avewright/proofwiki-math huggingface dataset:
Theorem: Linear Second Order ODE/y'' + 8 y = 0

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:
:$y = C_1 \cos 2 \sqrt 2 x + C_2 \sin 2 \sqrt 2 x$
{{qed}}
Theorem: Linear Second Order ODE/2 y'' - 4 y' + 8 y = 0

The second order ODE:
:$(1): \quad 2 y' ' - 4 y + 8 y = 0$
has the general solution:
:$y = e^x \paren {C_1 \cos \sqrt 3 x + C_2 \sin \sqrt 3 x}$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Let $(1)$ be written in the form:
:$y' ' - 2 y + 4 y = 0$
Its auxiliary equation is:
:$(2): \quad: m^2 - 2 m + 4 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 1 + \sqrt 3 i$
:$m_2 = 1 - \sqrt 3 i$
These are complex and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = e^x \paren {C_1 \cos \sqrt 3 x + C_2 \sin \sqrt 3 x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' - 9 y' + 20 y = 0

The second order ODE:
:$(1): \quad y' ' - 9 y' + 20 y = 0$
has the general solution:
:$y = C_1 e^{4 x} + C_2 e^{5 x}$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 9 m + 20 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 4$
:$m_2 = 5$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{4 x} + C_2 e^{5 x}$
{{qed}}
Theorem: Linear Second Order ODE/2 y'' + 2 y' + 3 y = 0

The second order ODE:
:$(1): \quad 2 y' ' + 2 y' + 3 y = 0$
has the general solution:
:$y = e^{-x/2} \paren {C_1 \cos \dfrac {\sqrt 5} 2 x + C_2 \sin \dfrac {\sqrt 5} 2 x}$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: 2 m^2 + 2 m + 3 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = -\dfrac 1 2 + \dfrac {\sqrt 5} 2 i$
:$m_2 = -\dfrac 1 2 - \dfrac {\sqrt 5} 2 i$
These are complex and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = e^{-x/2} \paren {C_1 \cos \dfrac {\sqrt 5} 2 x + C_2 \sin \dfrac {\sqrt 5} 2 x}$
{{qed}}
Theorem: Condition for Solutions to Constant Coefficient Homogeneous LSOODE to tend to Zero

Let:
:$(1): \quad y' ' + p y' + q y = 0$
be a constant coefficient homogeneous linear second order ODE.
Let the general solution to $(1)$ be $\map y {x, C_1, C_2}$.
Then:
:$\ds \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$
{{iff}}
:$p$ and $q$ are both strictly positive.

Proof:
By Solution of Constant Coefficient Homogeneous LSOODE, $y$ is in one of the following forms:
:<nowiki>$y = \begin{cases}
C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\
 & \\
C_1 e^{m_1 x} + C_2 x e^{m_2 x} & : p^2 = 4 q \\
 & \\
e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q
\end{cases}$</nowiki>
where:
:$m_1$ and $m_2$ are the roots of the auxiliary equation $m^2 + p m + q = 0$
:$a + i b = m_1$
:$a - i b = m_2$
Theorem: Derivative of Solution to Constant Coefficient Homogeneous LSOODE is also Solution

Let:
:$(1): \quad y' ' + p y' + q y = 0$
be a constant coefficient homogeneous linear second order ODE.
Let $\map y x$ be a particular solution of $(1)$.
Then its derivative $\map {y'} x$ is also a particular solution of $(1)$.

