hcju/proofwikiqa huggingface dataset:
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}}
Consider the functions: :$\map {y_1} x = e^x$ :$\map {y_2} x = e^{2 x}$ We have that: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} \, e^x       | r = e^x       | c = Power Rule for Derivatives }} {{eqn | l = \frac {\d} {\d x} \, e^{2 x}       | r = 2 e^{2 x}       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d^2} {\d x^2} \, e^x       | r = e^x       | c =  }} {{eqn | l = \frac {\d^2} {\d x^2} \, e^{2 x}       | r = 4 e^{2 x}       | c =   }} {{end-eqn}} Putting $e^x$ and $e^{2 x}$ into $(1)$ in turn: {{begin-eqn}} {{eqn | l = e^0 - 3 e^0 + 2 e^0       | r = 1 - 3 + 2       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 4 e^{2 x} - 3 \times 2 e^{2 x} + 2 e^{2 x}       | r = 4 - 6 + 2       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} Hence it can be seen that: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = e^x }} {{eqn | l = \map {y_2} x       | r = 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^x & e^{2 x} \\ e^x & 2 e^{2 x} \end {vmatrix}       | c =  }} {{eqn | r = e^x \times 2 e^{2 x} - e^{2 x} \times e^x       | c =  }} {{eqn | r = 2 e^{3 x} - e^{3 x}       | c =  }} {{eqn | r = e^{3 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 $\map P x = -3$ 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: ::$y = C_1 e^x + C_2 e^{2 x}$ {{qed}}
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: :$y = C_1 e^x + C_2 e^{2 x}$ {{qed}}
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 rwwnplqveq38970wxbsx99ciiyn8p2t
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 pd5h9hruyy2du3ixzw31q3zyp2m3zl4
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}}
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = {y_1}'       | r = 1       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_1}''       | r = 0       | c = Derivative of Constant }} {{end-eqn}} and so by inspection: :$y_1 = x$ is a particular solution of $(1)$. $(1)$ can be expressed as: :$(2): \quad y'' + \dfrac 2 x y' - \dfrac 2 {x^2} y = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = \dfrac 2 x$ 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 \dfrac 2 x \rd x       | c =  }} {{eqn | r = 2 \ln x       | c = Primitive of Reciprocal }} {{eqn | r = \ln {x^2}       | c = Logarithm of Power }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-\ln {x^2} }       | c =  }} {{eqn | r = \frac 1 {x^2}       | 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 {x^2} \frac 1 {x^2} \rd x       | c =  }} {{eqn | r = \int \dfrac 1 {x^4} \rd x       | c =  }} {{eqn | r = -\frac 1 {3 x^3}       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = \paren {-\frac 1 {3 x^3} } x       | c =  }} {{eqn | r = -\frac 1 {3 x^2}       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 x + k \paren {-\dfrac 1 {3 x^2} }$ where $k$ is arbitrary. Setting $C_2 = -\dfrac k 3$: :$y = C_1 x + \dfrac {C_2} {x^2}$ {{qed}}
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 kjp37andmikkls9xbinool6klbb8a0j
It can be seen that $(1)$ is an instance of the Cauchy-Euler Equation: :$x^2 y'' + p x y' + q y = 0$ where: :$p = 2$ :$q = -2$ By Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE, this can be expressed as: :$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t} + q y = 0$ by making the substitution: :$x = e^t$ Hence it can be expressed as: :$(2): \quad \dfrac {\d^2 y} {\d t^2} + \dfrac {\d y} {\d t} - 2 y = 0$ From Linear Second Order ODE: $y'' + y' - 2 y = 0$, this has the general solution: {{begin-eqn}} {{eqn | l = y       | r = C_1 e^t + C_2 e^{-2 t}       | c =  }} {{eqn | r = C_1 x + C_2 x^{-2}       | c = substituting $x = e^t$ }} {{eqn | r = C_1 x + \frac {C_2} {x^2}       | c =  }} {{end-eqn}} {{qed}}
