nyuuzyou/wikis huggingface dataset:
Linear Second Order ODE/y'' + 2 b y' + a^2 y = 0/b less than a

== Theorem ==
The second order ODE:
:$(1): \quad y + 2 b y' + a^2 y = 0$ where $b^2 < a^2$
has the general solution:
:$y = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + 2 b m + a^2 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = - b + \sqrt {b^2 - a^2}$
:$m_2 = - b - \sqrt {b^2 - a^2}$
As $b^2 < a^2$, this converts into:
:$m_1 = - b + i \sqrt {a^2 - b^2}$
:$m_2 = - b - i \sqrt {a^2 - b^2}$
These are complex and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + 2 b y' + a^2 y = K cosine omega x/b less than a

== Theorem ==
The second order ODE:
:$(1): \quad y + 2 b y' + a^2 y = K \cos \omega x$ where $b^2 < a^2$
has the general solution:
:$y = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
:$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
== 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:
:$p = 2 b$
:$q = a^2$
:$\map R x = K \cos \omega x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y + 2 b y' + a^2 y = 0$
From Linear Second Order ODE: $y + 2 b y' + a^2 y = 0: b < a$, this has the general solution:
:$y_g = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$
Substituting $\alpha = \sqrt {a^2 - b^2}$, this can be written more compactly as:
:$y_g = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x}$
We have that:
:$\map R x = K \cos \omega x$
It is noted that $K \cos \omega x$ is not itself a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for Sine and Cosine:
:$y_p = A \sin \omega x + B \cos \omega x$
where $A$ and $B$ are to be determined.
Hence:
Substituting into $(1)$:
Hence:
From Multiple of Sine plus Multiple of Cosine: Cosine Form:
and so:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
where:
:$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + 2 y' + 10 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + 2 y' + 10 y = 0$
has the general solution:
:$y = e^{-x} \paren {C_1 \cos 3 x + C_2 \sin 3 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + 2 m + 10 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = -1 + 3 i$
:$m_2 = -1 - 3 i$
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = e^{-x} \paren {C_1 \cos 3 x + C_2 \sin 3 x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + 2 y' + 2 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + 2 y' + 2 y = 0$
has the general solution:
:$y = e^{-x} \paren {A \cos x + B \sin x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + 2 m + 2 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = -1 + i$
:$m_2 = -1 - i$
These are complex conjugates.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = e^{-x} \paren {A \cos x + B \sin x}$
== Verification ==
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.1$ The reduced equation: Example $1$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Category:Linear Second Order ODE/y + 2 y' + 2 y = 0
Linear Second Order ODE/y'' + 2 y' + 2 y = 0/Verification

== Theorem ==
The equation:
:$(1): \quad y = e^{-x} \paren {A \cos x + B \sin x}$
is a set of solutions to the second order ODE:
:$y + 2 y' + 2 y = 0$
== Proof ==
Differentiating $(1)$ twice $x$ gives:
Then:
thus demonstrating that $(1)$ is indeed a solution to the second order ODE given.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.1$ The reduced equation: Example $1$
Category:Linear Second Order ODE/y + 2 y' + 2 y = 0
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 = exp -x secant 2 x

== Theorem ==
The second order ODE:
:$(1): \quad y + 2 y' + 5 y = e^{-x} \sec 2 x$
has the general solution:
:$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \dfrac {x e^{-x} \sin 2 x} 2 + \dfrac {e^{-x} \cos 2 x \ln \cos 2 x} 4$
== 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:
:$p = 2$
:$q = 5$
:$\map R x = e^{-x} \sec 2 x$
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:
:$y_g = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
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:
:$y = y_g + y_p = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \dfrac {x e^{-x} \sin 2 x} 2 + \dfrac {e^{-x} \cos 2 x \ln \cos 2 x} 4$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.19$: Problem $3 \ \text{(d)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Variation of Parameters
Linear Second Order ODE/y'' + 2 y' + 5 y = x sin x

== Theorem ==
The second order ODE:
:$(1): \quad y + 2 y' + 5 y = x \sin x$
has the general solution:
:$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \ldots$
== 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:
:$p = 2$
:$q = 5$
:$\map R x = x \sin x$
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:
:$y_g = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
It remains to find a particular solution $y_p$ to $(1)$.
We have that:
:$\map R x = x e^{-x}$
and so from the Method of Undetermined Coefficients for Product of Polynomial and Function of Sine and Cosine, we assume a solution:
Substituting in $(1)$:
Hence by equating coefficients:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \ldots$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $10$
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:
:$y = C_1 e^{-x} + C_2 x e^{-x}$
== Sources ==
* : $\S 3.17$: Problem $1 \ \text{(b)}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + 2 y' + y = exp -x log x

== Theorem ==
The second order ODE:
:$(1): \quad y + 2 y' + y = e^{-x} \ln x$
has the general solution:
:$y = C_1 e^{-x} + C_2 x e^{-x} - \dfrac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}$
== 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:
:$p = 2$
:$q = 1$
:$\map R x = e^{-x} \ln x$
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:
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:
:$y = y_g + y_p = C_1 e^{-x} + C_2 x e^{-x} - \dfrac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.19$: Problem $3 \ \text{(b)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Variation of Parameters
Linear Second Order ODE/y'' + 2 y' + y = x exp -x

