y'' + 4 y = 4 \cos 2 x$, this has the general solution: :$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x$ :$y'' + p y' + q y = \map {R_2} x$ From Linear Second Order ODE: $y'' + 4 y = 6 \cos x$, this has the general solution: :$y = C_1 \sin 2 x + C_2 \cos 2 x + 2 \cos x$ :$y'' + p y' + q y = \map {R_3} x$ From Linear Second Order ODE: $y'' + 4 y = 8 x^2 - 4 x$, this has the general solution: :$y = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$ It follows from Combination of Solutions to Non-Homogeneous LSOODE with same Homogeneous Part that the general solution to $(1)$ is: :$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x + 2 \cos x - 1 - x + 2 x^2$ {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = 4$ :$\map R x = 4 \cos 2 x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' + 4 y = 0$ From Linear Second Order ODE: $y'' + 4 y = 0$, this has the general solution: :$y_g = C_1 \sin 2 x + C_2 \cos 2 x$ It is noted that $4 \cos 2 x$ is a particular solution of $(2)$. So from the Method of Undetermined Coefficients for Sine and Cosine: :$y_p = A x \sin 2 x + B x \cos 2 x$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A x \sin 2 x + B x \cos 2 x       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = 2 A x \cos 2 x - 2 B x \sin 2 x + A \sin 2 x + B \cos 2 x       | c = Derivative of Sine Function, Derivative of Cosine Function, Product Rule }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = -4 A x \sin 2 x - 4 B x \cos 2 x + 2 A \cos 2 x - 2 B \sin 2 x + 2 A \cos 2 x - 2 B \sin 2 x       | c = Derivative of Sine Function, Derivative of Cosine Function, Product Rule }} {{eqn | r = -4 A x \sin 2 x - 4 B x \cos 2 x + 4 A \cos 2 x - 4 B \sin 2 x       | c =  }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = -4 A x \sin 2 x - 4 B x \cos 2 x + 4 A \cos 2 x - 4 B \sin 2 x + 4 \paren {A x \sin 2 x + B x \cos 2 x}       | r = 4 \cos 2 x       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {4 - 4} x \sin 2 x - 4 B \sin 2 x       | r = 0       | c = equating coefficients }} {{eqn | l = B \paren {4 - 4} x \cos 2 x + 4 A \cos 2 x       | r = 4 \cos 2 x       | c =  }} {{eqn | ll= \leadsto       | l = - 4 B       | r = 0       | c =  }} {{eqn | l = 4 A       | r = 4       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = 0       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + x \sin 2 x$ is the general solution to $(1)$. {{qed}} Category:Examples of Constant Coefficient LSOODEs Category:Examples of Method of Undetermined Coefficients ckxfh56cvg4csyr4ue05k2mmlea9987
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = 4$ :$\map R x = 6 \cos x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' + 4 y = 0$ From Linear Second Order ODE: $y'' + 4 y = 0$, this has the general solution: :$y_g = C_1 \sin 2 x + C_2 \cos 2 x$ It is noted that $\cos x$ is not a particular solution of $(2)$. So from the Method of Undetermined Coefficients for Sine and Cosine: :$y_p = A \sin x + B \cos x$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A \sin x + B \cos x       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = A \cos x - B \sin x       | c = Derivative of Sine Function, Derivative of Cosine Function }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = -A \sin x - B \cos x       | c = Derivative of Sine Function, Derivative of Cosine Function }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = -A \sin x - B \cos x + 4 \paren {A \sin x + B \cos x}       | r = 6 \cos x       | c =  }} {{eqn | ll= \leadsto       | l = 3 A \sin x       | r = 0       | c = equating coefficients }} {{eqn | l = 3 B \cos x       | r = 6 \cos x       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = 0       | c =  }} {{eqn | l = B       | r = 2       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + 2 \cos x$ is the general solution to $(1)$. {{qed}} Category:Examples of Constant Coefficient LSOODEs Category:Examples of Method of Undetermined Coefficients i707s896lgr763ay1nl3jlne0l5pbdl
