nyuuzyou/wikis huggingface dataset:
Linear Second Order ODE/y'' - 2 y' + y = 2 x/Proof 1

== Theorem ==
== 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:
:$p = -2$
:$q = 1$
:$\map R x = 2 x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$(2): \quad y' ' - 2 y' + y = 0$
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 $(1)$.
Taking the expression $\map R x = 2 x$ and differentiating twice $x$:
Trying out $y = 2 x$ in $(1)$:
:$0 - 4 + 2 x = 2 x - 4$
which is off by that constant of $4$.
But by Derivative of Constant:
:$4' = 0$
Hence setting $y = 2 x + 4$:
:$0 - 4 + \paren {2 x + 4} = 2 x$
and it can be seen that $y_p = 2 x + 4$ is indeed a particular solution to $(1)$.
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^x + C_2 x e^x + 2 x + 4$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.19$: Problem $1$
Category:Linear Second Order ODE/y - 2 y' + y = 2 x
Linear Second Order ODE/y'' - 2 y' + y = 2 x/Proof 2

== Theorem ==
== 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:
:$p = -2$
:$q = 1$
:$\map R x = 2 x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$(2): \quad y' ' - 2 y' + y = 0$
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:
:$y = y_g + y_p = C_1 e^x + C_2 x e^x + 2 x + 4$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.19$: Problem $1$
Category:Linear Second Order ODE/y - 2 y' + y = 2 x
Category:Examples of Method of Variation of Parameters
Linear Second Order ODE/y'' - 2 y' + y = 2 x/Proof 3

== 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:
:$p = -2$
:$q = 1$
:$\map R x = 2 x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$(2): \quad y' ' - 2 y' + y = 0$
From Linear Second Order ODE: $y - 2 y' + y = 0$, this has the general solution:
:$y_g = C_1 e^x + C_2 x e^x$
We have that:
:$R \left({x}\right) = 2 x$
So from the Method of Undetermined Coefficients for Polynomial:
:$y_p = A_0 + A_1 x$
Hence:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^x + C_2 x e^x + 2 x + 4$
is the general solution to $(1)$.
Category:Linear Second Order ODE/y - 2 y' + y = 2 x
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 2 y' + y = exp x

== Theorem ==
The second order ODE:
:$(1): \quad y - 2 y' + y = e^x$
has the general solution:
:$y = C_1 e^x + C_2 x e^x + \dfrac {x^2 e^x} 2$
== 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:
:$p = -2$
:$q = 1$
:$\map R x = e^x$
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:
:$y_g = C_1 e^x + C_2 x e^x$
We have that:
:$\map R x = e^x$
and so from the Method of Undetermined Coefficients for the Exponential function:
:$y_p = \dfrac {K x^2 e^{a x} } 2$
where:
:$K = 1$
:$a = 1$
Hence:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^x + C_2 x e^x + \dfrac {x^2 e^x} 2$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.5$ Particular solution: exponential $\map f x$: Example $5$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 2 y' - 3 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - 2 y' - 3 y = 0$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 2 m - 3 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 3$
:$m_2 = -1$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{3 x} + C_2 e^{-x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - 2 y' - 3 y = 64 x exp -x

== Theorem ==
The second order ODE:
:$(1): \quad y - 2 y' - 3 y = 64 x e^{-x}$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-x} - e^{-x} \paren {8 x^2 + 4 x + 1}$
== 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:
:$p = -2$
:$q = -3$
:$\map R x = 64 x e^{-x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y - 2 y' - 3 y = 0$
From Linear Second Order ODE: $y - 2 y' - 3 y = 0$, this has the general solution:
:$y_g = C_1 e^{3 x} + C_2 e^{-x}$
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:
:$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-x} - e^{-x} \paren {8 x^2 + 4 x + 1}$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.19$: Problem $3 \ \text{(c)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Variation of Parameters
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:
:$y = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x}$
== 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:
:$m_1 = 1 + \sqrt 6$
:$m_2 = 1 - \sqrt 6$
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x

== Theorem ==
The second order ODE:
:$(1): \quad y - 2 y' - 5 y = 2 \cos 3 x - \sin 3 x$
has the general solution:
:$y = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x} + \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$
== 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:
:$p = -2$
:$q = -5$
:$\map R x = 2 \cos 3 x - \sin 3 x$
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:
:$y_g = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x}$
We have that:
:$\map R x = 2 \cos 3 x - \sin 3 x$
and it is noted that $2 \cos 3 x - \sin 3 x$ is not itself a particular solution of $(2)$.
We then determine the particular solution:
=== Particular Solution ===
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x} + \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Category:Linear Second Order ODE/y - 2 y' - 5 y = 2 cos 3 x - sin 3 x
Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x/Particular Solution

