hcju/proofwikiqa huggingface dataset:
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: :$m_1 = 1 + \sqrt 6$ :$m_2 = 1 - \sqrt 6$ So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 \, \map \exp {\paren {1 + \sqrt 6} x} + C_2 \, \map \exp {\paren {1 - \sqrt 6} x}$ {{qed}} Category:Examples of Constant Coefficient Homogeneous LSOODEs 4x3uefk7dz7qyoq1ta764cjytli7mrl
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: :$p = -2$ :$q = -5$ :$\map R x = 2 \cos 3 x - \sin 3 x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - 2 y' - 5 y = 0$ From Linear Second Order ODE: $y'' - 2 y' - 5 y = 0$, this has the general solution: :$y_g = C_1 \, \map \exp {\paren {1 + \sqrt 6} x} + C_2 \, \map \exp {\paren {1 - \sqrt 6} x}$ We have that: :$\map R x = 2 \cos 3 x - \sin 3 x$ and it is noted that $2 \cos 3 x - \sin 3 x$ is not itself a particular solution of $(2)$. We then determine the particular solution: === Particular Solution === {{:Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x/Particular Solution}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \, \map \exp {\paren {1 + \sqrt 6} x} + C_2 \, \map \exp {\paren {1 - \sqrt 6} x} + \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$ {{qed}} Category:Examples of Constant Coefficient LSOODEs Category:Examples of Method of Undetermined Coefficients Category:Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x qzfckbucapgfbjsdr9zjnj493l10hje
From Linear Second Order ODE: $y'' - 2 y' - 5 y = 0$, we have established that the general solution to $(1)$ is: :$y_g = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x}$ === Solution using Trigonometric Form === {{:Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x/Particular Solution/Trigonometric Form}} === Solution using Exponential Form === {{:Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x/Particular Solution/Exponential Form}} Category:Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x Category:Examples of Method of Undetermined Coefficients gd3j1neepe2jlry4mk53bj7k6cpwbj4
We note that $2 \cos 3 x - \sin 3 x$ is not itself a particular solution of $(2)$. From the Method of Undetermined Coefficients for Sine and Cosine: :$y_p = A \cos 3 x + B \sin 3 x$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A \cos 3 x + B \sin 3 x       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = -3 A \sin 3 x + 3 B \cos 3 x       | c = Derivative of Sine Function, Derivative of Cosine Function }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = -9 A \cos 3 x - 9 B \sin 3 x       | c = Power Rule for Derivatives }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = -9 A \cos 3 x - 9 B \sin 3 x - 2 \paren {-3 A \sin 3 x + 3 B \cos 3 x} - 5 \paren {A \cos 3 x + B \sin 3 x}       | r = 2 \cos 3 x - \sin 3 x       | c =  }} {{eqn | ll= \leadsto       | l = -9 B \sin 3 x + 6 A \sin 3 x - 5 B \sin 3 x       | r = \sin 3 x       | c = equating coefficients }} {{eqn | l = -9 A \cos 3 x - 6 B \cos 3 x - 5 A \cos 3 x       | r = 2 \cos 3 x       | c =  }} {{eqn | ll= \leadsto       | l = 6 A - 14 B       | r = 1       | c =  }} {{eqn | l = -6 B - 14 A       | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {14^2 + 6^2} A       | r = -28 - 6       | c =  }} {{eqn | l = \paren {14^2 + 6^2} B       | r = -12 + 14       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = -\dfrac {17} {116}       | c =  }} {{eqn | l = B       | r = \dfrac 1 {116}       | c =  }} {{end-eqn}} Hence the result: :$y_p = \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$ {{qed}}