It can be seen that $\paren 1$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = 1$ :$\map R x = \csc x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$\paren 2: \quad y'' + y = 0$ From Linear Second Order ODE: $y'' + y = 0$, this has the general solution: :$y_g = C_1 \sin x + C_2 \cos x$ It remains to find a particular solution $y_p$ to $\paren 1$. 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 = \sin x       | c =  }} {{eqn | l = \map {y_2} x       | r = \cos x       | c =  }} {{eqn | ll= \leadsto       | l = \map {y_1'} x       | r = \cos x       | c = Derivative of Sine Function }} {{eqn | l = \map {y_2'} x       | r = -\sin 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 = \sin x \paren {- \sin x} - \cos x \cos x       | c =  }} {{eqn | r = -\paren {\sin^2 x + \cos^2 x}       | c =  }} {{eqn | r = -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 {\cos x \csc x} {-1} \rd x       | c =  }} {{eqn | r = \int \frac {\cos x} {\sin x} \rd x       | c = {{Defof|Cosecant}} }} {{eqn | r = \map \ln {\sin x}       | c = Primitive of $\cot x$ }} {{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 {\sin x \csc x} {-1} \rd x       | c =  }} {{eqn | r = -\int \frac {\sin x} {\sin x} \rd x       | c = {{Defof|Cosecant}} }} {{eqn | r = -x       | c =  }} {{end-eqn}} It follows that: :$y_p = \sin x \map \ln {\sin x} - x \cos x$ So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \sin x + C_2 \cos x + \sin x \map \ln {\sin x} - x \cos x$ is the general solution to $\paren 1$. {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 - 2 m + 1 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = m_2 = 1$ These are real and equal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 e^x + C_2 x e^x$ {{qed}} Category:Examples of Constant Coefficient Homogeneous LSOODEs c8k5v3yna87ldu4rn2mriauu6cb43cu
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: :$p = -2$ :$q = 1$ :$\map R x = 2 x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' - 2 y' + y = 0$ 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 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 2 x} {e^{2 x} } \rd x       | c =  }} {{eqn | r = \int -2 x^2 e^{- x} \rd x       | c =  }} {{eqn | r = 2 e^{-x} \paren {x^2 + 2 x + 2}       | c = Primitive of $x^2 e^{a x}$ }} {{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 2 x} {e^{2 x} } \rd x       | c =  }} {{eqn | r = \int 2 x e^{-x} \rd x       | c =  }} {{eqn | r = -2 e^{- x} \paren {x + 1}       | c = Primitive of $x e^{a x}$ }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = y_p       | r = 2 e^{-x} \paren {x^2 + 2 x + 2} e^x - 2 e^{-x} \paren {x + 1} x e^x       | c =  }} {{eqn | r = 2 x^2 + 4 x + 4 - 2 x^2 - 2 x       | c = simplifying }} {{eqn | r = 2 x + 4       | c = further simplifying }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^x + C_2 x e^x + 2 x + 4$ 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: :$p = -2$ :$q = 1$ :$\map R x = 2 x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' - 2 y' + y = 0$ 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 $(1)$. Taking the expression $\map R x = 2 x$ and differentiating twice {{WRT|Differentiation}} $x$: {{begin-eqn}} {{eqn | l = \map R x       | r = 2 x       | c =  }} {{eqn | ll= \leadsto       | l = \map {R'} x       | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = \map {R''} x       | r = 0       | c =  }} {{end-eqn}} Trying out $y = 2 x$ in $(1)$: :$0 - 4 + 2 x = 2 x - 4$ which is off by that constant of $4$. But by Derivative of Constant: :$4' = 0$ Hence setting $y = 2 x + 4$: :$0 - 4 + \paren {2 x + 4} = 2 x$ and it can be seen that $y_p = 2 x + 4$ is indeed a particular solution to $(1)$. So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^x + C_2 x e^x + 2 x + 4$ is the general solution to $(1)$. {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = -2$ :$q = 1$ :$\map R x = 2 x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' - 2 y' + y = 0$ From Linear Second Order ODE: $y'' - 2 y' + y = 0$, this has the general solution: :$y_g = C_1 e^x + C_2 x e^x$ We have that: :$R \left({x}\right) = 2 x$ So from the Method of Undetermined Coefficients for Polynomial: :$y_p = A_0 + A_1 x$ Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A_0 + A_1 x       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = A_1       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = 0       | c = Derivative of Constant }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = 0 - 2 A_1 + \paren {A_0 + A_1 