D-2-2D-1-y-x-ln-x-

Question Number 103201 by bobhans last updated on 13/Jul/20

(D^{2} - 2 D + 1) y = x \ln (x)

Commented by bobhans last updated on 14/Jul/20

t h a n k y o u b o t h

Answered by bramlex last updated on 14/Jul/20

homogenous solution

η^{2} - 2 η + 1 = 0; {(η - 1)}^{2} = 0

y_{h} = C_{1} e^{x} + C_{2} {xe}^{x}

now we use variation of

parameters y = u . e^{x} + v . {xe}^{x}

differentiating yields

y^{'} = u^{'} e^{x} + {ue}^{x} + v^{'} {xe}^{x} + v (x + 1) e^{x}

setting u^{'} e^{x} + v^{'} {xe}^{x} = 0

y^{″}

= u^{'} e^{x} + {ue}^{x} + v^{'} (x + 1) e^{x} + v (x + 2) e^{x}

substituting y, y^{'}, y^{″} into original

differential equation gives us

(u^{'} e^{x} + {ue}^{x} + v^{'} (x + 1) e^{x} + v (x + 2) e^{x})

- 2 ({ue}^{x} + v (x + 1) e^{x}) + ({ue}^{x} + {vxe}^{x}) = xln (x)

which simplifies to

u^{'} + (x + 1) v^{'} = {xe}^{- x} \ln (x)

now we have two equations for

u^{'} and v^{'} \Rightarrow u^{'} e^{x} + v^{'} {xe}^{x} = 0

and u^{'} + (x + 1) v^{'} = {xe}^{- x} \ln (x)

solving this system of equation

u^{'} = - x^{2} e^{- x} \ln (x) and

v^{'} = {xe}^{- x} \ln (x) . integrating gives

us u = - \underset{1}{\int^{x}} t^{2} e^{- t} \ln (t) dt and

v = \underset{1}{\int^{x}} {te}^{- t} \ln (t) dt

hence we conclude that a

general solution is given by

y = C_{1} e^{x} + C_{2} {xe}^{x} - e^{x} \underset{1}{\int^{x}} t^{2} e^{- t} \ln (t) dt + {xe}^{x} \underset{1}{\int^{x}} {te}^{- t} \ln (t) dt

Answered by abdomsup last updated on 13/Jul/20

(h e) \to y^{^{″}} - 2 y^{^{'}} + y = 0 \Rightarrow

r^{2} - 2 r + 1 = 0 \Rightarrow {(r - 1)}^{2} = 0 \Rightarrow r = 1

\Rightarrow y_{h} = (a x + b) e^{x} = {a x e}^{x} + {b e}^{x}

= {a u}_{1} + {b u}_{2}

W (u_{1}, u_{2}) = | \begin{matrix} {x e}^{x} e^{x} \\ (x + 1) e^{x} e^{x} \end{matrix} |

= {x e}^{2 x} - (x + 1) e^{2 x}

= - e^{2 x} \neq 0

w_{1} = | \begin{matrix} 0 e^{x} \\ x l n x e^{x} \end{matrix} | = - {x e}^{x} l n x

w_{2} = | \begin{matrix} {x e}^{x} o \\ (x + 1) e^{x} x l n x \end{matrix} | =

= x^{2} e^{x} l n x

v_{1} = \int \frac{w_{1}}{w} d x = \int \frac{{x e}^{x} l n x}{e^{2 x}} d x

= \int x e^{- x} l n x d x

v_{2} = \int \frac{w_{2}}{w} d x = - \int \frac{x^{2} e^{x} l n x}{e^{2 x}} d x

= - \int x^{2} e^{- x} l n x d x

\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2}

= {x e}^{x} \int^{x} t e^{- t} l n t d t - e^{x} \int^{x} t^{2} e^{- t} l n t d t

y = y_{h} + y_{p}

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