nyuuzyou/wikis huggingface dataset:
Linear Second Order ODE/y'' + 4 y = 6 cosine x

== Theorem ==
The second order ODE:
:$(1): \quad y + 4 y = 6 \cos x$
has the general solution:
:$y = C_1 \sin 2 x + C_2 \cos 2 x + 2 \cos x$
== Proof ==
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:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + 2 \cos x$
is the general solution to $(1)$.
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x

== Theorem ==
The second order ODE:
:$(1): \quad y + 4 y = 8 x^2 - 4 x$
has the general solution:
:$y = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Constant Coefficient LSOODEs
Category:Linear Second Order ODE/y + 4 y = 8 x^2 - 4 x
Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x/Proof 1

== Theorem ==
== Proof ==
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:
:$p = 0$
:$q = 4$
:$\map R x = 8 x^2 - 4 x$
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:
:$y_g = C_1 \sin 2 x + C_2 \cos 2 x$
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:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$
is the general solution to $(1)$.
Category:Examples of Method of Undetermined Coefficients
Category:Linear Second Order ODE/y + 4 y = 8 x^2 - 4 x
Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x/Proof 2

== Theorem == 
== Proof ==
Taking Laplace transforms: 
:$\laptrans {y + 4 y} = \laptrans {8 x^2 - 4 x}$
We have: 
We also have: 
So:
:$\paren {s^2 + 4} \laptrans y = s \map y 0 + \map {y'} 0 + \dfrac {16 - 4 s} {s^3}$
Giving: 
So:
Setting $C_1 = \dfrac {\paren {\map {y'} 0 + 1} } 2$ and $C_2 = \map y 0 + 1$ gives the result.
Category:Linear Second Order ODE/y + 4 y = 8 x^2 - 4 x
Linear Second Order ODE/y'' + 4 y = tan 2 x

== Theorem ==
The second order ODE:
:$(1): \quad y + 4 y = \tan 2 x$
has the general solution:
:$y = C_1 \cos 2 x + C_2 \sin 2 x - \dfrac 1 4 \cos 2 x \map \ln {\sec 2 x + \tan 2 x}$
== Proof ==
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:
By the Method of Variation of Parameters, we have that:
:$y_p = v_1 y_1 + v_2 y_2$
where:
where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.
We have that:
Hence:
It follows that:
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 $(1)$.
== Sources ==
* : $\S 3.19$: Problem $3 \ \text{(a)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Variation of Parameters
Linear Second Order ODE/y'' + 8 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + 8 y = 0$
has the general solution:
:$y = C_1 \cos 2 \sqrt 2 x + C_2 \sin 2 \sqrt 2 x$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + 8 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 2 \sqrt 2 i$
:$m_2 = -2 \sqrt 2 i$
These are complex and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 \cos 2 \sqrt 2 x + C_2 \sin 2 \sqrt 2 x$
== Sources ==
* : $\S 3.17$: Problem $1 \ \text{(c)}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + k^2 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + k^2 y = 0$
has the general solution:
:$y = A \, \map \sin {k x + B}$
or can be expressed as:
:$y = C_1 \sin k x + C_2 \cos k x$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
This second order ODE can also be presented in the form:
:$\dfrac {\d^2 y} {\d x^2} = -k^2 y$
== Also see ==
* Linear Second Order ODE: $y - k^2 y = 0$
== Sources ==
* : Chapter $\text I$: Introduction and Definitions. Elimination. Graphical Representation: $1 \ (1)$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Category:Linear Second Order ODE/y + k^2 y = 0
Linear Second Order ODE/y'' + k^2 y = 0/Proof 1

