I am trying to solve the integral-differential equation: $$x'(t) + \int_0 ^{t} (t-s)x(s) ds = t + \frac{1}{2}t^2 + \frac{1}{24}t^4$$ With $x(0) = 1$ Taking the Laplace transform of this and using the convolution theorem I obtain: $$\bar{x}(p) (p^3+1) = p^2 + p + \frac{1}{p} + \frac{1}{p^2}$$ And now I am really struggling to invert this to find $x(t)$, so any help is very much appreciated, Thanks
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1 Answer
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I only know how to solve it directly. You decide if it is useful to you or not.
Differentiating (1) wrt $t$ twice and we obtain:
$$x^{(3)}(t)+x(t)-1-t^2/2=0$$
The solution (via Mathematica) is given by:
$$x(t)=1+t^2/2+c_1e^{-t}+c_2 e^t \sin\left(\frac{\sqrt{3}}{2}t\right)+c_3 e^t \cos\left(\frac{\sqrt{3}}{2}t\right)$$
$$1=x(0)=1+c_1+c_3$$ Therefore $c_3=-c_1$.
