Method of Undetermined Coefficients/Exponential
== Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be of the form of an exponential function: :$\map R x = K e^{a x}$ The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$.  Let $\map R x = K e^{a x}$. Consider the auxiliary equation to $(1)$: :$(2): \quad m^2 + p m + q = 0$ There are three cases to consider. ;$a$ is not a root of $(2)$ Assume that there is a particular solution to $(1)$ of the form: :$y_p = A e^{a x}$ We have: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} y_p       | r = a A e^{a x}       | c = Derivative of Exponential Function }} {{eqn | l = \frac {\d^2} {\d x^2} y_p       | r = a^2 A e^{a x}       | c = Derivative of Exponential Function }} {{end-eqn}} Inserting into $(1)$: {{begin-eqn}} {{eqn | l = a^2 A e^{a x} + a p A e^{a x} + q A e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A \left({a^2 + p a + q}\right) e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \frac K {a^2 + p a + q}       | c =  }} {{end-eqn}} Hence: :$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$ From Solution of Constant Coefficient Homogeneous LSOODE, $y_g$ depends on whether $(2)$ has equal or unequal roots. Let $m_1$ and $m_2$ be the roots of $(2)$. Then: :$y = \begin{cases} C_1 e^{m_1 x} + C_2 e^{m_2 x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 \ne m_2: m_1, m_2 \in \R \\ C_1 e^{m_1 x} + C_2 x e^{m_1 x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 = m_2 \\ e^{r x} \paren {C_1 \cos s x + C_2 \sin s x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 = r + i s, m_2 = r - i s \end{cases}$ is the general solution to $(1)$. {{qed|lemma}} ;$a$ is a root of $(2)$ If $a$ is a root of $(2)$, then $a^2 + p a + q = 0$ and so $\dfrac {K e^{a x} } {a^2 + p a + q}$ is undefined. Let the auxiliary equation to $(2)$ have two unequal real roots $a$ and $b$. Assume that there is a particular solution to $(1)$ of the form: :$y_p = A x e^{a x}$ We have: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} y_p       | r = a A x e^{a x} + A e^{a x}       | c = Product Rule and Derivative of Exponential Function }} {{eqn | l = \frac {\d^2} {\d x^2} y_p       | r = a^2 A x e^{a x} + a A e^{a x} + a A e^{a x}       | c = Product Rule and Derivative of Exponential Function }} {{eqn | r = a^2 A x e^{a x} + 2 a A e^{a x}       | c =  }} {{end-eqn}} Inserting into $(1)$: {{begin-eqn}} {{eqn | l = a^2 A x e^{a x} + 2 a A e^{a x} + p \paren {a A x e^{a x} + A e^{a x} } + q A x e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {a^2 + p a + q} x e^{a x} + A \paren {2 a + p} e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {2 a + p} e^{a x}       | r = K e^{a x}       | c = as $a^2 + p a + q = 0$ }} {{eqn | ll= \leadsto       | l = A       | r = \frac K {2 a + p}       | c =  }} {{end-eqn}} Hence: :$y_p = \dfrac {K x e^{a x} } {2 a + p}$ and so from Solution of Constant Coefficient Homogeneous LSOODE: :$y = C_1 e^{a x} + C_2 e^{b x} + \dfrac {K x e^{a x} } {2 a + p}$ is the general solution to $(1)$. {{qed|lemma}} ;$a$ is a repeated root of $(2)$ If the auxiliary equation to $(2)$ has two equal real roots $a$, then: :$a = - \dfrac p 2$ and so not only: :$a^2 + p a + q = 0$ but also: :$2 a + p = 0$ and so neither of the above expressions involving $e^{a x}$ and $x e^{a x}$ will work as particular solution to $(1)$. So, assume that there is a particular solution to $(1)$ of the form: :$y_p = A x^2 e^{a x}$ We have: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} y_p       | r = a A x^2 e^{a x} + 2 A x e^{a x}       | c = Product Rule and Derivative of Exponential Function }} {{eqn | l = \frac {\d^2} {\d x^2} y_p       | r = a^2 A x^2 e^{a x} + 2 a x A e^{a x} + 2 a x A e^{a x} + 2 A e^{a x}       | c = Product Rule and Derivative of Exponential Function }} {{eqn | r = a^2 A x e^{a x} + 4 a x A e^{a x} + 2 A e^{a x}       | c = Derivative of Exponential Function }} {{end-eqn}} Inserting into $(1)$: {{begin-eqn}} {{eqn | l = a^2 A x e^{a x} + 4 a x A e^{a x} + 2 A e^{a x} + p \paren {a A x^2 e^{a x} + 2 A x e^{a x} } + q A x^2 e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {a^2 + p a + q} x e^{a x} + 2 A \paren {2 a + p} x e^{a x} + 2 A e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = 2 A e^{a x}       | r = K e^{a x}       | c = as $a^2 + p a + q = 0$ and $2 a + p = 0$ }} {{eqn | ll= \leadsto       | l = A       | r = \frac K 2       | c =  }} {{end-eqn}} Hence: :$y_p = \dfrac {K x^2 e^{a x} } 2$ and so: :$y = y_g + \dfrac {K x^2 e^{a x} } 2$ and so from Solution of Constant Coefficient Homogeneous LSOODE: :$y = C_1 e^{a x} + C_2 x e^{a x} + \dfrac {K x^2 e^{a x} } 2$ is the general solution to $(1)$.  == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Method of Undetermined Coefficients/Polynomial|next = Linear Second Order ODE/y'' - 3 y' + 2 y = 5 exp 3 x}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.4$ Particular solution: exponential $\map f x$ * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Method of Undetermined Coefficients|next = Method of Undetermined Coefficients/Sine and Cosine}}: $\S 3.18$: The Method of Undetermined Coefficients