Initial Value problem

**Altair** · April 11th 2008, 05:57 AM

The question asks to "solve" the following differential equation.

$D^2(D-3)y = 6; Y(1) = 1, y'(1) = -1, y''(1)=-2$

We have not been told yet how to use the Method Of Undetermined Coefficients or by variable coefficients.

We have, however, solved similar questions with partial fractions.

I therefore tried the question in the same way.

$D^2(D-3)y = 6e^{0x}$

$y = 6\frac{e^{0x}}{(D^2)(D-3)}$

$\frac{1}{(D^2)(D-3)} = \frac{A}{D} + \frac{B}{D^2} + \frac{C}{D-3}$

After solving I get,

$A = -\frac{2}{3}$

$B = -\frac{1}{3}$

$c = -\frac{1}{9}$

Then after solving (as D is the differential operator) I get the following.

$(\frac{-2x}{3} -\frac{-x^2}{3} - \frac{1}{27}) * 6$

Where as the answer is simply $-x^2 + x + 1$

**Altair** · April 11th 2008, 07:17 AM

Can't do it by Binomial expansion as well.

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