Differential equation solved by Laplace transform

Let begin with an example.
We want to solve initial value problem for the differential equation

$$y"(x)+6y'(x)+25y(x)=f(x)u(x)$$

with y(+0)=a and y'(+0)=b
Let

$$Y(s)=\mathcal{L}(y(x))(s)=\int_{0}^{+\infty}y(x)e^{-sx}dx$$

We have

$$F(s)=\mathcal{L}(f(x)u(x))(s)$$
$$\mathcal{L}(y'(x))(s)=sY(s)-a;\:\mathcal{L}(y"(x))(s)=s^2Y(s)-as-b$$

We have then

$$s^2Y(s)-as-b+6(sY(s)-a)+25Y(s)=F(s)$$

or

$$(s^2+6s+25)Y(s)=as+6a+b+F(s);\:Y(s)=\frac{as+6a+b}{s^2+6s+25}+\frac{1}{s^2+6s+25}F(s)$$

Now we can write

$$Y(s)=\frac{a(s+3)}{(s+3)^2+16}+\frac{3a+b}{(s+3)^2+16}+\frac{1}{(s+3)^2+16}F(s)$$

so

$$y(x)=ae^{-3x}\cos4x\: u(x)+(3a+b)\frac{1}{4}e^{-3x}\sin4x\: u(x)+\frac{1}{4}\int_{0}^{x}e^{-3t}\sin4tf(x-t)\:u(x-t)dt$$
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