We note that $2 \cos 3 x - \sin 3 x$ is not itself a particular solution of $(2)$. From the Method of Undetermined Coefficients for Sine and Cosine: :$y_p = A \cos 3 x + B \sin 3 x$ where $A$ and $B$ are to be determined. The {{RHS}} of $(1)$ is the real part of $\paren {2 + i} e^{3 i x}$. Thus, to find a particular solution of $(1)$ when the {{RHS}} is $\paren {2 + i} e^{3 i x}$, we substitute $y = A e^{3 i x}$. Hence: {{begin-eqn}} {{eqn | l = A \paren {\paren {3 i}^2 - 2 \paren {3 i} - 5}       | r = 2 + i       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = -\dfrac {2 + i} {14 + 6 i}       | c =  }} {{eqn | r = -\dfrac {\paren {2 + i} \paren {14 - 6 i} } {14^2 + 6^2}       | c =  }} {{eqn | r = -\dfrac {34 + 2 i} {232}       | c =  }} {{eqn | r = -\dfrac 1 {116} \paren {17 + i}       | c =  }} {{end-eqn}} Taking the real part of the corresponding solution $A e^{3 i x}$, we get: {{begin-eqn}} {{eqn | l = y       | r = \map \Re {-\dfrac 1 {116} \paren {17 + i} \paren {\cos 3 x + i \sin 3 x} }       | c =  }} {{eqn | r = -\dfrac 1 {116} \paren {17 \cos 3 x - \sin 3 x}       | c =  }} {{end-eqn}} Hence the result: :$y_p = \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$ {{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: :$p = 0$ :$q = 4$ :$\map R x = 3 \sin 2 x$ 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 $3 \sin 2 x$ 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 = 3 \sin 2 x       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {4 - 4} x \sin 2 x - 4 B \sin 2 x       | r = 3 \sin 2 x       | c = equating coefficients }} {{eqn | l = B \paren {4 - 4} x \cos 2 x + 4 A \cos 2 x       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = -4 B       | r = 3       | c =  }} {{eqn | l = 4 A       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = -\dfrac 3 4       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x - \dfrac 3 4 x \cos 2 x$ is the general solution to $(1)$. {{qed}}
Let $\sequence {x_n} $ be a Cauchy sequence in $V$. Then by definition: :$\forall \epsilon \in \R_{\gt 0}: \exists N \in \N : \forall n, m \ge N: \norm {x_n - x_m} < \epsilon$ Let $N$ satisfy: :$\forall n, m \ge N: \norm {x_n - x_m} < 1$ Let $m = N + 1 > N$ Then $\forall n \ge N$: {{begin-eqn}} {{eqn | l = \norm {x_n}       | r = \norm {x_n - x_{N + 1} + x_{N + 1} } }} {{eqn | o = \le       | r = \norm {x_n - x_{N + 1} } + \norm {x_{N + 1} }       | c = Norm axiom (N3) (Triangle Inequality). }} {{eqn | o = \le       | r = 1 + \norm {x_{N + 1} } }} {{end-eqn}} Let $M = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_N}, 1 + \norm {x_{N + 1} } }$. Then: :$\forall n < N: \norm {x_n} \le M$ :$\forall n \ge N: \norm {x_n} \le 1 + \norm {x_{N + 1} } \le M$ It follows by definition that $\sequence {x_n}$ is bounded. {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 - 4 m - 5 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = 5$ :$m_2 = -1$ These are real and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 e^{5 x} + C_2 e^{-x}$ {{qed}} Category:Examples of Constant Coefficient Homogeneous LSOODEs ejqnhyp24d5wga3b2167bk1yoapt1u4