== Theorem ==
The second order ODE:
:$(1): \quad y - 2 y' - 5 y = 2 \cos 3 x - \sin 3 x$
has a particular solution:
:$y_p = \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$
== Proof ==
From Linear Second Order ODE: $y - 2 y' - 5 y = 0$, we have established that the general solution to $(1)$ is:
:$y_g = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x}$
=== Solution using Trigonometric Form ===
=== Solution using Exponential Form ===
Category:Linear Second Order ODE/y - 2 y' - 5 y = 2 cos 3 x - sin 3 x
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x/Particular Solution/Exponential Form

== Theorem ==
== Proof ==
From Linear Second Order ODE: $y - 2 y' - 5 y = 0$, we have established that the general solution to $(1)$ is:
:$y_g = C_1 \, \map \exp {\paren {1 + \sqrt 6} x} + C_2 \, \map \exp {\paren {1 - \sqrt 6} x}$
We note that $2 \cos 3 x - \sin 3 x$ is not itself a particular solution of $(2)$.
From the Method of Undetermined Coefficients for Sine and Cosine:
:$y_p = A \cos 3 x + B \sin 3 x$
where $A$ and $B$ are to be determined.
The of $(1)$ is the real part of $\paren {2 + i} e^{3 i x}$.
Thus, to find a particular solution of $(1)$ when the is $\paren {2 + i} e^{3 i x}$, we substitute $y = A e^{3 i x}$.
Hence:
Taking the real part of the corresponding solution $A e^{3 i x}$, we get:
Hence the result:
:$y_p = \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$
== 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 $7$
Category:Linear Second Order ODE/y - 2 y' - 5 y = 2 cos 3 x - sin 3 x
Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x/Particular Solution/Trigonometric Form

== Theorem ==
== Proof ==
From Linear Second Order ODE: $y - 2 y' - 5 y = 0$, we have established that the general solution to $(1)$ is:
:$y_g = C_1 \, \map \exp {\paren {1 + \sqrt 6} x} + C_2 \, \map \exp {\paren {1 - \sqrt 6} x}$
We note that $2 \cos 3 x - \sin 3 x$ is not itself a particular solution of $(2)$.
From the Method of Undetermined Coefficients for Sine and Cosine:
:$y_p = A \cos 3 x + B \sin 3 x$
where $A$ and $B$ are to be determined.
Hence:
Substituting into $(1)$:
Hence the result:
:$y_p = \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$
== 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 $6$
Category:Linear Second Order ODE/y - 2 y' - 5 y = 2 cos 3 x - sin 3 x
Linear Second Order ODE/y'' - 2 y' = 0

== Theorem ==
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:
:$y = C_1 e^{0 x} + C_2 e^{2 x} = C_1 + C_2 e^{2 x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - 2 y' = 12 x - 10

== Theorem ==
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:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 + C_2 e^{2 x} + 2 x - 3 x^2$
== Sources ==
* : $\S 3.18$: Problem $1 \ \text{(h)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 3 y' + 2 y = 0

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

== Theorem ==
== Proof ==
Consider the functions:
:$\map {y_1} x = e^x$
:$\map {y_2} x = e^{2 x}$
We have that:
Putting $e^x$ and $e^{2 x}$ into $(1)$ in turn:
Hence it can be seen that:
are particular solutions to $(1)$.
Calculating the Wronskian of $y_1$ and $y_2$:
So the Wronskian of $y_1$ and $y_2$ is never zero.
Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$y_1$ and $y_2$ are linearly independent everywhere on $\R$.
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 $\map P x = -3$ and $\map Q x = 2$.
So by Constant Real Function is Continuous:
:$P$ and $Q$ are continuous on $\R$.
Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$(1)$ has the general solution:
::$y = C_1 e^x + C_2 e^{2 x}$
== Sources ==
* : $\S 3.15$: Problem $3$
Category:Linear Second Order ODE/y - 3 y' + 2 y = 0
Linear Second Order ODE/y'' - 3 y' + 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:
:$(2): \quad: m^2 - 3 m + 2 = 0$
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:Linear Second Order ODE/y - 3 y' + 2 y = 0
Linear Second Order ODE/y'' - 3 y' + 2 y = 0/y(0) = -1, y'(0) = 1

== Theorem ==
The second order ODE:
:$(1): \quad y - 3 y' + 2 y = 0$
with initial conditions:
:$\map y 0 = -1$
:$\map {y'} 0 = 1$
has the particular solution:
:$y = -3 e^x + 2 e^{2 x}$
== Proof ==
From Linear Second Order ODE: $y - 3 y' + 2 y = 0$, the general solution of $(1)$ is:
:$y = C_1 e^x + C_2 e^{2 x}$
Differentiating $x$:
:$y' = C_1 e^x + 2 C_2 e^{2 x}$
Thus for the initial conditions:
and:
Hence the result.
== Sources ==
* : $\S 3.15$: Problem $3$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - 3 y' + 2 y = 14 sine 2 x - 18 cosine 2 x