Proof:
By Solution of Constant Coefficient Homogeneous LSOODE, $y$ is in one of the following forms:
:<nowiki>$y = \begin{cases}
C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\
 & \\
C_1 e^{m_1 x} + C_2 x e^{m_2 x} & : p^2 = 4 q \\
 & \\
e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q
\end{cases}$</nowiki>
Let:
:$y = A e^{m_1 x} + B e^{m_2 x}$
where $A, B \in \R$.
Then:
:$y' = m_1 A e^{m_1 x} + m_2 B e^{m_2 x}$
which is also in the form:
:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
and so is also a particular solution of $(1)$.
{{qed|lemma}}
Let:
:$y = A e^{m_1 x} + B x e^{m_1 x}$
where $A, B \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = y'
      | r = m_1 A e^{m_1 x} + B e^{m_1 x} + B m_1 x e^{m_1 x}
      | c = 
}}
{{eqn | r = \paren {m_1 A + B} e^{m_1 x} + B m_1 x e^{m_1 x}
      | c = 
}}
{{end-eqn}}
which is also in the form:
:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
and so is also a particular solution of $(1)$.
{{qed|lemma}}
Let:
:$y = e^{a x} \left({A \sin b x + B \cos b x}\right)$
where $A, B \in \R$.
Then:
:$y' = m_1 A e^{m_1 x} + m_2 B e^{m_2 x}$
{{begin-eqn}}
{{eqn | l = y'
      | r = a e^{a x} \paren {A \sin b x + B \cos b x} + e^{a x} \paren {A b \cos b x - B b \sin b x}
      | c = 
}}
{{eqn | r = e^{a x} \paren {a A \sin b x + a B \cos b x + A b \cos b x - B b \sin b x}
      | c = 
}}
{{eqn | r = e^{a x} \paren {\paren {a B + A b} \cos b x + \paren {a A - b B} \sin b x}
      | c = 
}}
{{end-eqn}}
which is also in the form:
:$y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$
and so is also a particular solution of $(1)$.
{{qed|lemma}}
Hence the result.
{{qed}}
Theorem: Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE

Consider the Cauchy-Euler equation:
:$(1): \quad x^2 \dfrac {\d^2 y} {\d x^2} + p x \dfrac {\d y} {\d x} + q y = 0$
By making the substitution:
:$x = e^t$
it is possible to convert $(1)$ into a constant coefficient homogeneous linear second order ODE:
:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t} + q y = 0$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = x
      | r = e^t
      | c = Derivative of Exponential Function
}}
{{eqn | ll=\leadsto
      | l = \frac {\d x} {\d t}
      | r = e^t
      | c = 
}}
{{eqn | n = 1
      | r = x
      | c = 
}}
{{eqn | ll=\leadsto
      | l = \frac {\d t} {\d x}
      | r = \frac 1 x
      | c = Derivative of Inverse Function
}}
{{eqn | n = 2
      | r = e^{-t}
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x}
      | r = \frac {\d y} {\d t} \frac {\d t} {\d x}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = \frac {\d y} {\d t} e^{-t}
      | c = from $(2)$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \frac {\d^2 y} {\d x^2}
      | r = \frac {\d} {\d x} \paren {\frac {\d y} {\d x} }
      | c = 
}}
{{eqn | r = \frac {\d} {\d t} \paren {\frac {\d y} {\d x} } \frac {\d t} {\d x}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = \frac {\d} {\d t} \paren {\frac {\d y} {\d t} e^{-t} } e^{-t}
      | c = from $(2)$
}}
{{eqn | r = e^{-t} \paren {e^{-t} \frac {\d^2 y} {\d t^2} - e^{-t} \frac {\d y} {\d t} }
      | c = Product Rule for Derivatives
}}
{{eqn | r = e^{- 2 t} \paren {\frac {\d^2 y} {\d t^2} - \frac {\d y} {\d t} }
      | c = 
}}
{{end-eqn}}
Substituting back into $(1)$:
{{begin-eqn}}
{{eqn | l = x^2 \dfrac {\d^2 y} {\d x^2} + p x \dfrac {\d y} {\d x} + q y
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{2 t} \paren {e^{- 2 t} \paren {\frac {\d^2 y} {\d t^2} - \frac {\d y} {\d t} } } + p e^t \paren {\frac {\d y} {\d t} e^{-t} } + q y
      | r = 0
      | c = 
}}
{{eqn | l = \frac {\d^2 y} {\d t^2} - \frac {\d y} {\d t} + p \frac {\d y} {\d t} + q y
      | r = 0
      | c = 
}}
{{eqn | l = \frac {\d^2 y} {\d t^2} + \paren {p - 1} \frac {\d y} {\d t} + q y
      | r = 0
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Second Order ODE/x^2 y'' + 3 x y' + 10 y = 0

The second order ODE:
:$(1): \quad x^2 y'' + 3 x y' + 10 y = 0$
has the general solution:
:$y = \dfrac 1 x \paren {C_1 \map \cos {\ln x^3} + C_2 \map \sin {\ln x^3} }$