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: :$p = -4$ :$q = -5$ :$\map R x = x^2$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' - 4 y' - 5 y = 0$ From Linear Second Order ODE: $y'' - 4 y' - 5 y = 0$, this has the general solution: :$y_g = C_1 e^{5 x} + C_2 e^{-x}$ We have that: :$\map R x = x^2$ and it is noted that $x^2$ is not itself a particular solution of $(2)$. So from the Method of Undetermined Coefficients for Polynomials: :$y_p = A_0 + A_1 x + A_2 x^2$ for $A_n$ 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 - 4 \paren {A_1 + 2 A_2 x} - 5 \paren {A_0 + A_1 x + A_2 x^2}       | r = x^2       | c =  }} {{eqn | ll= \leadsto       | l = -5 A_2 x^2       | r = x^2       | c = equating coefficients of powers }} {{eqn | l = \paren {-8 A_2 - 5 A_1} x       | r = 0       | c =  }} {{eqn | l = 2 A_2 - 4 A_1 - 5 A_0       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A_2       | r = -\dfrac 1 5       | c =  }} {{eqn | l = 5 A_1       | r = \dfrac 8 5       | c =  }} {{eqn | ll= \leadsto       | l = A_1       | r = \dfrac 8 {25}       | c =  }} {{eqn | l = 5 A_0       | r = -\dfrac 2 5 - \dfrac {4 \times 8} {25}       | c =  }} {{eqn | ll= \leadsto       | l = A_0       | r = -\dfrac {42} {125}       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^{5 x} + C_2 e^{-x} - \dfrac 1 5 x^2 + \dfrac 8 {25} x - \dfrac {42} {125}$ Differentiating {{WRT|Differentiation}} $x$: :$y' = 5 C_1 e^{5 x} - C_2 e^{-x} - \dfrac 2 5 x + \dfrac 8 {25}$ Substituting the initial conditions, remembering that $e^0 = 1$: {{begin-eqn}} {{eqn | l = 1       | r = C_1 + C_2 - \dfrac {42} {125}       | c = setting $y = 1$ when $x = 0$ }} {{eqn | l = -1       | r = 5 C_1 - C_2 + \dfrac 8 {25}       | c = setting $y' = -1$ when $x = 0$ }} {{eqn | ll= \leadsto       | l = -\dfrac 2 {125} + 6 C_1       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = C_1       | r = \dfrac 1 {375}       | c =  }} {{eqn | ll= \leadsto       | l = C_2       | r = 1 - 5 A + \dfrac 8 {25}       | c =  }} {{eqn | r = 1 + \dfrac 1 {75} + \dfrac 8 {27}       | c =  }} {{eqn | r = \dfrac 4 3       | c =  }} {{end-eqn}} The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \paren {D^2 - 2 D + 1} y       | r = \dfrac 1 {1 + e^x}       | c = expressing $(1)$ in a different form }} {{eqn | ll= \leadsto       | l = \paren {D^2 - 1}^2 y       | r = \dfrac 1 {1 + e^x}       | c = and in a different form again }} {{eqn | ll= \leadsto       | l = \paren {D - 1} z       | r = \dfrac 1 {1 + e^x}       | c = putting $\paren {D - 1} y = z$ }} {{eqn | ll= \leadsto       | l = \map D {z e^{-x} }       | r = \dfrac {e^{-x} } {1 + e^x}       | c = Solution to Linear First Order ODE with Constant Coefficients: integrating factor $e^x$ }} {{eqn | r = \dfrac {e^{-2 x} } {1 + e^{-x} }       | c = multiplying top and bottomof {{RHS}} by $e^{-x}$ }} {{eqn | ll= \leadsto       | l = z e^{-x}       | r = \int \dfrac {e^{-2 x} } {1 + e^{-x} } \rd x       | c = integrating both sides {{WRT|Integration}} $x$ }} {{eqn | r = -\int \dfrac u {1 + u} \rd u       | c = Integration by Substitution: setting $e^{-x} = u$ }} {{eqn | r = -\int 1 - \dfrac 1 {1 + u} \rd u       | c =  }} {{eqn | r = -u + \map \ln {1 + u}       | c = Primitive of Constant, Primitive of Reciprocal }} {{eqn | r = -e^{-x} + \map \ln {1 + e^{-x} }       | c = substituting back for $u$ }} {{end-eqn}} Then we need to solve: {{begin-eqn}} {{eqn | l = \paren {D - 1}y       | r = z       | c =  }} {{eqn | ll= \leadsto       | l = \map D {y e^{-x} }       | r = z e^{-x}       | c =  }} {{eqn | ll= \leadsto       | l = y e^{-x}       | r = e^{-x} + \int \map \ln {1 + e^{-x} } \rd x       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = 1 + e^x \int \map \ln {1 + e^{-x} } \rd x       | c =  }} {{end-eqn}} and so on. {{Qed}}
First we solve the reduced equation of $(1)$: :$(2): \quad \paren {D^4 - 1} y' = 0$ The auxiliary equation of $(1)$ is: :$(3): \quad: m^4 - 1 = 0$ From Complex $4$th Roots of Unity, the roots of $(2)$ are: :$m_1 = 1$ :$m_2 = i$ :$m_3 = -1$ :$m_4 = -i$ So from Solution of Constant Coefficient Linear nth Order ODE, the general solution of $(2)$ is: :$y_g = C_1 e^x + C_2 e^{-x} + C_3 \sin x + C_4 \cos x$ Because $\sin x$ is already a solution of $(2)$, we try: :$y = A x e^{i x}$ with a view to taking the real part in due course. {{begin-eqn}} {{eqn | l = y       | r = A x e^{i x}       | c =  }} {{eqn | ll= \leadsto       | l = y'       | r = A e^{i x} + A i x e^{i x}       | c =  }} {{eqn | r = A \paren {1 + i x} e^{i x}       | c =  }} {{eqn | ll= \leadsto       | l = y''       | r = A \paren {1 + i x} i e^{i x} + A i e^{i x}       | c =  }} {{eqn | r = A \paren {2 i - x} e^{i x}       | c =  }} {{eqn | ll= \leadsto       | l = y^{\paren 3}       | r = A \paren {2 i - x} i e^{i x} - A e^{i x}       | c =  }} {{eqn | r = A \paren {-3 - i x} e^{i x}       | c =  }} {{eqn | ll= \leadsto       | l = y^{\paren 4}       | r = A \paren {-3 - i x} i e^{i x} - i e^{i x}       | c =  }} {{eqn | r = A \paren {x - 4 i} e^{i x}       | c =  }} {{end-eqn}} It follows that :$-4 i A - 1$ and so: :$A = -\dfrac 1 4$ Thus a particular solution to $(1)$ can be given as: {{begin-eqn}} {{eqn | l = y_p       | r = \map \Im {\dfrac 1 4 i x e^{i x} }       | c =  }} {{eqn | r = \dfrac {x \cos x} 4       | c =  }} {{end-eqn}} and the general solution is: :$y = C_1 e^x + C_2 e^{-x} + C_3 \sin x + C_4 \cos x + \dfrac {x \cos x} 4$ {{Qed}}
=== Necessary Condition === Let $x$ be a $p$-adic unit. Then: :$x \in \Z_p$ :$x^{-1} \in \Z_p$ By definition of the $p$-adic integers: :$\norm x_p \le 1$ :$\norm {x^{-1} }_p \le 1$ From Norm of Inverse in Division Ring: :$\norm x_p \ge 1$ It follows that: :$\norm x_p = 1$ {{qed|lemma}} === Sufficient Condition === Let $\norm x_p = 1$. From Norm of Inverse in Division Ring: :$\norm {x^{-1}}_p = 1$ By definition of the $p$-adic integers: :$x \in \Z_p$ :$x^{-1} \in \Z_p$ It follows that $x$ is invertible elements in $\Z_p$. That is, x is a $p$-adic unit. {{qed}}
