avewright/proofwiki-math huggingface dataset:
Theorem: Linear Second Order ODE/y'' + y' - 12 y = 0

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:
:$y = C_1 e^{3 x} + C_2 e^{-4 x}$
{{qed}}
Category:Examples of Constant Coefficient Homogeneous LSOODEs
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Theorem: Linear Second Order ODE/y'' + 2 y' + 10 y = 0

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:
:$y = e^{-x} \paren {C_1 \cos 3 x + C_2 \sin 3 x}$
{{qed}}
Category:Examples of Constant Coefficient Homogeneous LSOODEs
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Theorem: Linear Second Order ODE/y'' + 4 y' + 4 y = 0

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:
:$y = C_1 e^{-2 x} + C_2 x e^{-2 x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' + y' - 2 y = 0

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:
:$y = C_1 e^x + C_2 e^{-2 x}$
{{qed}}
Category:Examples of Constant Coefficient Homogeneous LSOODEs
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Theorem: Linear Second Order ODE/y'' - 4 y = 0

The second order ODE:
:$(1): \quad y' ' - 4 y = 0$
has the general solution:
:$y = C_1 e^{2 x} + C_2 e^{-2 x}$

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

The second order ODE:
:$(1): \quad y' ' - 4 y = 0$
has the general solution:
:$y = C_1 e^{2 x} + C_2 e^{-2 x}$

Proof:
This is an instance of:
:Linear Second Order ODE: $y'' - k^2 y = 0$
which yields:
:$y = C_1 e^{k x} + C_2 e^{-k x}$
where $k = 2$.
Hence the result.
{{qed}}
Theorem: Linear Second Order ODE/y'' - y' - 6 y = 0

The second order ODE:
:$(1): \quad y' ' - y' - 6 y = 0$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-2 x}$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - m - 6 = 0$
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:
:$y = C_1 e^{3 x} + C_2 e^{-2 x}$
{{qed}}
Category:Examples of Constant Coefficient Homogeneous LSOODEs
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Theorem: Linear Second Order ODE/y'' - y' - 6 y = 20 exp -2 x

The second order ODE:
:$(1): \quad y'' - y' - 6 y = 20 e^{-2 x}$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$

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:
:$p = -1$
:$q = -6$
:$\map R x = 20 e^{-2 x}$
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:
:$\map R x = 20 e^{-2 x}$
and so from the Method of Undetermined Coefficients for the Exponential function:
:$y_p = \dfrac {K x e^{a x} } {2 a + p}$
where:
:$K = 20$
:$a = -2$
:$p = -1$
Hence:
{{begin-eqn}}
{{eqn | l = y_p
      | r = \dfrac {20 x e^{-2 x} } {2 \times \paren {-2} - 1}
      | c = 
}}
{{eqn | r = \dfrac {20 x e^{-2 x} } {-5}
      | c = 
}}
{{eqn | r = -4 x e^{-2 x}
      | c = 
}}
{{end-eqn}}
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' - 3 y' + 2 y = 14 sine 2 x - 18 cosine 2 x

The second order ODE:
:$(1): \quad y'' - 3 y' + 2 y = 14 \sin 2 x - 18 \cos 2 x$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$