Proof:
$(1)$ is an instance of the Cauchy-Euler Equation:
:$x^2 y'' + p x y' + q y = 0$
where:
:$p = 3$
:$q = 10$
By Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE, this can be expressed as:
:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$
by making the substitution:
:$x = e^t$
Hence $(1)$ can be expressed as:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} + 2 \dfrac {\d y} {\d t^2} + 10 y = 0$
It can be seen that $(2)$ is a constant coefficient homogeneous linear second order ODE.
From Linear Second Order ODE: $y'' + 2 y' + 10 y = 0$, this has the general solution:
{{begin-eqn}}
{{eqn | l = y
      | r = e^{-t} \paren {C_1 \cos 3 t + C_2 \sin 3 t}
      | c = 
}}
{{eqn | r = \dfrac 1 x \paren {C_1 \cos 3 t + C_2 \sin 3 t}
      | c = substituting $x = e^t$
}}
{{eqn | r = \dfrac 1 x \paren {C_1 \map \cos {3 \ln x} + C_2 \map \sin {3 \ln x} }
      | c = substituting $t = \ln x$
}}
{{eqn | r = \dfrac 1 x \paren {C_1 \map \cos {\ln x^3} + C_2 \map \sin {\ln x^3} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Second Order ODE/2 x^2 y'' + 10 x y' + 8 y = 0

The second order ODE:
:$(1): \quad 2 x^2 y' ' + 10 x y' + 8 y = 0$
has the general solution:
:$y = C_1 x^{-2} + C_2 x^{-2} \ln x$

Proof:
Let $(1)$ be rewritten as:
:$x^2 y' ' + 5 x y' + 4 y = 0$
It can be seen to be an instance of the Cauchy-Euler Equation:
:$x^2 y' ' + p x y' + q y = 0$
where:
:$p = 5$
:$q = 4$
By Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE, this can be expressed as:
:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$
by making the substitution:
:$x = e^t$
Hence it can be expressed as:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} + 4 \dfrac {\d y} {\d t^2} + 4 y = 0$
From Linear Second Order ODE: $y' ' + 4 y' + 4 y = 0$, this has the general solution:
{{begin-eqn}}
{{eqn | l = y
      | r = C_1 e^{-2 t} + C_2 t e^{-2 t}
      | c = 
}}
{{eqn | r = C_1 x^{-2} + C_2 t x^{-2}
      | c = substituting $x = e^t$
}}
{{eqn | r = C_1 x^{-2} + C_2 x^{-2} \ln x
      | c = substituting $t = \ln x$
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Second Order ODE/x^2 y'' + 2 x y' - 12 y = 0

The second order ODE:
:$(1): \quad x^2 y'' + 2 x y' - 12 y = 0$
has the general solution:
:$y = C_1 x^3 + C_2 x^{-4}$

Proof:
It can be seen that $(1)$ is an instance of the Cauchy-Euler Equation:
:$x^2 y'' + p x y' + q y = 0$
where:
:$p = 2$
:$q = -12$
By Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE, this can be expressed as:
:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$
by making the substitution:
:$x = e^t$
Hence it can be expressed as:
:$(2): \quad \dfrac {\d^2 y} {\d t^2} + \dfrac {\d y} {\d t^2} - 12 y = 0$
From Linear Second Order ODE: $y'' + y' - 12 y = 0$, this has the general solution:
{{begin-eqn}}
{{eqn | l = y
      | r = C_1 e^{3 t} + C_2 e^{-4 t}
      | c = 
}}
{{eqn | r = C_1 x^3 + C_2 x^{-4}
      | c = substituting $x = e^t$
}}
{{end-eqn}}
{{qed}}
Theorem: Cardinality of Image of Set not greater than Cardinality of Set

Let $X, Y$ be sets.
Let $f: X \to Y$ be a mapping.
Let $A$ be a subset of $X$.
Then:
:$\card {\map {f^\to} A} \le \card A$
where $\card A$ denotes the cardinality of $A$.

