One last question

Well, thanks for another one! I just saw this problem and it is little late, I know.

$y'' - y' + y = x$ , so this is our DE.

Now, look at this equation - $y'' - y' + y = 0$ . It will be easy to find solution of this DE by just substituting $y = {e}^{\lambda x}$ and forming auxiliary equation as -

${\lambda}^{2} - \lambda + 1 = 0$ , and so $\lambda = \frac{1 + \sqrt{3}\iota}{2}, \frac{1 - \sqrt{3}\iota}{2}$

So our solution would be - $y = {C}_{1}{e}^{x/2}cos(\frac{\sqrt{3}x}{2}) + {C}_{2}{e}^{x/2}sin(\frac{\sqrt{3}x}{2})$ , which can easily be seen as we have 2 solutions of the quadratic and so in DE, we will just add them(and forget not to "de-substitute") and yeah trigonometric ratios are just a result of ${e}^{\iota x} = cosx + isinx$

So, we have found the solution to $y'' - y' + y = 0$ and we have to find solution to $y'' - y' + y = x$ . Well, just "add" $x$ , right?

Yes. Adding $x$ in $y$ will make it for the ${y}^{(0)}$ term but will add an extra $1$ on differentiating once. To remove it, we will add another term i.e additive inverse of that $1$ in $y$ whose differentiation would be just $0$ .

Therefore, our final answer finally becomes -

$y = {C}_{1}{e}^{x/2}cos(\frac{\sqrt{3}x}{2}) + {C}_{2}{e}^{x/2}sin(\frac{\sqrt{3}x}{2}) + x + 1$

Rest is just trivial!

We'd like to solve this ODE via a Laplace Transform, right? Only after performing a simple substitution first! Let $g(t) = f(x)$ (where $t = x-1$ ) so that $g(0) = f(1) = 1, g'(0) = f'(1) = 0$ are our initial conditions. We now obtain:

$g''(t) - g'(t) + g(t) = t+1$

whose Laplace Transform computes to:

$[s^{2}G(s) - s \cdot g(0) - g'(0)] - [sG(s) - g(0)] + G(s) = \frac{1}{s^{2}} + \frac{1}{s}$ ;

or $(s^2 - s + 1) \cdot G(s) = \frac{1+s}{s^2} + s - 1;$

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

whose inverse Laplace Transform computes to:

$g(t) = L^{-1} [G(s)] = \boxed{t + 2 - [\frac{1}{\sqrt{3}} sin(\frac{\sqrt{3}t}{2}) + cos(\frac{\sqrt{3}t}{2})] \cdot e^{t/2} }.$

Finally, we calculate $f(1.21) = g(0.21) \approx \boxed{1.001624}.$

VIOLA!!!!!!!

The solution I'm gonna post here is not a rigorous maths, so, for those who don't like it. I do apologise.

Since the question has already given the value at x=1 and its derivative a t the same point, so, why not just try a series solution.(I'm not gonna prove that this series is a solution nor whether the value we want to find is in the radius of convergent or not.)

By differentiating repeatedly, one should see a pattern of the derivatives at x=1 that

$\displaystyle y=1,{ y }^{ ' }=0,{ y }^{ " }=0,{ y }^{ (3) }=1,{ y }^{ (4) }=1,{ y }^{ (5) }=0,{ y }^{ (6) }=-1,{ y }^{ (7) }=-1,{ y }^{ (8) }=0\quad and\quad { y }^{ (9) }=1$ and so on.

so $\displaystyle f(1.21)=1+\frac { { (0.21) }^{ 3 } }{ 3! } +\frac { { (0.21) }^{ 4 } }{ 4! } -\frac { { (0.21) }^{ 6 } }{ 6! } +...$

$\displaystyle f(1.21)=\boxed{1.001624}$