== Theorem ==
== Proof ==
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
:$p \dfrac {\d p} {\d y} = -k^2 y$
where $p = \dfrac {\d y} {\d x}$.
From:
:First Order ODE: $y \rd y = k x \rd x$
this has the general solution:
:$p^2 = -k^2 y^2 + C$
or: 
:$p^2 + k^2 y^2 = C$
As the is the sum of squares, $C$ has to be positive for this to have any solutions.
Thus, let $C = \alpha^2$.
Then:
From Multiple of Sine plus Multiple of Cosine: Sine Form, this can be expressed as:
:$y = C_1 \sin k x + C_2 \cos k x$
== Sources ==
* : $\S 2.11$: Reduction of Order: Example $2$
Category:Linear Second Order ODE/y + k^2 y = 0
Linear Second Order ODE/y'' + k^2 y = 0/Proof 2

== Theorem ==
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + k^2 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = k i$
:$m_2 = -k i$
These are complex and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 \cos k x + C_2 \sin k x$
or, by disposing the constants differently:
:$y = C_1 \sin k x + C_2 \cos k x$
Category:Linear Second Order ODE/y + k^2 y = 0
Linear Second Order ODE/y'' + k^2 y = sine b x

== Theorem ==
The second order ODE:
:$(1): \quad y + k^2 y = \sin b x$
has the general solution:
:$y = \begin{cases} C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2} & : b \ne k \\
C_1 \sin k x + C_2 \cos k x - \dfrac {x \cos k x} {2 k} & : b = k 
\end{cases}$
== Proof ==
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:
:$p = 0$
:$q = k^2$
:$\map R x = \sin b x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$(2): \quad y + k^2 y = 0$
From Linear Second Order ODE: $y + k^2y = 0$, this has the general solution:
:$y_g = C_1 \sin k x + C_2 \cos k x$
We have that:
:$\map R x = \sin b x$
There are two cases to address:
:$b = k$
:$b \ne k$
First suppose that $b = k$.
It is noted that $\sin b x = \sin k x$ is a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for Sine and Cosine:
:$y_p = A x \sin k x + B x \cos k x$
where $A$ and $B$ are to be determined.
Hence:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x - \dfrac {x \cos k x} {2 k}$
Now suppose that $b \ne k$.
It is noted that $\sin b x$ is not a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for Sine and Cosine:
:$y_p = A \sin b x + B \cos b x$
where $A$ and $B$ are to be determined.
Hence:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2}$
== Sources ==
* : $\S 3.18$: Problem $2$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + y' - 12 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + y' - 12 y = 0$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-4 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + m - 12 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 3$
:$m_2 = -4$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{3 x} + C_2 e^{-4 x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + y' - 2 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + y' - 2 y = 0$
has the general solution:
:$y = C_1 e^x + C_2 e^{-2 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + m - 2 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 1$
:$m_2 = -2$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^x + C_2 e^{-2 x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + y' - 6 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + y' - 6 y = 0$
has the general solution:
:$y = C_1 e^{2 x} + C_2 e^{-3 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + m - 6 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = -3$
:$m_2 = 2$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{2 x} + C_2 e^{-3 x}$
== Sources ==
* : $\S 3.17$: Problem $1 \ \text{(a)}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + y = 0$
has the general solution:
:$y = C_1 \sin x + C_2 \cos x$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Category:Linear Second Order ODE/y + y = 0
Linear Second Order ODE/y'' + y = 0/Proof 1

== Theorem ==
== Proof ==
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
:$p \dfrac {\d p} {\d y} = -y$
where $p = \dfrac {\d y} {\d x}$.
From:
:First Order ODE: $y \rd y = k x \rd x$
with $k = 1$, this has the solution:
:$p^2 = -y^2 + C$
or: 
:$p^2 + y^2 = C$
As the is the sum of squares, $C$ has to be positive for this to have any solutions.
Thus, let $C = \alpha^2$.
Then:
From Multiple of Sine plus Multiple of Cosine: Sine Form, this can be expressed as:
:$y = C_1 \sin x + C_2 \cos x$
Category:Linear Second Order ODE/y + y = 0
Linear Second Order ODE/y'' + y = 0/Proof 2

