y-2y-y-x-2-e-x-cos-x-what-is-particular-solution-

Question Number 92852 by i jagooll last updated on 09/May/20

y^{″} + {2 y}^{'} + y = x^{2} e^{- x} \cos x

what is particular solution

Commented by i jagooll last updated on 09/May/20

Mr Niroj's favorite ��

Answered by Joel578 last updated on 09/May/20

∙ Homogeneous solution with char . eq .

λ^{2} + 2 λ + 1 = 0 \to λ_{1, 2} = - 1

\Rightarrow y_{h} (x) = c_{1} e^{- x} + c_{2} {x e}^{- x}

∙ Particular solution for r (x) = x^{2} e^{- x} \cos x

using variation of parameters

Let y_{p} (x) = u (x) y_{1} (x) + v (x) y_{2} (x)

with y_{1} (x) = e^{- x}, y_{2} (x) = {x e}^{- x}

and u (x) = - \int \frac{r (x) y_{2} (x)}{W (x)} d x, v (x) = \int \frac{r (x) y_{1} (x)}{W (x)} d x

Now,

W (x) = | \begin{matrix} y_{1} & y_{2} \\ y_{1}^{'} & y_{2}^{'} \end{matrix} | = | \begin{matrix} e^{- x} & {x e}^{- x} \\ - e^{- x} & e^{- x} - {x e}^{- x} \end{matrix} |

= e^{- 2 x}

Hence,

u (x) = - \int \frac{(x^{2} e^{- x} \cos x) ({x e}^{- x})}{e^{- 2 x}} d x = - \int x^{3} \cos x d x

v (x) = \int \frac{(x^{2} e^{- x} \cos x) (e^{- x})}{e^{- 2 x}} d x = \int x^{2} \cos x d x

\Rightarrow y_{p} (x) = \dots

Commented by i jagooll last updated on 09/May/20

tobe continue sir

Commented by Joel578 last updated on 09/May/20

I think the integrals are easy for someone that asking

an ODE problem, so I leave the rest for you

Commented by Joel578 last updated on 09/May/20

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Commented by i jagooll last updated on 09/May/20

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Commented by i jagooll last updated on 09/May/20

v (x) = x^{2} \sin x + 2 xcos x - 2 \sin x

u (x) = - x^{3} \sin x - {3 x}^{2} \cos x + 6 xsin x + 6 \cos x

Answered by niroj last updated on 09/May/20

PI = \frac{x^{2} e^{- x} \cos x}{D^{2} + 2 D + 1}

= e^{- x} \frac{x^{2} \cos x}{- 1 + 2 D + 1}

= e^{- x} \frac{x^{2} \cos x}{2 D}

= \frac{e^{- x}}{2} \int x^{2} \cos x dx

x^{2} (\sin x) - \int 2 x (\sin x) dx

x^{2} \sin x - 2 [x (- \cos x) - \int 1 . (- cosx) dx]

x^{2} \sin x + 2 x cosx + 2 \sin x .

PI = \frac{e^{- x}}{2} (x^{2} \sin x + 2 xcos x + 2 sinx)

= e^{- x} (\frac{1}{2} x^{2} \sin x + x \cos x + \sin x) .

Commented by i jagooll last updated on 09/May/20

I have a lot to learn in this way sir ��

Commented by niroj last updated on 09/May/20

you can solve by other method also

like using by variation of parameter .

Answered by niroj last updated on 09/May/20

By the help of variation parametrs,

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

A . E ., m^{2} + 2 m + 1 = 0

{(m + 1)}^{2} = 0

m = - 1, - 1

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

= C_{1} e^{- x} + C_{2} {xe}^{- x}

PI = - y_{1} \int \frac{y_{2} xdx}{w} + y_{2} \int \frac{y_{1} xdx}{w}

y_{1} = e^{- x}, y_{2} = x e^{- x},

{Wronskian}^{^{'}} s determinants :

W = | \begin{matrix} y_{1} y_{2} \\ y_{1}^{^{'}} y_{2}^{^{'}} \end{matrix} |

= | \begin{matrix} e^{- x} e^{- x} x \\ - e^{- x} e^{- x} - {xe}^{- x} \end{matrix} |

w = e^{- x} (e^{- x} - {xe}^{- x}) + (e^{- x} x . e^{- x})

= e^{- 2 x} - {xe}^{- 2 x} + {xe}^{- 2 x}

= e^{- 2 x}

PI = - e^{- x} \int \frac{x e^{- x} x^{2} e^{- x} \cos x dx}{e^{- 2 x}} + {xe}^{- x} \int \frac{e^{- x} x^{2} e^{- x} \cos x dx}{e^{- 2 x}}

= - e^{- x} \int x^{3} \cos x dx + x e^{- x} \int x^{2} \cos xdx

now can get easily \dots .

Commented by john santu last updated on 09/May/20

great...cheer ��