solve-y-y-x-

Question Number 97797 by abdomathmax last updated on 09/Jun/20

solve y^{^{″}} - y = x

Answered by niroj last updated on 09/Jun/20

y^{^{″}} - y = x

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

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

m = \bar{+} 1

C F = C_{1} e^{x} + C_{2} e^{- x}

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

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

W = - e^{- x} . e^{x} - e^{- x} . e^{x} = - 2

PI = - y_{1} \int \frac{y_{2} Q}{W} dx + y_{2} \int \frac{y_{1} Q}{W} dx

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

= \frac{e^{x}}{2} {x . (- e^{- x}) - \int 1 . (- e^{x}) dx} - \frac{e^{- x}}{2} {x . e^{x} - \int 1 . e^{x} dx}

= \frac{e^{x}}{2} (- {xe}^{- x} + e^{x}) - \frac{e^{- x}}{2} ({xe}^{x} - e^{x})

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

= \frac{1}{2} ({2 e}^{2 x} - 2 x) = e^{2 x} - x

∴ y = C_{1} e^{x} + C_{2} e^{- x} + e^{2 x} - x / / .

Answered by mathmax by abdo last updated on 09/Jun/20

(he) \to y^{^{″}} - y = 0 \to r^{2} - 1 = 0 \Rightarrow r = \bar{+} 1 \Rightarrow y_{h} = α e^{x} + β e^{- x} = α u_{1} + β u_{2}

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

W_{1} = | \begin{matrix} 0 e^{- x} \\ x - e^{- x} \end{matrix} | = - {xe}^{- x} and W_{2} = | \begin{matrix} e^{x} 0 \\ e^{x} x \end{matrix} | = {xe}^{x}

v_{1} = \int \frac{w_{1}}{w} dx = \int \frac{- {xe}^{- x}}{- 2} dx = \frac{1}{2} \int {xe}^{- x} dx = \frac{1}{2} {- {xe}^{- x} + \int e^{- x} dx}

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

v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{{xe}^{x}}{- 2} dx = - \frac{1}{2} \int {xe}^{x} dx = - \frac{1}{2} {{xe}^{x} - e^{x}} = \frac{1}{2} (1 - x) e^{x}

so the particular solution is

y_{p} = u_{1} v_{1} + u_{2} v_{2} = - \frac{1}{2} (x + 1) + \frac{1}{2} (1 - x) = - x the general solution is

y (x) = y_{h} + y_{p} = α e^{x} + β e^{- x} - x

Answered by mathmax by abdo last updated on 10/Jun/20

let solve it by laplace

e \Rightarrow L (y^{^{″}}) - L (y) = L (x) \Rightarrow x^{2} L (y) - xy (o) - y^{^{'}} (0) - L (y) = L (x) \Rightarrow

(x^{2} - 1) L (y) = xy (0) + y^{^{'}} (0) + L (x) we have L (x) = \int_{0}^{\infty} t e^{- xt} dt

= {[- \frac{t}{x} e^{- xt}]}_{t = 0}^{\infty} + \frac{1}{x} \int_{0}^{\infty} e^{- xt} dt = - \frac{1}{x^{2}} {[e^{- xt}]}_{t = 0}^{\infty} = \frac{1}{x^{2}}

(also we can use L (x^{n}) = \frac{n!}{x^{n + 1}}) e \Rightarrow (x^{2} - 1) L (y) = xy (o) + y^{^{'}} (0) + \frac{1}{x^{2}} \Rightarrow

L (y) = \frac{x}{x^{2} - 1} y (0) + \frac{y^{^{'}} (0)}{x^{2} - 1} + \frac{1}{x^{2} (x - 1)} \Rightarrow y (x) = y (0) L^{- 1} (\frac{x}{x^{2} - 1}) + y^{^{'}} (0) L^{- 1} (\frac{1}{x^{2} - 1})

+ L^{- 1} (\frac{1}{x^{2} (x - 1)}) we havef (x) \frac{x}{x^{2} - 1} = \frac{1}{2} x (\frac{1}{x - 1} - \frac{1}{x + 1})

= \frac{1}{2} (\frac{x}{x - 1} - \frac{x}{x + 1}) = \frac{1}{2} (\frac{x - 1 + 1}{x - 1} - \frac{x + 1 - 1}{x + 1}) = \frac{1}{2} (\frac{1}{x - 1} + \frac{1}{x + 1}) \Rightarrow

L^{- 1} (f) = \frac{1}{2} e^{x} + \frac{1}{2} e^{- x}

g (x) = \frac{1}{x^{2} - 1} = \frac{1}{2} (\frac{1}{x - 1} - \frac{1}{x + 1}) \Rightarrow L^{- 1} (g) = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x}

h (x) = \frac{1}{x^{2} (x - 1)} = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{x - 1}

b = - 1, c = 1 \lim_{x \to + \infty} xh (x) = 0 = a + c \Rightarrow a = - 1 \Rightarrow

h (x) = - \frac{1}{x} - \frac{1}{x^{2}} + \frac{1}{x - 1} \Rightarrow L^{- 1} (h) = - 1 - x + e^{x} \Rightarrow

y (x) = y (0) {\frac{e^{x} + e^{- x}}{2}} + y^{^{'}} (0) {\frac{e^{x} - e^{- x}}{2}} - 1 - x + e^{x}

= (\frac{y (o) + y^{^{'}} (0)}{2} + 1) e^{x} + (\frac{y (0) - y^{^{'}} (0)}{2}) e^{- x} - x - 1