Solve a second order differential equation $ax''(t) +bx'(t) + cx(t) = d$

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We Know that if

$$ax'' +bx' + cx = g(t)$$

We have to find the solution of homogeneous case $x_h(t)$ and a particular solution $x_p(t)$. So the general solution will be

$$x(t) = x_h(t) + x_p(t)$$

But, I saw that if $g(t)= d$ is constant, then the general solution is

$$x(t) = x_h(t) + d$$

I dont understand it. It is true?

If it is not true, there is some way to find a particular solution more easy in case $g(t) = d$ constant?

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May be, to make it clearer, considering $$ax'' +bx' + cx = d$$ let $$cx-d=y \implies x=\frac{y+d}c\implies x'=\frac{y'}c\implies x''=\frac{y''}c$$ Replace to get $$ay'' +by' + cy = 0$$