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

i) Check if $y_1(x)=e^x$ and $y_2(x) = x$ is a solution to the differential equation (homogeneous).

ii) Use variation of parameters to find a general solution to the inhomogeneous equation. :)

Any tips/solution on this one? :D

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  • $\begingroup$ yeah.. thats the one :P $\endgroup$ – wellalright Mar 31 '14 at 23:56
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  • Test $y_1$: $y_1 = e^x, y'_1 = e^x, y''_1 = e^x \rightarrow (1-x) e^x + x e^x - e^x = 0$
  • Test $y_2$: $y_2 = x, y'_2 = 1, y''_2 = 0 \rightarrow (1-x)0 + x \times 1 - x = 0$

Thus both $y_1(x)$ and $y_2(x)$ are solutions to the homogeneous equation.

We are asked to solve this using Variation of Parameters (VoP), given:

$$\tag 1 y'' + \dfrac{x}{1-x} y' -\dfrac{1}{1-x} y = 1$$

Note the restrictions on $x$.

Step 1

Find the homogeneous solution to $(1)$, so we have:

$$\tag 2 y'' + \dfrac{x}{1-x} y' -\dfrac{1}{1-x} y = 0$$

This yields (the problem provided this and we verified it above):

$$y_h = c_1 e^x + c_2 x$$

Step 2

We now make use of VoP, so we set: $y_1 = e^x$ and $y_2 = x$ from $y_h$ and $f = 1$ from $(1)$.

We calculate the Wronskian of $y_1$ and $y_2$, yielding $W(e^x, x) = -e^x (x-1)$.

Using VoP, we have:

$\displaystyle u_1 = \int \dfrac{-y_2 f}{W(e^x, x)} dx = \int \dfrac{-x(1)}{-e^x (x-1)} dx = \dfrac{\text{Ei}(1-x)}{e}-e^{-x}$, $~E_i$ is the Elliptic Integral

$\displaystyle u_2 = \int \dfrac{y_1 f}{W(e^x, x)} dx = \int \dfrac{e^x(1)}{-e^x (x-1)} dx = -\ln (1-x)$

Now, $y_p$ is given by:

$$y_p = y_1 u_1 + y_2 u_2 = e^{x-1} \text{Ei}(1-x)-x \ln (x-1)-1$$

Step 3

Our final solution is given by:

$$y(x) = y_h(x) + y_p(x) = c_1e^x + c_2 x + e^{x-1} \text{Ei}(1-x)-x \ln (x-1)-1$$

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  • $\begingroup$ Elliptic Integral? mmh I've never heard that before :O $\endgroup$ – wellalright Apr 1 '14 at 1:04
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Firstly sub in $ y_1 $ and $y_2$ into the homogeneous version of (i) and hopefully you'll get zero. Then a general solution for the homogeneous equation would be $A y_1+By_2$

So if we let $ y = e ^ x $, then: $(1-x)y''+xy''-y=(1-x)e ^ x+x e ^ x-e ^ x=0$

Now try the same for $y=x $

edit (not so sure on the in homogenous part) And the general in homogenous solution would be $y=Ay_1+By_2+Cx+Dx ^ 2 $

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  • $\begingroup$ so the homogenous version is that (1-x)y'' + xy'-y=0 ?? $\endgroup$ – wellalright Mar 31 '14 at 23:34
  • $\begingroup$ Indeed it is, you just let the right hand sideway zero $\endgroup$ – Ellya Mar 31 '14 at 23:35
  • $\begingroup$ hmm Im not sure how to sub that in :(( do I just find the derivative of y_1(x) and then sub it in for y or something? $\endgroup$ – wellalright Mar 31 '14 at 23:40
  • $\begingroup$ See my edit above $\endgroup$ – Ellya Mar 31 '14 at 23:45
  • $\begingroup$ Sorry! I should mention that the Derivative of $e ^ x $ is $e ^ x $ (it does not change) $\endgroup$ – Ellya Mar 31 '14 at 23:58

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