From P-adic Norm of p-adic Number is Power of p, there exists $v \in \Z$ such that $\norm a_p = p^{-v}$. Let $n = -v$. Now: {{begin-eqn}} {{eqn | l = \norm{p^n a}_p       | r = \norm{p^n }_p \norm a_p       | c = Norm axiom (N2) (Multiplicativity) }} {{eqn | r = p^{-n} \norm a_p       | c = {{Defof|P-adic Norm|$p$-adic norm}} on the rational numbers }} {{eqn | r = p^{-n} p^{n}       | c = Definition of $n$ }} {{eqn | r = 1 }} {{eqn | ll= \leadsto       | l = p^n a       | o = \in       | r = \Z_p^\times       | c = P-adic Unit has Norm Equal to One }} {{end-eqn}} {{qed}} Category:P-adic Number Theory bvsxpju69y6jresixmmdtu81uqybml4
We look for solutions to $(1)$ and $(2)$ of the form: {{begin-eqn}} {{eqn | n = 3       | l = y       | r = A e^{k x} }} {{eqn | n = 4       | l = z       | r = B e^{k x} }} {{end-eqn}} We do of course have the Trivial Solution of Homogeneous Linear 1st Order ODE: :$y = z = 0$ which happens when $A = B = 0$. So let us investigate solutions where either or both of $A$ and $B$ are non-zero. Substituting $(3)$ and $(4)$ into $(1)$ and $(2)$ and cancelling $e^{k x}$, we get:: {{begin-eqn}} {{eqn | n = 5       | l = \paren {k + a} A + b B       | r = 0 }} {{eqn | n = 6       | l = c A + \paren {k + d} B       | r = 0 }} {{end-eqn}} From $(5)$ and $(6)$ we get: {{begin-eqn}} {{eqn | n = 7       | o =        | r = \paren {\paren {k + a} \paren {k + d} - b c} A       | c =  }} {{eqn | r = \paren {\paren {k + a} \paren {k + d} - b c} B       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So $A = B = 0$ unless $k$ is a root of the quadratic equation: :$\paren {k + a} \paren {k + d} - b c = 0$ That is: :$(8): \quad \begin {vmatrix} k + a & b \\ c & k + d \end {vmatrix} = 0$ where the above notation denotes the determinant. Assume $(8)$ has distinct roots $k_1$ and $k_2$. Taking $k = k_1$ and $k = k_2$ in $(7)$, we can obtain ratios $A_1 : B_1$ and $A_2 : B_2$ such that: {{begin-eqn}} {{eqn | l = y       | r = A_1 e^{k_1 x} }} {{eqn | l = z       | r = B_1 e^{k_1 x} }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y       | r = A_2 e^{k_2 x} }} {{eqn | l = z       | r = B_2 e^{k_2 x} }} {{end-eqn}} are solutions of $(1)$ and $(2)$. By taking arbitrary linear combinations of these, we obtain the general solution. {{finish|Cover the case where $k_1 {{=}} k_2$. The source work is vague on this subject. Recommend this solution be reworked, preferably in conjunction with a more rigorous and thorough source work than the one used here.}}
Using the technique of Solution to Simultaneous Homogeneous Linear First Order ODEs with Constant Coefficients, we calculate the roots of the quadratic equation: :$\paren {k + a} \paren {k + d} - b c = 0$ where: {{begin-eqn}} {{eqn | l = a       | r = -3 }} {{eqn | l = b       | r = 2 }} {{eqn | l = c       | r = 4 }} {{eqn | l = d       | r = -1 }} {{end-eqn}} That is: :$\paren {k - 3} \paren {k - 1} - 8 = 0$ or: :$k^2 - 4 k - 5 = 0$ This has roots: {{begin-eqn}} {{eqn | l = k_1       | r = 5 }} {{eqn | l = k_2       | r = -1 }} {{end-eqn}} We also obtain: {{begin-eqn}} {{eqn | l = \paren {k - 3} A  + 2 B       | r = 0 }} {{eqn | l = 4 A + \paren {k - 1} B       | r = -1 }} {{end-eqn}} When $k = 5$ we get that $A + B = 0$. When $k = -1$ we get that $2 A - B = 0$. This provides us with the solutions: {{begin-eqn}} {{eqn | l = y       | r = e^{5 x} }} {{eqn | l = z       | r = e^{-5 x} }} {{end-eqn}} or: {{begin-eqn}} {{eqn | l = y       | r = e^{-x} }} {{eqn | l = z       | r = 2 e^{-x} }} {{end-eqn}} From these, the general solution is constructed: {{begin-eqn}} {{eqn | l = y       | r = C_1 e^{5 x} + C_2 e^{-x} }} {{eqn | l = z       | r = -C_1 e^{5 x} + 2 C_2 e^{-x} }} {{end-eqn}} {{qed}}