== Theorem ==
== Proof ==
$(1)$ can be seen to be a special case of:
:$(2): \quad$ Linear Second Order ODE: $y + k^2 y = 0$
with $k = 1$.
$(2)$ has the solution:
:$y = C_1 \sin k x + C_2 \cos k x$
Hence setting $k = 1$:
:$y = C_1 \sin x + C_2 \cos x$
Category:Linear Second Order ODE/y + y = 0
Linear Second Order ODE/y'' + y = 0/Proof 3

== Theorem ==
== Proof ==
We have that:
Hence it can be seen by inspection that:
are particular solutions to $(1)$.
It is noted that:
:$\dfrac {y_1} {y_2} = \tan x$
which is not a constant function on any closed interval $\closedint a b$.
Calculating the Wronskian of $y_1$ and $y_2$:
So the Wronskian of $y_1$ and $y_2$ is never zero.
We have that $(1)$ is a homogeneous linear second order ODE in the form:
:$y + \map P x y' + \map Q x y = 0$
where $\map P x = 0$ and $\map Q x$ are constant functions.
So by Constant Real Function is Continuous: $P$ and $Q$ are continuous on $\closedint a b$
Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$(1)$ has the general solution:
::$y = C_1 \sin x + C_2 \cos x$
:on $\closedint a b$.
Since $\closedint a b$ can be extended indefinitely without introducing points where $P$ or $Q$ are discontinuous, the general solution is valid for all $x$.
== Sources ==
* : $\S 3.15$: The General Solution of the Homogeneous Equation: Example $1$
Category:Linear Second Order ODE/y + y = 0
Linear Second Order ODE/y'' + y = 0/Proof 4

== Theorem ==
== Proof ==
Note that:
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:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
Hence:
and so:
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:
:$y = C_1 \sin x + C_2 \cos x$
== Sources ==
* : $\S 3.16$: Problem $2 \ \text{(a)}$
Category:Linear Second Order ODE/y + y = 0
Linear Second Order ODE/y'' + y = 0/Proof 5

== Theorem ==
== Proof ==
Taking Laplace transforms, we have: 
:$\laptrans {y + y} = \laptrans 0$
From Laplace Transform of Constant Mapping, we have: 
:$\laptrans 0 = 0$
We also have: 
So: 
:$\paren {s^2 + 1} \laptrans y = s \map y 0 + \map {y'} 0$
Giving: 
:$\laptrans y = \map y 0 \dfrac s {s^2 + 1} + \map {y'} 0 \dfrac 1 {s^2 + 1}$
So: 
Setting $C_1 = \map {y'} 0$ and $C_2 = \map y 0$ gives the result.
Category:Linear Second Order ODE/y + y = 0
Linear Second Order ODE/y'' + y = 0/y(0) = 2, y'(0) = 3

== Theorem ==
The second order ODE:
:$(1): \quad y + y = 0$
with initial conditions:
:$\map y 0 = 2$
:$\map {y'} 0 = 3$
has the particular solution:
:$y = 3 \sin x + 2 \cos x$
== Proof ==
From Linear Second Order ODE: $y + y = 0$, the general solution of $(1)$ is:
:$y = C_1 \sin x + C_2 \cos x$
Differentiating $x$:
:$y' = C_1 \cos x - C_2 \sin x$
Thus for the initial conditions:
and:
Hence the result.
== Sources ==
* : $\S 3.15$: The General Solution of the Homogeneous Equation: Example $1$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + y = 2 cosine x

== Theorem ==
The second order ODE:
:$(1): \quad y + y = 2 \cos x$
has the general solution:
:$y = C_1 \sin x + C_2 \cos x + x \cos x$
== Proof ==
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:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \sin x + C_2 \cos x + x \sin x$
== Sources ==
* : $\S 3.18$: Problem $1 \ \text{(g)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + y = K