Note that: {{begin-eqn}} {{eqn | l = y_1       | r = x^2       | c =  }} {{eqn | ll= \leadsto       | l = y'       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = y''       | r = 2       | c = Power Rule for Derivatives }} {{end-eqn}} and so by inspection: :$y_1 = x^2$ is a particular solution of $(1)$. $(1)$ can be expressed as: :$(2): \quad y'' + \dfrac 1 x y' - \dfrac 4 {x^2} y = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = -\dfrac 4 {x^2}$ 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 \dfrac 1 x \rd x       | c =  }} {{eqn | r = \ln x       | c = Primitive of Reciprocal }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-\ln x}       | c =  }} {{eqn | r = \frac 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 {x^4} \frac 1 x \rd x       | c =  }} {{eqn | r = \int \dfrac 1 {x^5} \rd x       | c =  }} {{eqn | r = - \frac 1 {4 x^4}       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = -\frac 1 {4 x^4} x^2       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 x^2 + k \paren {-\dfrac 1 {4 x^2} }$ where $k$ is arbitrary. Setting $C_2 = - \dfrac k 4$ yields the result: :$y = C_1 x^2 + \dfrac {C_2} {x^2}$ {{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: :$p = 1$ :$q = -4$ 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: :$(2): \quad \dfrac {\d^2 y} {\d t^2} - 4 y = 0$ From Linear Second Order ODE: $y'' - 4 y = 0$, this has the general solution: {{begin-eqn}} {{eqn | l = y       | r = C_1 e^{2 t} + C_2 e^{-2 t}       | c =  }} {{eqn | r = C_1 x^2 + C_2 x^{-2}       | c = substituting $x = e^t$ }} {{eqn | r = C_1 x^2 + \frac {C_2} {x^2}       | c =  }} {{end-eqn}} {{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: :$m_1 = 2$ :$m_2 = -2$ 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}}
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}}
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = p \frac {\d p} {\d y}        | r = k^2 y       | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto       | l = p^2       | r = k^2 y^2 + k^2 \alpha       | c = First Order ODE: $y \rd y = k x \rd x$ }} {{eqn | ll= \leadsto       | l = p = \dfrac {\d y} {\d x}       | r = \pm k \sqrt {y^2 + k^2 \alpha}       | c =  }} {{eqn | ll= \leadsto       | l = \int \dfrac {\d y} {\sqrt {y^2 + \alpha} }       | r = \int \pm k \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \map \ln {y + \sqrt{y^2 + \alpha} }       | r = \pm k x + \beta       | c = Primitive of $\dfrac 1 {\sqrt {x^2 + k} }$ }} {{eqn | ll= \leadsto       | l = y + \sqrt {y^2 + \alpha}       | r = e^{\pm k x + \beta}       | c =  }} {{eqn | r = C e^{\pm k x}       | c = where $C = e^\beta$ }} {{eqn | ll= \leadsto       | l = y^2 + \alpha       | r = \paren {C e^{\pm k x} - y}^2       | c =  }} {{eqn | r = C^2 e^{\pm 2 k x} - 2 C e^{\pm k x} + y^2       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \frac {C^2 e^{\pm 2 k x} - \alpha} {2 C e^{\pm k x} }       | c = Quadratic Formula }} {{eqn | r = \frac {C e^{\pm k x} - \frac \alpha C e^{\mp k x} } 2       | c = Quadratic Formula }} {{end-eqn}} Setting $C_1 = \dfrac C 2$ and $C_2 = - \dfrac \alpha {2 C}$: :$y = C_1 e^{\pm k x} + C_2 e^{\mp k x}$ which is the same thing as: :$y = C_1 e^{k x} + C_2 e^{-k x}$ by allowing for the constants to be interchanged. {{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: :$m_1 = k$ :$m_2 = -k$ These are real and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 e^{k x} + C_2 e^{-k x}$ {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 - 1 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = 1$ :$m_2 = -1$ These are complex and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 e^x + C_2 e^{- x}$ {{qed}}
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 $k = 1$. Hence the result. {{qed}}