== Theorem ==
The second order ODE:
:$(1): \quad y + 2 y' + y = x e^{-x}$
has the general solution:
:$y = e^{-x} \paren {C_1 + C_2 x + \dfrac {x^3} 6}$
== 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:
:$p = 2$
:$q = 1$
:$\map R x = x e^{-x}$
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:
:$y_g = \paren {C_1 + C_2 x} e^{-x}$
It remains to find a particular solution $y_p$ to $(1)$.
We have that:
:$\map R x = x e^{-x}$
and so from the Method of Undetermined Coefficients for Product of Polynomial and Exponential, we assume a solution:
Substituting in $(1)$:
Hence by equating coefficients:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = e^{-x} \paren {C_1 + C_2 x + \dfrac {x^3} 6}$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $9$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + 3 y' - 10 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + 3 y' - 10 y = 0$
has the general solution:
:$y = C_1 e^{2 x} + C_2 e^{-5 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + 3 m - 10 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 2$
:$m_2 = -5$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{2 x} + C_2 e^{-5 x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + 3 y' - 10 y = 6 exp 4 x

== Theorem ==
The second order ODE:
:$(1): \quad y' ' + 3 y' - 10 y = 6 e^{4 x}$
has the general solution:
:$y = C_1 e^{2 x} + C_2 e^{-5 x} + \dfrac {e^{4 x} } 3$
== 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:
:$p = 3$
:$q = -10$
:$\map R x = 6 e^{4 x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y' ' + 3 y' - 10 y = 0$
From Linear Second Order ODE: $y' ' + 3 y' - 10 y = 0$, this has the general solution:
:$y_g = C_1 e^{2 x} + C_2 e^{-5 x}$
We have that:
:$\map R x = 6 e^{4 x}$
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:
:$K = 6$
:$a = 4$
:$p = 3$
:$q = -10$
Hence:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{2 x} + C_2 e^{-5 x} + \dfrac {e^{4 x} } 3$
== Sources ==
* : $\S 3.18$: Problem $1 \ \text{(a)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + 4 y' + 4 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + 4 y' + 4 y = 0$
has the general solution:
:$y = \paren {C_1 + C_2 x} e^{-2 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + 4 m + 4 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = m_2 = -2$
These are real and equal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{-2 x} + C_2 x e^{-2 x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + 4 y' + 5 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y + 4 y' + 5 y = 0$
has the general solution:
:$y = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad m^2 + 4 m + 5 = 0$
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:
:$y = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' + 4 y' + 5 y = 2 exp -2 x

== Theorem ==
The second order ODE:
:$(1): \quad y + 4 y' + 5 y = 2 e^{-2 x}$
has the general solution:
:$y = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x + 2}$
== 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:
:$p = 4$
:$q = 5$
:$\map R x = 2 e^{-2 x}$
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:
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 $(1)$.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $5$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + 4 y = 0

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

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

== Theorem ==
== Proof ==
This is an instance of:
:Linear Second Order ODE: $y + k^2 y = 0$
which yields:
:$y = C_1 \cos k x + C_2 \sin k x$
where $k = 2$.
Hence the result.
Category:Linear Second Order ODE/y + 4 y = 0
Linear Second Order ODE/y'' + 4 y = 3 sin 2 x

== Theorem ==
The second order ODE:
:$(1): \quad y + 4 y = 3 \sin 2 x$
has the general solution:
:$y = C_1 \sin k x + C_2 \cos k x - \dfrac 3 4 x \cos 2 x$
== 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:
:$p = 0$
:$q = 4$
:$\map R x = 3 \sin 2 x$
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:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x - \dfrac 3 4 x \cos 2 x$
is the general solution to $(1)$.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.5$ Particular solution: trigonometric $\map f x$: Example $8$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + 4 y = 3 sine x

== Theorem ==
The second order ODE:
:$(1): \quad y' ' + 4 y = 3 \sin x$
has the general solution:
:$y = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$
== 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:
:$p = 0$
:$q = 4$
:$\map R x = 3 \sin x$
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:
:$y_g = C_1 \cos 2 x + C_2 \sin 2 x$
We have that:
:$\map R x = 3 \sin x$
and it is noted that $\sin x$ is not itself a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for the Sine and Cosine functions:
:$y_p = A \sin x + B \cos x$
for some $A$ and $B$ to be determined.
Hence:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$
== Sources ==
* : $\S 3.18$: Problem $1 \ \text{(b)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' + 4 y = 4 cosine 2 x

== Theorem ==
The second order ODE:
:$(1): \quad y' ' + 4 y = 4 \cos 2 x$
has the general solution:
:$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x$
== 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:
:$p = 0$
:$q = 4$
:$\map R x = 4 \cos 2 x$
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:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + x \sin 2 x$
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 = 4 cosine 2 x + 6 cosine x + 8 x^2 - 4 x

== Theorem ==
The second order ODE:
:$(1): \quad y + 4 y = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$
has the general solution:
:$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x + 2 \cos x - 1 - x + 2 x^2$
== 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:
:$p = 0$
:$q = 4$
:$\map R x = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$
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:
:$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x + 2 \cos x - 1 - x + 2 x^2$
== Sources ==
* : $\S 3.18$: Problem $3$
Category:Examples of Constant Coefficient LSOODEs