x}       | r = 2 x       | c =  }} {{eqn | ll= \leadsto       | l = -2 A_1 + A_0       | r = 0       | c = equating coefficients }} {{eqn | ll= \leadsto       | l = A_1       | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = A_0       | r = 4       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^x + C_2 x e^x + 2 x + 4$ is the general solution to $(1)$. {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = -1$ :$q = -6$ :$\map R x = e^{-x}$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - y' - 6 y = 0$ From Linear Second Order ODE: $y'' - y' - 6 y = 0$, this has the general solution: :$y_g = C_1 e^{3 x} + C_2 e^{-2 x}$ We have that: :$\map R x = e^{-x}$ and so from the Method of Undetermined Coefficients for the Exponential function: :$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$ where: :$K = 1$ :$a = -1$ :$p = -1$ :$q = -6$ Hence: {{begin-eqn}} {{eqn | l = y_p       | r = \dfrac {e^{-x} } {1 + 1 - 6}       | c =  }} {{eqn | r = -\dfrac {e^{-x} } 4       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$ {{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: :$p = -1$ :$q = -6$ :$\map R x = e^{-x}$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - y' - 6 y = 0$ From Linear Second Order ODE: $y'' - y' - 6 y = 0$, this has the general solution: :$y_g = C_1 e^{3 x} + C_2 e^{-2 x}$ 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^{3 x}       | c =  }} {{eqn | l = \map {y_2} x       | r = e^{-2 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 = -2 e^{-2 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 {-2 e^{-2 x} } - e^{-2 x} \paren {3 e^{3 x} }       | c =  }} {{eqn | r = -2 e^x - 3 e^x       | c =  }} {{eqn | r = -5 e^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^{-2 x} \cdot e^{-x} } {-5 e^x} \rd x       | c =  }} {{eqn | r = \int \frac {e^{-4 x} } 5 \rd x       | c =  }} {{eqn | r = -\frac 1 {20} e^{-4 x}       | c = Primitive of $e^{a x}$ }} {{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^{3 x} \cdot e^{-x} } {-5 e^x} \rd x       | c =  }} {{eqn | r = \int -\frac {e^x} 5 \rd x       | c =  }} {{eqn | r = -\frac 1 5 e^x       | c = Primitive of $e^{a x}$ }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = y_p       | r = -\frac 1 {20} e^{-4 x} e^{3 x} - \frac 1 5 e^x e^{-2 x}       | c =  }} {{eqn | r = -\frac 1 {20} e^{-x} - \frac 1 5 e^{-x}       | c = simplifying }} {{eqn | r = -\frac 1 4 e^{-x}       | c = further simplifying }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$ is the general solution to $(1)$. {{qed}}
This follows by mutatis mutandis of the proof of Directed iff Finite Subsets have Upper Bounds. {{qed}}
Firstly, we have: {{begin-eqn}} {{eqn | n = 1       | l = \tan x       | r = \frac {2 \tan \frac x 2} {1 - \tan ^2 \frac x 2}       | c = Double Angle Formula for Tangent }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map \tan {\frac x 2 + \frac \pi 4}       | r = \frac {\tan \frac x 2 + \tan \frac \pi 4} {1 - \tan \frac x 2 \tan \frac \pi 4}       | c = Tangent of Sum }} {{eqn | r = \frac {\tan \frac x 2 + 1} {1 - \tan \frac x 2}       | c = Tangent of $45 \degrees$ }} {{eqn | r = \frac {\paren {\tan \frac x 2 + 1} \paren {\tan \frac x 2 + 1} } {\paren {1 - \tan \frac x 2} \paren {\tan \frac x 2 + 1} } }} {{eqn | r = \frac {\tan^2 \frac x 2 + 2 \tan \frac x 2 + 1} {1 - \tan^2 \frac x 2}       | c = Difference of Two Squares, Square of Sum }} {{eqn | r = \frac {2 \tan \frac x 2} {1 - \tan^2 \frac x 2} + \frac {\tan^2 \frac x 2 + 1} {1 - \tan^2 \frac x 2} }} {{eqn | r = \tan x + \frac {\tan^2 \frac x 2 + 1} {1 - \tan^2 \frac x 2}       | c = Double Angle Formula for Tangent: see $(1)$ above }} {{eqn | r = \tan x + \frac {\sin^2 \frac x 2 + \cos^2 \frac x 2} {\cos^2 \frac x 2 - \sin^2 \frac x 2}       | c = multiplying Denominator and Numerator by $\cos^2 \frac x 2$ }} {{eqn | r = \tan x + \frac {\sin^2 \frac x 2 + \cos^2 \frac x 2} {\cos 2 \frac x 2}       | c = Double Angle Formula for Cosine }} {{eqn | r = \tan x + \frac 1 {\cos x}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \tan x + \sec x       | c = Secant is Reciprocal of Cosine }} {{end-eqn}} {{qed}} Category:Trigonometric Identities Category:Tangent Function Category:Secant Function 0wwp1p2ickye49m0s6kfoq8lv5b8409