The particular solution: :$y_1 = x$ can be found by inspection. Let $(1)$ be written as: :$(2): \quad y'' + \dfrac {y'} x - \dfrac y {x^2} = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = \dfrac 1 {x^2}$ 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 \frac 1 x \rd x       | c =  }} {{eqn | r = \ln x       | c =  }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{-\ln x}       | c =  }} {{eqn | r = \frac 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 {x^2} \frac 1 x \rd x       | c =  }} {{eqn | r = \int \dfrac {\d x} {x^3}       | c =  }} {{eqn | r = -\dfrac 1 {2 x^2}       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = x \paren {-\dfrac 1 {2 x^2} }       | c =  }} {{eqn | r = -\dfrac 1 {2 x}       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 x + k \paren {-\dfrac 1 {2 x} }$ where $k$ is arbitrary. Setting $C_2 = -\dfrac k 2$ yields the result: :$y = C_1 x + \dfrac {C_2} x$ {{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: :$p = 1$ :$q = -1$ 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: :$(2): \quad \dfrac {\d^2 y} {\d t^2} - y = 0$ From Linear Second Order ODE: $y'' - y = 0$, this has the general solution: {{begin-eqn}} {{eqn | l = y       | r = C_1 e^t + C_2 e^{-t}       | c =  }} {{eqn | r = C_1 x + C_2 x^{-1}       | c = substituting $x = e^t$ }} {{eqn | r = C_1 x + \frac {C_2} x       | c =  }} {{end-eqn}} {{qed}}
Consider the functions: :$\map {y_1} x = x$ :$\map {y_2} x = x^2$ We have that: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \, x       | r = 1       | c = Power Rule for Derivatives }} {{eqn | l = \frac \d {\d x} \, x^2       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d^2} {\d x^2} \, x       | r = 0       | c =  }} {{eqn | l = \frac {\d^2} {\d x^2} \, x^2       | r = 2       | c =   }} {{end-eqn}} Putting $x$ and $x^2$ into $(1)$ in turn: {{begin-eqn}} {{eqn | l = x^2 \cdot 0 - 2 x \cdot 1 + 2 x       | r = 2 x - 2 x       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = x^2 \cdot 2 - 2 x \cdot 2 x + 2 x^2       | r = 2 x^2 - 4 x^2 + 2 x^2       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} Hence it can be seen that: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = x }} {{eqn | l = \map {y_2} x       | r = x^2 }} {{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} x & x^2 \\ 1 & 2 x \end{vmatrix}       | c =  }} {{eqn | r = 2 x^2 - x^2       | c =  }} {{eqn | r = x^2       | c =  }} {{end-eqn}} So the Wronskian of $y_1$ and $y_2$ is zero only when $x = 0$. Let $\Bbb I = \closedint a b$ be a closed real interval such that $0 \notin \Bbb I$. Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent: :$y_1$ and $y_2$ are linearly independent on $\Bbb I$. We can manipulate $(1)$ is a homogeneous linear second order ODE in the form: :$y'' + \map P x y' + \map Q x y = 0$ where $\map P x = \dfrac {-2 x} {x^2}$ and $\map Q x = \dfrac 2 {x^2}$. So by Real Rational Function is Continuous: $P$ and $Q$ are continuous on $\closedint a b$ Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$(1)$ has the general solution: ::$y = C_1 x + C_2 x^2$ :on any closed real interval $\Bbb I$ which does not include $0$. {{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: :$p = -2$ :$q = 2$ 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: :$(2): \quad \dfrac {\d^2 y} {\d t^2} - \dfrac {\d y} {\d t^2} + 2 y = 0$ From Linear Second Order ODE: $y'' - 3 y' + 2 y = 0$, this has the general solution: {{begin-eqn}} {{eqn | l = y       | r = C_1 e^t + C_2 e^{2 t}       | c =  }} {{eqn | r = C_1 x + C_2 x^2       | c = substituting $x = e^t$ }} {{end-eqn}} {{qed}} Note that when $x = 0$, $e^t = x$ has no solution for $t$, and so is excluded from this solution.
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 q6aiorh940jz104l1655sf7955s36e4
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}}
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}}