solve-y-2y-x-2-sinx-and-y-0-0-y-0-1-

Question Number 96772 by abdomathmax last updated on 04/Jun/20

solve y^{^{″}} - 2 y = x^{2} sinx and y (0) = 0, y^{^{'}} (0) = 1

Answered by mathmax by abdo last updated on 04/Jun/20

(he) \to y^{^{″}} - 2 y = 0 \to r^{2} - 2 = 0 \Rightarrow r = \bar{+} \sqrt{2} \Rightarrow y_{h} = a e^{\sqrt{2} x} + b e^{- \sqrt{2} x} = {au}_{1} + {bu}_{2}

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

W_{1} = | \begin{matrix} o e^{- x \sqrt{2}} \\ x^{2} sinx - \sqrt{2} e^{- \sqrt{2} x} \end{matrix} | = - x^{2} sinx e^{- x \sqrt{2}}

W_{2} = | \begin{matrix} e^{x \sqrt{2}} 0 \\ \sqrt{2} e^{x \sqrt{2}} x^{2} sinx \end{matrix} | = x^{2} sinx e^{x \sqrt{2}}

v_{1} = \int \frac{W_{1}}{W} dx = \frac{1}{2 \sqrt{2}} \int x^{2} sinx e^{- x \sqrt{2}} dx = \frac{1}{2 \sqrt{2}} Im (\int x^{2} e^{ix - x \sqrt{2}} dx)

\int x^{2} e^{(i - \sqrt{2}) x} dx =_{byparts} \frac{1}{i - \sqrt{2}} x^{2} e^{(i - \sqrt{2}) x} - \frac{1}{i - \sqrt{2}} \int 2 x e^{(i - \sqrt{2}) x} [dx

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

= \frac{x^{2}}{i - \sqrt{2}} e^{(i - \sqrt{2}) x} - \frac{2 x}{{(i - \sqrt{2})}^{2}} e^{(i - \sqrt{2}) x} + \frac{2}{{(i - \sqrt{2})}^{2}} e^{(i - \sqrt{2}) x} = \dots

v_{2} = \int \frac{W_{2}}{W} dx = - \frac{1}{2 \sqrt{2}} \int x^{2} sinx e^{x \sqrt{2}} = \dots . (we follow the same way)

\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} and general solution is y = y_{h} + y_{p}

Answered by mathmax by abdo last updated on 05/Jun/20

let use Laplace transform

(e) \Rightarrow L (y^{(2)}) - 2 L (y) = L (x^{2} sinx) \Rightarrow

x^{2} L (y) - y (0) - y^{^{'}} (0) - 2 L (y) = L (x^{2} sinx) \Rightarrow

(x^{2} - 2) L (y) = 1 + L (x^{2} sinx) we have L (x^{2} sinx) = \int_{0}^{\infty} t^{2} sint e^{- xt} dt

= Im (\int_{0}^{\infty} t^{2} e^{it - xt} dt) but \int_{0}^{\infty} t^{2} e^{(i - x) t} dt =_{by parts}

{[\frac{t^{2}}{i - x} e^{(i - x) t}]}_{0}^{\infty} - \int_{0}^{\infty} \frac{2 t}{i - x} e^{(i - x) t} dt = - \frac{2}{i - x} \int_{0}^{\infty} t e^{(i - x) t} dt

= \frac{2}{x - i} {{[\frac{t}{i - x} e^{(i - x) t}]}_{0}^{\infty} - \int_{0}^{\infty} \frac{1}{i - x} e^{(i - x) t} dt}

= \frac{2}{{(x - i)}^{2}} \times {[\frac{1}{i - x} e^{(i - x) t}]}_{0}^{\infty} = \frac{2}{{(x - i)}^{3}} = \frac{2 {(x + i)}^{3}}{{(x^{2} + 1)}^{3}} = \frac{2 (x^{3} + {3 x}^{2} i - 3 x - i)}{{(x^{2} + 1)}^{3}}

= \frac{{2 x}^{3} + {6 x}^{2} i - 6 x - 2 i}{{(x^{2} + 1)}^{3}} \Rightarrow L (x^{2} sinx) = \frac{{6 x}^{2} - 2}{{(x^{2} + 1)}^{3}}

(e) \Rightarrow (x^{2} - 2) L (y) = 1 + \frac{{6 x}^{2} - 2}{{(x^{2} + 1)}^{3}} \Rightarrow L (y) = \frac{1}{x^{2} - 2} + \frac{{6 x}^{2} - 2}{(x^{2} - 2) {(x^{2} + 1)}^{3}} \Rightarrow

y (x) = L^{- 1} (\frac{1}{x^{2} - 2}) + L^{- 1} (\frac{{6 x}^{2} - 2}{(x^{2} - 2) {(x^{2} + 1)}^{3}})

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

let decompose F (x) = \frac{{6 x}^{2} - 2}{(x^{2} - 2) {(x^{2} + 1)}^{3}}

F (x) = \frac{a}{x - \sqrt{2}} + \frac{b}{x + \sqrt{2}} + \frac{a_{1} x + b_{1}}{x^{2} + 1} + \frac{a_{2} x + b_{2}}{{(x^{2} + 1)}^{2}} + \frac{a_{3} x + b_{3}}{{(x^{2} + 1)}^{3}}

\dots be continued \dots

Commented by mathmax by abdo last updated on 05/Jun/20

error at line 2 \to x^{2} L (y) - xy (o) - y^{^{'}} (0) - 2 L (y) = L (x^{2} sinx)

but this dont change the result because y (0) = 0!