== Theorem ==
The second order ODE:
:$(1): \quad y' ' + y = K$
has the general solution:
:$y = C_1 \sin x + C_2 \cos x + K$
== Proof ==
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:
:$p = 0$
:$q = 1$
:$\map R x = K$
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:
:$\map R x = K$
So from the Method of Undetermined Coefficients for Polynomial:
:$y_p = A K$
where $A$ is to be determined.
Hence:
Substituting into $(1)$:
Our work is done.
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \sin x + C_2 \cos x + K$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + y = cosecant x

== Theorem ==
The second order ODE:
:$(1): \quad y + y = \csc x$
has the general solution:
:$y = C_1 \sin x + C_2 \cos x - x \cos x + \sin x \map \ln {\sin x}$
== Proof ==
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:
By the Method of Variation of Parameters, we have that:
:$y_p = v_1 y_1 + v_2 y_2$
where:
where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.
We have that:
Hence:
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 $\paren 1$.
== Sources ==
* : $\S 3.19$: The Method of Variation of Parameters: Example $1$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Variation of Parameters
Linear Second Order ODE/y'' + y = exp -x cos x

== Theorem ==
The second order ODE:
:$(1): \quad y + y = e^{-x} \cos x$
has the general solution:
:$y = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x} + C_1 \sin x + C_2 \cos x$
== Proof ==
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:
:$p = 0$
:$q = 1$
:$\map R x = e^{-x} \cos x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y + y = 0$
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 $(1)$.
We have that:
:$\map R x = e^{-x} \cos x$
From the Method of Undetermined Coefficients for Exponential of Sine and Cosine:
:$y_p = e^{-x} \paren {A \cos x + B \sin x}$
where $A$ and $B$ are to be determined.
Hence:
Substituting into $(1)$:
Hence the result:
:$y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x}$
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x} + C_1 \sin x + C_2 \cos x$
is the general solution to $(1)$.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $6$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 2 y' + 5 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - 2 y' + 5 y = 0$
has the general solution:
:$y = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
== Proof ==
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:
:$y = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - 2 y' + 5 y = 25 x^2 + 12

== Theorem ==
The second order ODE:
:$(1): \quad y - 2 y' + 5 y = 25 x^2 + 12$
has the general solution:
:$y = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x} + 2 + 4 x + 5 x^2$
== Proof ==
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:
:$p = -2$
:$q = 5$
:$\map R x = 25 x^2 + 12$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$(2): \quad y - 2 y' + 5 y = 0$
From Linear Second Order ODE: $y - 2 y' + 5 y = 0$, this has the general solution:
:$y_g = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
We have that:
:$\map R x = 25 x^2 + 12$
and it is noted that $25 x^2 + 12$ 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:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x} + 2 + 4 x + 5 x^2$
== Sources ==
* : $\S 3.18$: Problem $1 \ \text{(d)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 2 y' + y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - 2 y' + y = 0$
has the general solution:
:$y = C_1 e^x + C_2 x e^x$
== Proof ==
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:
:$y = C_1 e^x + C_2 x e^x$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - 2 y' + y = 1 over 1 + e^x

== Theorem ==
The second order ODE:
:$(1): \quad y - 2 y' + y = \dfrac 1 {1 + e^x}$
has a particular solution:
:$y = 1 + e^x \ds \int \map \ln {1 + e^{-x} } \rd x$
== Proof ==
Then we need to solve:
and so on.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.8$ Recapitulation: Example $10$
Category:Examples of Constant Coefficient LSOODEs
Linear Second Order ODE/y'' - 2 y' + y = 2 x

== Theorem ==
The second order ODE:
:$(1): \quad y' ' - 2 y' + y = 2 x$
has the general solution:
:$y = C_1 e^x + C_2 x e^x + 2 x + 4$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Examples of Constant Coefficient LSOODEs
Category:Linear Second Order ODE/y - 2 y' + y = 2 x