Proof:
By definitions of surjection and restriction of mapping:
:$F \restriction_A: A \to \map {f^\to} A$ is a surjection
Thus by Surjection iff Cardinal Inequality:
:$\card {\map {f^\to} A} \le \card A$
{{qed}}
Theorem: Projector has Norm 1

An idempotent operator $P$ is a projector on the Hilbert Space $H$ {{iff}} $P$ has norm $1$:
:$\ds \norm P \equiv \sup_{x \mathop \in H} \frac {\norm P} {\norm x} = 1$

Proof:
For all $x \in \Rng P$:
:$\norm {P \dfrac x {\norm x} } = \dfrac {\norm x} {\norm x} = 1$
so $\norm P \ge 1$.
It remains to show that this holds with equality {{iff}} $P$ is a projector.
First, suppose $P$ is a projector.
Let $\set {p_1, p_2, \ldots}$ be an orthonormal basis for $\Rng P$.
Let $\set {q_1, q_2, \ldots}$ be an orthonormal basis for $\Rng P_\perp$.
Then for any $x \in H$, we can choose scalars $\set {\alpha_1, \alpha_2, \ldots}$ and $\set {\beta_1, \beta_2, \ldots}$ so that:
:$\ds x = \sum_{i \mathop = 1}^\infty \alpha_i p_i + \sum_{i \mathop = 1}^\infty \beta_i q_i$
Because the basis vectors are orthogonal, Pythagoras's theorem shows that:
:$\ds \norm x^2 = \sum_{i \mathop = 1}^\infty \norm {\alpha_i}^2 + \norm {\beta_i}^2$
Then $\norm {P x}$ can be expanded thus:
{{begin-eqn}}
{{eqn | l = \norm {P x}^2
      | r = \norm {\sum_{i \mathop = 1}^\infty \alpha_i P p_i + \sum_{i \mathop = 1}^\infty \beta_i P q_i}^2
}}
{{eqn| r = \norm {\sum_{i \mathop = 1}^\infty \alpha_i p_i + 0}^2
}}
{{eqn | r = \sum_{i \mathop = 1}^\infty \size {\alpha_i}^2
      | c = Pythagoras's theorem again
}}
{{eqn | o = \le
      | r = \norm x^2
}}
{{end-eqn}}
Hence $\norm P \le 1$.
Since it was already shown $\norm P \ge 1$, it follows that $\norm P = 1$.
Now suppose $P$ is not a projector.
Then there exists $x \in H$ so that $P x - x$ is not orthogonal to $\Rng P$.
By writing $x = p + q$ with $p \in \Rng P$ and $q \in \Rng P_\perp$, it follows that:
:$P x - x = P \paren {p + q} - \paren {p + q} = P q - q$
By rescaling $x$, we can assume $\norm q = 1$.
Since $q \in \Rng P_\perp$ but
:$P q - q = P x - x = \notin \Rng P_\perp$
it follows that $P q \ne 0$.
Let $\norm {P q} = c \ne 0$.
It will be shown that
:$y = c q + \dfrac 1 c P q$
satisfies
:$\dfrac {\norm {P y} } {\norm y} > 1$
Notice first that:
{{begin-eqn}}
{{eqn | l = \norm y^2
      | r = \norm {c q + \dfrac 1 c P q}^2
}}
{{eqn | r = \norm {c q}^2 + \norm {\dfrac 1 c P q}^2
      | c = Pythagoras's Theorem ($q$ and $P q$ are orthogonal)
}}
{{eqn | r = c^2 + 1
      | c = since $\norm {P q} = c$ and $\norm q = 1$.
}}
{{end-eqn}}
Now:
{{begin-eqn}}
{{eqn | l = \norm {P \frac y {\norm y} }^2
      | r = \frac 1 {\norm y^2} \norm {P \paren {c q + \frac 1 c P q} }^2
}}
{{eqn | r = \frac 1 {c^2 + 1} \norm {c P q + \frac 1 c P^2 q}^2
}}
{{eqn | r = \frac 1 {c^2 + 1} \norm {P q \paren {c + \frac 1 c} }^2
      | c = $P$ is idempotent
}}
{{eqn | r = \frac {\norm {P q}^2} {c^2 + 1} \paren {c + \frac 1 c}^2
}}
{{eqn | r = \frac {c^2 \paren {c + \frac 1 c}^2} {c^2 + 1}
}}
{{eqn | r = c^2 + 1
}}
{{eqn | o = >
      | r = 1
}}
{{end-eqn}}
{{qed}}
Category:Hilbert Spaces
kyzassxt02w069d7uv2uyhb3jg9ruhc
Theorem: Linear Second Order ODE/y'' + 3 y' - 10 y = 0

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:
:$y = C_1 e^{2 x} + C_2 e^{-5 x}$
{{qed}}
Category:Examples of Constant Coefficient Homogeneous LSOODEs
sslbnnyxvfqrhqphqk97n2kvjnz6f1w
Theorem: Linear Second Order ODE/y'' + 3 y' - 10 y = 6 exp 4 x

