Can somebody help me with the beginning of Laplace transform?

Question

$ty''(t)+(t-1)y'(t)-y(t)=0$

$y(0)=5$

$y(+\infty)=0$

Can somebody tell me what $y(+\infty)=0$ represents?

Answer 1

$$ty''(t)+(t-1)y'(t)-y(t)=0$$ $$t(y''(t)+y'(t))-y'(t)-y(t)=0$$ Apply Laplace Transform now: $$-\dfrac {d}{ds}(s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0))-sY(s)+y(0)-Y(s)=0$$ $$\dfrac {d}{ds}(s^2Y(s)-5s+sY(s))+Y(s)(s+1)=5$$ $$s(s+1)Y'(s)+(3s+2)Y(s)=10$$

First solve the differential equation. To do so multiply by $s$ both sides: $$(s^3+s^2)Y'(s)+(3s^2+2s)Y(s)=10s$$ $$((s^3+s^2)Y(s))'=10s$$ Integrate: $$(s^3+s^2)Y(s)=5s^2+C_1$$ $$\implies Y(s)=\dfrac {5s^2+C_1}{s^3+s^2}$$

Apply inverse Laplace transform.

You may have to use $y(\infty)=\lim\limits_{t \to \infty }y(t)=0$ to determine the constants.