Can Laplace transformation be used here?

Question

I would really appreciate it if someone can tell me what i got wrong below or any other method that could solve this easier.

The problem:

$(1-x)y''+xy'-y=0$

$y(0)=2$
$y'(0)=-1$

I tried solving it with laplace:

$$(\frac{1}{s}-\frac{1}{s^2})(Ys^2-2s+1)+\frac{1}{s^2}(Ys-2)-Y=0$$ simplifying: $$Ys-2+\frac{1}{s}-Y+\frac{2}{s}-\frac{1}{s^2}+\frac{Y}{s}-\frac{2}{s^2}-Y=0$$ $$Ys+\frac{3}{s}-2Y-\frac{3}{s^2}+\frac{Y}{s}=2 $$ $$Ys-2Y+\frac{Y}{s}=2-\frac{3}{s}+\frac{3}{s^2}$$ $$Y(s-2+\frac{1}{s})=2-\frac{3}{s}+\frac{3}{s^2}$$ $$Y\frac{(s-1)^2}{s}=2-\frac{3}{s}+\frac{3}{s^2}$$ $$Y = \frac{2s}{(s-1)^2}-\frac{3}{(s-1)^2}+\frac{3}{s(s-1)^2}$$ $$ Y=\frac{2s-3}{(s-1)^2}+\frac{3}{s(s-1)^2}$$

And I got for each rational fraction respectively:

$$\frac{2}{s-1}-\frac{1}{(s-1)^2}$$ $$\frac{3}{s}-\frac{3}{s-1}+\frac{3}{(s-1)^2}$$

and the final when combined:

$$\frac{3}{s}-\frac{1}{s-1}+\frac{2}{(s-1)^2}=3-e^x+2xe^x$$

Problem is that for $y(0)=2$ stands but for $y'(0)$ I get $1$ $$y'=2xe^x+e^x$$

I tried to guess the answer and it seems that : $$y=2e^x-3x$$ $$y'=2e^x-3$$ $$y''=2e^x$$

is the correct answer.

laplace table :http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx

Answer 1

As Lutzl pointed out you have no $F'(s)$. With Laplace transform you should get an ODE of first order.. You can solve it with reduction of order Since an obvious particular solution is $y=x$. Try $y=xv(x)$

Answer 2

I suspect your method is wrong, there should be derivatives of $Y(s)$ in the formula to account for the factors $x$.

Putting clues together like a detective you can sort the terms in the equation to reverse-engineer its construction. Separating the terms with factor $x$ and applying binomial theorems gives $$ 0=y''-y-x(y''-y')=(D+1)(D-1)y-xD(D-1)y=((1-x)D+1)(D-1)y. $$ An adapted first order system is $$ v=(D-1)y=y'-y\\ (1-x)v'+v=0 $$ where the second equation has solutions $v=C_1(x-1)$ and then $$ y'-y=C_1(x-1) $$ has solutions $$ y=C_2e^x-C_1x. $$

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Applying the Laplace transform should result in the equation \begin{align} 0&=[s^2Y(s)-y'(0)s-y(0)]-\frac{d}{ds}[s^2Y(s)-y'(0)s-y(0)]+\frac{d}{ds}[sY(s)-y(0)]-Y(s) \\ &= [s^2Y(s)+s-2]-[s^2Y'(s)+2sY(s)+1]+[sY'(s)+Y(s)]-Y(s) \\ &= (s-s^2)Y'(s)+(s^2-2s)Y(s)+(s-3) \\[1em] \frac{d}{ds}[e^{-s}(s-1)Y(s)]&=e^{-s}(1-3/s) \end{align}