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:
{{begin-eqn}}
{{eqn | l = y_p
      | r = \dfrac {6 e^{4 x} } {4^2 + 3 \times 4 - 10}
      | c = 
}}
{{eqn | r = \dfrac {6 e^{4 x} } {16 + 12 - 10}
      | c = 
}}
{{eqn | r = \dfrac {e^{4 x} } 3
      | c = 
}}
{{end-eqn}}
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{2 x} + C_2 e^{-5 x} + \dfrac {e^{4 x} } 3$
{{qed}}
Theorem: Linear Second Order ODE/y'' + 4 y = 0

The second order ODE:
:$(1): \quad y' ' + 4 y = 0$
has the general solution:
:$y = C_1 \cos 2 x + C_2 \sin 2 x$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + 4 = 0$
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:
:$y = C_1 \cos 2 x + C_2 \sin 2 x$
{{qed}}
Theorem: Linear Second Order ODE/y'' + 4 y = 0

The second order ODE:
:$(1): \quad y' ' + 4 y = 0$
has the general solution:
:$y = C_1 \cos 2 x + C_2 \sin 2 x$

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.
{{qed}}
Theorem: Linear Second Order ODE/y'' + 4 y = 3 sine x

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:
{{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 = 3 \sin x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {-A + 4 A} \sin x
      | r = 3 \sin x
      | c = 
}}
{{eqn | l = -B \cos x + B \cos x
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 3 A
      | r = 3
      | c = 
}}
{{eqn | l = -B \cos x + 4 B \cos x
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A
      | r = 1
      | c = 
}}
{{eqn | l = B
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y_p
      | r = \sin x
      | c = 
}}
{{end-eqn}}
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$
{{qed}}
Theorem: Linear Second Order ODE/y'' + 10 y' + 25 y = 0

The second order ODE:
:$(1): \quad y' ' + 10 y' + 25 y = 0$
has the general solution:
:$y = C_1 e^{-5 x} + C_2 x e^{-5 x}$

Proof:
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 + 10 m + 25 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = m_2 = -5$
These are real and equal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{-5 x} + C_2 x e^{-5 x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' + 10 y' + 25 y = 14 exp -5 x

The second order ODE:
:$(1): \quad y'' + 10 y' + 25 y = 14 e^{-5 x}$
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:
:$p = 10$
:$q = 25$
:$\map R x = 14 e^{-5 x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$(2): \quad y'' + 10 y' + 25 y = 0$
From Linear Second Order ODE: $y'' + 10 y' + 25 y = 0$, this has the general solution:
:$y_g = C_1 e^{-5 x} + C_2 x e^{-5 x}$
We have that:
:$\map R x = 14 e^{-5 x}$
and it is noted that $14 e^{-5 x}$ is a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for the Exponential function:
:$y_p = A x^2 e^{-5 x}$
where:
:$A = \dfrac {14} 2$
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{-5 x} + C_2 x e^{-5 x} + 7 x^2 e^{-5 x}$
{{qed}}
Theorem: Linear Second Order ODE/y'' - 2 y' + 5 y = 0

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:
:$y = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
{{qed}}
Category:Examples of Constant Coefficient Homogeneous LSOODEs
bnd6l4fb9mqwavl1sj421hzoe1gylyv
Theorem: Linear Second Order ODE/y'' - 2 y' + 5 y = 25 x^2 + 12

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:
{{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} + 5 \paren {A_0 + A_1 x + A_2 x^2}
      | r = 25 x^2 + 12
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 5 A_2 x^2
      | r = 25 x^2
      | c = equating powers
}}
{{eqn | l = -4 A_2 x + 5 A_1 x
      | r = 0
      | c = 
}}
{{eqn | l = 2 A_2 - 2 A_1 + 5 A_0
      | r = 12
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A_2
      | r = 5
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -4 \times 5 x + 5 A_1 x
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -4 + A_1
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A_1
      | r = 4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 \times 5 - 2 \times 4 + 5 A_0
      | r = 12
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 10 - 8 + 5 A_0
      | r = 12
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 5 A_0
      | r = 10
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A_0
      | r = 2
      | c = 
}}
{{end-eqn}}
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x} + 2 + 4 x + 5 x^2$
{{qed}}