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:
:$p = -3$
:$q = 2$
:$\map R x = 14 \sin 2 x - 18 \cos 2 x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y'' - 3 y' + 2 y = 0$
From Linear Second Order ODE: $y'' - 3 y' + 2 y = 0$, this has the general solution:
:$y_g = C_1 e^x + C_2 e^{2 x}$
We have that:
:$\map R x = 14 \sin 2 x - 18 \cos 2 x$
and it is noted that $14 \sin 2 x - 18 \cos 2 x$ is not itself a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for Sine and Cosine:
:$y_p = A \sin 2 x + B \cos 2 x$
where $A$ and $B$ are to be determined.
Hence:
{{begin-eqn}}
{{eqn | l = y_p
      | r = A \sin 2 x + B \cos 2 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = {y_p}'
      | r = 2 A \cos 2 x - 2 B \sin 2 x
      | c = Derivative of Sine Function, Derivative of Cosine Function
}}
{{eqn | ll= \leadsto
      | l = {y_p}''
      | r = -4 A \sin 2 x - 4 B \cos 2 x
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
Substituting into $(1)$:
{{begin-eqn}}
{{eqn | l = -4 A \sin 2 x - 4 B \cos 2 x - 3 \paren {2 A \cos 2 x - 2 B \sin 2 x} + 2 \paren {A \sin 2 x + B \cos 2 x}
      | r = 14 \sin 2 x - 18 \cos 2 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -4 A \sin 2 x + 6 B \sin 2 x + 2 A \sin 2 x
      | r = 14 \sin 2 x
      | c = equating coefficients
}}
{{eqn | l = -4 B \cos 2 x - 6 A \cos 2 x + 2 B \cos 2 x
      | r = - 18 \cos 2 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -2 A + 6 B
      | r = 14
      | c = 
}}
{{eqn | l = -2 B - 6 A
      | r = -18
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -A + 3 B
      | r = 7
      | c = 
}}
{{eqn | l = 3 A + B
      | r = 9
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -A + 3 \paren {-3 A + 9}
      | r = 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -10 A
      | r = -20
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A
      | r = 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -2 + 3 B
      | r = 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = B
      | r = 3
      | c = 
}}
{{end-eqn}}
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^x + C_2 e^{2 x} + 2 \sin 2 x + 3 \cos 2 x$
{{qed}}
Theorem: Linear Second Order ODE/y'' + y = 2 cosine x

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:
{{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}}
Theorem: Linear Second Order ODE/y'' - 2 y' = 0

The second order ODE:
:$(1): \quad y' ' - 2 y' = 0$
has the general solution:
:$y = C_1 + C_2 e^{2 x}$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 2 m = 0$
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
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Theorem: Linear Second Order ODE/y'' - 2 y' = 12 x - 10

The second order ODE:
:$(1): \quad y'' - 2 y' = 12 x - 10$
has the general solution:
:$y = C_1 + C_2 e^{2 x} + 2 x - 3 x^2$

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:
:$p = 2$
:$q = 0$
:$\map R x = 12 x - 10$
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}}
Theorem: Directed iff Finite Subsets have Upper Bounds

Let $\struct {S, \precsim}$ be a preordered set.
Let $H$ be a non-empty subset of $S$.
Then $H$ is directed {{iff}}:
:for every finite subset $A$ of $H$:
::$\exists h \in H: \forall a \in A: a \precsim h$

Proof:
=== Sufficient Condition ===
Let $R$ be directed.
We will prove by induction of the cardinality of finite subset of $H$.
Theorem: Combination of Solutions to Non-Homogeneous LSOODE with same Homogeneous Part

Let $\map {y_1} x$ be a particular solution of the linear second order ODE:
:$(1): \quad y' ' + \map P x y' + \map Q x y = \map {R_1} x$
Let $\map {y_2} x$ be a particular solution of the linear second order ODE:
:$(2): \quad y' ' + \map P x y' + \map Q x y = \map {R_2} x$
Then $\map y x = \map {y_1} x + \map {y_2} x$ is a particular solution of the linear second order ODE:
:$(3): \quad y' ' + \map P x y' + \map Q x y = \map {R_1} x + \map {R_2} x$

Proof:
{{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}}
Theorem: Linear Second Order ODE/y'' + 4 y = 4 cosine 2 x + 6 cosine x + 8 x^2 - 4 x

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:
:$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x + 2 \cos x - 1 - x + 2 x^2$
{{qed}}
Theorem: Linear Second Order ODE/y'' + 4 y = 4 cosine 2 x

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:
{{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 for Derivatives
}}
{{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 for Derivatives
}}
{{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
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Theorem: Linear Second Order ODE/y'' + 4 y = 6 cosine x

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:
{{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
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Theorem: Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x

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:
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:
{{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_0 + A_1 x + A_2 x^2}
      | r = 8 x^2 - 4 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 A_2 + 4 A_0
      | r = 0
      | c = equating coefficients
}}
{{eqn | l = 4 A_1
      | r = -4
      | c = 
}}
{{eqn | l = 4 A_2
      | r = 8
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A_1
      | r = -1
      | c = 
}}
{{eqn | l = A_2
      | r = 2
      | c = 
}}
{{eqn | l = A_0
      | r = -1
      | c = 
}}
{{end-eqn}}
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$
is the general solution to $(1)$.
{{qed}}