== Theorem ==
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:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^x + C_2 e^{2 x} + 2 \sin 2 x + 3 \cos 2 x$
== Sources ==
* : $\S 3.18$: Problem $1 \ \text{(f)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 3 y' + 2 y = 5 exp 3 x

== Theorem ==
The second order ODE:
:$(1): \quad y - 3 y' + 2 y = 5 e^{3 x}$
has the general solution:
:$y = C_1 e^x + C_2 e^{2 x} + \dfrac {5 e^{3 x} } 2$
== 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:
:$p = -3$
:$q = 2$
:$\map R x = 5 e^{3 x}$
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:
:$\map R x = 5 e^{3 x}$
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:
:$K = 5$
:$a = 3$
:$p = -3$
:$q = 2$
Hence:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^x + C_2 e^{2 x} + \dfrac {5 e^{3 x} } 2$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.4$ Particular solution: exponential $\map f x$: Example $3$
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:
:$y = C_1 e^{2 x} + C_2 x e^{2 x}$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Category:Linear Second Order ODE/y - 4 y' + 4 y = 0
Linear Second Order ODE/y'' - 4 y' + 4 y = 0/Proof 1

== Theorem ==
== Proof ==
Consider the functions:
:$\map {y_1} x = e^{2 x}$
:$\map {y_2} x = x e^{2 x}$
We have that:
Putting $e^{2 x}$ and $x e^{2 x}$ into $(1)$ in turn:
Hence it can be seen that:
are particular solutions to $(1)$.
Calculating the Wronskian of $y_1$ and $y_2$:
So the Wronskian of $y_1$ and $y_2$ is never zero.
Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$y_1$ and $y_2$ are linearly independent everywhere on $\R$.
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 $\map P x = -4$ and $\map Q x = 4$.
So by Constant Real Function is Continuous:
:$P$ and $Q$ are continuous on $\R$.
Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$(1)$ has the general solution:
::$y = C_1 e^{2 x} + C_2 x e^{2 x}$
== Sources ==
* : $\S 3.15$: Problem $4$
Category:Linear Second Order ODE/y - 4 y' + 4 y = 0
Linear Second Order ODE/y'' - 4 y' + 4 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:
:$(2): \quad: m^2 - 4 m + 4 = 0$
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:
:$y = C_1 e^{2 x} + C_2 x e^{2 x}$
== Sources ==
* : $\S 3.17$: Problem $1 \ \text{(e)}$
Category:Linear Second Order ODE/y - 4 y' + 4 y = 0
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:
:$y = C_1 e^{5 x} + C_2 e^{-x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 4 m - 5 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 5$
:$m_2 = -1$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{5 x} + C_2 e^{-x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - 4 y' - 5 y = x^2/y(0) = 1, y'(0) = -1

== Theorem ==
Consider the second order ODE:
:$(1): \quad y - 4 y' - 5 y = x^2$
whose initial conditions are:
:$y = 1$ when $x = 0$
:$y' = -1$ when $x = 0$
$(1)$ has the particular solution:
:$y = \dfrac {e^{5 x} } {375} + \dfrac {4 e^{-x} } 3 - \dfrac {x^2} 5 + \dfrac {8 x} {25} - \dfrac {42} {125}$
== 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:
:$p = -4$
:$q = -5$
:$\map R x = x^2$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$(2): \quad y - 4 y' - 5 y = 0$
From Linear Second Order ODE: $y - 4 y' - 5 y = 0$, this has the general solution:
:$y_g = C_1 e^{5 x} + C_2 e^{-x}$
We have that:
:$\map R x = x^2$
and it is noted that $x^2$ 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:
:$y = y_g + y_p = C_1 e^{5 x} + C_2 e^{-x} - \dfrac 1 5 x^2 + \dfrac 8 {25} x - \dfrac {42} {125}$
Differentiating $x$:
:$y' = 5 C_1 e^{5 x} - C_2 e^{-x} - \dfrac 2 5 x + \dfrac 8 {25}$
Substituting the initial conditions, remembering that $e^0 = 1$:
The result follows.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.7$ Arbitrary constants and initial conditions: Example $9$
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:
:$y = C_1 e^{2 x} + C_2 e^{-2 x}$
== 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:
:$m_1 = 2$
:$m_2 = -2$
These are complex and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{2 x} + C_2 e^{-2 x}$
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:
:$y = C_1 e^{k x} + C_2 e^{-k x}$
where $k = 2$.
Hence the result.
Category:Linear Second Order ODE/y - 4 y = 0