Ordinary Differential Equations 1

We group the terms to get:

$(x-2)^2{y}''+4(x-2)y'+2y=0$

We then make the substitution $z=x-2$ to get: $z^2{y}''+4zy'+2y=0$ .

This can further be split as: $(z^2y''+2zy')+2(zy'+y)=0$ ie. $\frac{\mathrm{d} }{\mathrm{d} z}(z^2y')+2\frac{\mathrm{d} }{\mathrm{d} z}(zy)=0$

Integrating with respect to z, we get: $z^2y'+2zy=C$

Again, this can be written as, $\frac{\mathrm{d} }{\mathrm{d} z}(z^2y)=C$

Integrating again with respect to z, we get: $z^2y=Cz+D$ We re-substitute $z$ in terms of $x$ to get:

$(x-2)^2y=C(x-2)+D$ ...(i)

And by differentiating (i) w.r.t to $x$ we also have:

$(x-2)^2y'+2(x-2)y=C$ ...(ii)

To determine the constants C and D, make the substitution $x=0$ in (i) and (ii) and use the given data.

The required expression comes out to be: $C=-1$ and $D=0$ which then gives: $(x-2)^2y+(x-2)=0$

If $x\neq2$ , we can write: $y= \frac{-1}{x-2}= \frac{1}{2}+\frac{x}{4}+\frac{x^2}{8}...\frac{x^9}{1024}...$

The sum of the first ten coefficients is then: $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...\frac{1}{1024}$ = $\frac{1023}{1024}$

Thus, $a+b=2047$