hcju/proofwikiqa huggingface dataset:
Consider the auxiliary equation of $(1)$: :$(2): \quad m^2 + p m + q$ Let $p^2 = 4 q$. From Solution to Quadratic Equation with Real Coefficients, $(2)$ has one (repeated) root, that is: :$m_1 = m_2 = -\dfrac p 2$ From Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation: :$y_1 = e^{m_1 x}$ is a particular solution to $(1)$. From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int p \rd x       | c =  }} {{eqn | r = p x       | c =  }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-p x}       | c =  }} {{eqn | r = e^{2 m_1 x}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int \dfrac 1 {e^{2 m_1 x} } e^{2 m_1 x} \rd x       | c = as $y_1 = e^{m_1 x}$ }} {{eqn | r = \int \rd x       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = x e^{m_1 x}       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 + m - 6 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = -3$ :$m_2 = 2$ These are real and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 e^{2 x} + C_2 e^{-3 x}$ {{qed}}
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: :$y = C_1 e^{-x} + C_2 x e^{-x}$ {{qed}}
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}}
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}}
Consider the functions: :$\map {y_1} x = e^{2 x}$ :$\map {y_2} x = x e^{2 x}$ We have that: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \, e^{2 x}       | r = 2 e^{2 x}       | c = Power Rule for Derivatives }} {{eqn | l = \frac \d {\d x} \, x e^{2 x}       | r = 2 x e^{2 x} + e^{2 x}       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d^2} {\d x^2} \, e^{2 x}       | r = 4 e^{2 x}       | c =  }} {{eqn | l = \frac {\d^2} {\d x^2} \, e^{2 x}       | r = 4 x e^{2 x} + 2 e^{2 x} + 2 e^{2 x}       | c =   }} {{eqn | r = 4 x e^{2 x} + 4 e^{2 x}       | c =   }} {{end-eqn}} Putting $e^{2 x}$ and $x e^{2 x}$ into $(1)$ in turn: {{begin-eqn}} {{eqn | l = 4 e^0 - 4 \times 2 e^0 + 4 e^0       | r = 8 - 8 + 4       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 4 \times 0 e^0 + 4 e^0 - 4 \paren {2 \times 0 e^0 + e^0} + 4 \paren {0 e^0}       | r = 0 + 4 - 4 + 0       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} Hence it can be seen that: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = e^{2 x} }} {{eqn | l = \map {y_2} x       | r = x e^{2 x} }} {{end-eqn}} are particular solutions to $(1)$. Calculating the Wronskian of $y_1$ and $y_2$: {{begin-eqn}} {{eqn | l =\map  W {y_1, y_2}       | r = \begin{vmatrix} e^{2 x} & x e^{2 x} \\ 2 e^{2 x} & \paren {2 x + 1} e^{2 x} \end{vmatrix}       | c =  }} {{eqn | r = e^{2 x} \times \paren {2 x + 1} e^{2 x} - x e^{2 x} \times 2 e^{2 x}       | c =  }} {{eqn | r = \paren {2 x + 1} e^{4 x} - 2 x e^{4 x}       | c =  }} {{eqn | r = e^{4 x}       | c =  }} {{end-eqn}} 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: ::$y = C_1 e^{2 x} + C_2 x e^{2 x}$ {{qed}}
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}}
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}}
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}}
By Solution of Constant Coefficient Homogeneous LSOODE, $y$ is in one of the following forms: :$y = \begin{cases} C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\  & \\ C_1 e^{m_1 x} + C_2 x e^{m_2 x} & : p^2 = 4 q \\  & \\ e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q \end{cases}$ 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}}
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 }} {{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}}
$(1)$ 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}}
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}}
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}}
By definitions of surjection and restriction of mapping: :$F \restriction_A: A \to f^\to\left({A}\right)$ is a surjection Thus by Surjection iff Cardinal Inequality: :$\left\vert{f^\to\left({A}\right)}\right\vert \le \left\vert{A}\right\vert$ {{qed}}
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: :$\displaystyle x = \sum_{i \mathop = 1}^\infty \alpha_i p_i + \sum_{i \mathop = 1}^\infty \beta_i q_i$ Because the basis vectors are orthogonal, Pythagoras's theorem shows that: :$\displaystyle \norm x^2 = \sum_{i \mathop = 1}^\infty \norm {\alpha_i}^2 + \norm {\beta_i}^2$ 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}} mdztc7gcmgwuwht6tbkijp9rc08rkpb
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 a1l28sxz22kzbjvcnihq84ksi3eenj6
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}}
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}}
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: :$p \dfrac {\d p} {\d y} = -k^2 y$ where $p = \dfrac {\d y} {\d x}$. From: :First Order ODE: $y \rd y = k x \rd x$ this has the solution: :$p^2 = -k^2 y^2 + C$ or:  :$p^2 + k^2 y^2 = C$ As the {{LHS}} is the sum of squares, $C$ has to be positive for this to have any solutions. Thus, let $C = \alpha^2$. Then: {{begin-eqn}} {{eqn | l = p^2 + k^2 y^2       | r = k^2 \alpha^2       | c =  }} {{eqn | ll= \leadsto       | l = p = \dfrac {\d y} {\d x}       | r = \pm k \sqrt {\alpha^2 - y^2}       | c =  }} {{eqn | ll= \leadsto       | l = \int \dfrac {\d y} {\sqrt {\alpha^2 - y^2} }       | r = \int \pm k \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \arcsin \dfrac y \alpha       | r = \int \pm k x + \beta       | c = Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ }} {{eqn | ll= \leadsto       | l = y       | r = \alpha \, \map \sin {\pm k x + \beta}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = A \, \map \sin {k x + B}       | c =  }} {{end-eqn}} From Multiple of Sine plus Multiple of Cosine: Sine Form, this can be expressed as: :$y = C_1 \sin k x + C_2 \cos k x$ {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 + k^2 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = k i$ :$m_2 = -k i$ These are complex and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 \cos k x + C_2 \sin k x$ or, by disposing the constants differently: :$y = C_1 \sin k x + C_2 \cos k x$ {{qed}}
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}}
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}}
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}}
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}}
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 jzicno37hdhhn46n8hu6kaerroomr72