Why operator method for differential equation gives wrong result?

Question

I'm confused with the result of operator method for equation: $$y''-a^2y=b,$$ where b is a constant. Right answer is: $$ y(t) = y(0)\cosh(at)+y'(0)\sinh(at)-\dfrac b {a^2}.$$ But when I apply Laplace transform I've got: $$p^2Y-py(0)-y'(0)-a^2Y=\dfrac bp ... \\ y(t)=(y(0)+b/a^2)\cosh(at)+(y'(0)/a)\sinh(at)-\frac b {a^2}.$$ So here a strange member $\dfrac b {a^2}\cosh(at)$ appears. Where did I make a mistake?

Answer 1

$$p^2Y-py(0)-y'(0)-a^2Y=b/p $$ $$Y(p^2-a^2)=py(0)+y'(0)+ \frac b p $$ $$Y=\dfrac p {(p^2-a^2)}y(0)+\dfrac {y'(0)}{(p^2-a^2)}+ \frac b {{(p^2-a^2)}p} $$ You have to decompose the last fraction: $$\frac b {{(p^2-a^2)}p}=\frac {bp}{a^2} \left (\frac 1{p^2-a^2}-\frac 1 {p^2} \right )$$ $$\boxed {y(t)=(y(0)+\frac {b}{a^2} )\cosh(at)+\dfrac {y'(0)}a\sinh(at)-\frac {b}{a^2}}$$

You can also write the solution as: $$y(t)=c_1\cosh(at)+c_2\sinh(at)-\frac {b}{a^2}$$ Or also; $$y(t)=c_1e^{at}+c_2e^{-at}-\frac {b}{a^2}$$ All these formulations are coorect.