$(1)$ is a linear first order ODE in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = -1$ :$\map Q x = x^2$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = -\int \rd x       | c =  }} {{eqn | r = -x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = e^{-x}       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: {{begin-eqn}} {{eqn | l = \map {\dfrac {\d} {\d x} } {y e^{-x} }       | r = x^2 e^{-x}       | c =  }} {{eqn | ll= \leadsto       | l = y e^{-x}       | r = \int x^2 e^{-x} \rd x       | c =  }} {{eqn | r = \frac {e^{-x} } {-1} \paren {x^2 - \frac {2 x} {-1} + \frac 2 {\paren {-1}^2} } + C       | c = Primitive of $x^2 e^{a x}$ for $a = -1$ }} {{eqn | r = -e^{-x} \paren {x^2 + 2 x + 2} + C       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = C e^x - \paren {x^2 + 2 x + 2}       | c =  }} {{end-eqn}} {{qed}}
$(1)$ is a linear first order ODE in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = -1$ :$\map Q x = e^x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x       | r = -\int \rd x       | c =  }} {{eqn | r = -x       | c =  }} {{eqn | ll= \leadsto       | l = e^{\int P \rd x}       | r = e^{-x}       | c =  }} {{end-eqn}} Thus from Solution by Integrating Factor, $(1)$ can be rewritten as: {{begin-eqn}} {{eqn | l = \map {\dfrac {\d} {\d x} } {y e^{-x} }       | r = e^x e^{-x}       | c =  }} {{eqn | ll= \leadsto       | l = y e^{-x}       | r = \int 1 \rd x       | c =  }} {{eqn | r = x + C       | c = Primitive of Constant }} {{eqn | n = 2       | ll= \leadsto       | l = y       | r = e^x \paren {x + C}       | c =  }} {{end-eqn}} Substituting the initial condition into $(2)$: {{begin-eqn}} {{eqn | l = 0       | r = e^0 \paren {0 + C}       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = C       | c =  }} {{end-eqn}} which leads to the particular solution: :$y = x e^x$ {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad m^2 + 4 m + 5 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = -2 + i$ :$m_2 = -2 - i$ So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x}$ {{qed}} Category:Examples of Constant Coefficient Homogeneous LSOODEs 9rdn4vnv8cesz9ynydv5pgja0y0y0ax
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: :$p = 4$ :$q = 5$ :$\map R x = 2 e^{-2 x}$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' + 4 y' + 5 y = 0$ From Linear Second Order ODE: $y'' + 4 y' + 5 y = 0$, this has the general solution: :$y_g = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x}$ It remains to find a particular solution $y_p$ to $(1)$. We have that: :$\map R x = 2 e^{-2 x}$ 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: :$K = 2$ :$a = -2$ :$p = 4$ :$q = 5$ Hence: {{begin-eqn}} {{eqn | l = y_p       | r = \dfrac {2 e^{-2 x} } {\paren {-2}^2 + 4 \times \paren {-2} + 5}       | c =  }} {{eqn | r = \dfrac {2 e^{-2 x} } {4 - 8 + 5}       | c =  }} {{eqn | r = 2 e^{-2 x}       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x + 2}$ is the general solution to $(1)$. {{qed}}