Answer to Question #135969 in Differential Equations for Ritika Lad

Question

Find the Laplace transforms of

1.(t^2 e^2t)

2.[1+e^(-t)]^3

3. [t√(1+sin t)]

Answer 1

$$F(s)=\displaystyle\int_{0}^\infin e^{-ts}f(t)dt$$

1. $f(t)=t^2e^{2t}$

$$F(s)=\displaystyle\int_{0}^\infin e^{-ts}t^2e^{2t}dt$$ $$\int t^2 e^{(2-s)t}dt=\dfrac{1}{2-s}t^2e^{(2-s)t}-\dfrac{2}{2-s}\int te^{(2-s)t}dt=$$

$$=\dfrac{1}{2-s}t^2e^{(2-s)t}-\dfrac{2}{(2-s)^2}te^{(2-s)t}+$$

$$+\dfrac{2}{(2-s)^2}\int e^{(2-s)t}dt=\dfrac{t^2}{2-s}e^{(2-s)t}-$$

$$-\dfrac{2t}{(2-s)^2}e^{(2-s)t}+\dfrac{2}{(2-s)^3}e^{(2-s)t}+C$$

$F(s)=\displaystyle\int_{0}^\infin e^{-ts}t^2e^{2t}dt=$

$=\lim\limits_{b\to\infin}\bigg[-\dfrac{t^2e^{(s-2)t}}{s-2}-\dfrac{2te^{(s-2)t}}{(s-2)^2}-\dfrac{2e^{(s-2)t}}{(s-2)^3}\bigg]\begin{matrix} b \\ 0 \end{matrix}=$

$=-0-0-0+0+0+\dfrac{2}{(s-2)^3}=\dfrac{2}{(s-2)^3}, s\not=2$

$L(t^2e^{2t})=\dfrac{2}{(s-2)^3}, s\not=2$

2. $f(t)=(1+e^{-t})^3=1+3e^{-t}+3e^{-2t}+e^{-3t}$

$$F(s)=\displaystyle\int_{0}^\infin e^{-ts}(1+3e^{-t}+3e^{-2t}+e^{-3t})dt=$$

$$=\displaystyle\int_{0}^\infin (e^{-ts}+3e^{-(s+1)t}+3e^{-(s+2)t}+e^{-(s+3)t})dt=$$

$$=\lim\limits_{b\to\infin}\bigg[-\dfrac{e^{-t}}{s}-\dfrac{3e^{-(s+1)t}}{s+1}-\dfrac{3e^{(s+2)t}}{s+2}-\dfrac{e^{(s+3)t}}{s+3}\bigg]\begin{matrix} b \\ 0 \end{matrix}=$$

$$=-0-0-0-0+\dfrac{1}{s}+\dfrac{3}{s+1}+\dfrac{3}{s+2}+\dfrac{1}{s+3}$$

$$L((1+e^{-t})^3)=\dfrac{1}{s}+\dfrac{3}{s+1}+\dfrac{3}{s+2}+\dfrac{1}{s+3},$$

$s\not=-3, s\not=-2, s\not=-1, s\not=0$

3.

$f(t)=t\sqrt{1+\sin{t}}=$

$=t\sqrt{\sin^2{(t/2)}+2\sin{(t/2)}\cos{(t/2)}+\cos^2{(t/2)}}=$

$=t|\sin{(t/2)}+\cos{(t/2)}|$

Let $\sin{(t/2)}+\cos{(t/2)}\geq0.$ Then

$f(t)=t(\sin{(t/2)}+\cos{(t/2)})$

$L(\sin{(t/2)}+\cos{(t/2)})=L(\sin{(t/2)})+L(\cos{(t/2)})=$

$=\dfrac{\dfrac{1}{2}}{s^2+\dfrac{1}{4}}+\dfrac{s}{s^2+\dfrac{1}{4}}$

By multiplication by $t$ property

$L(t(\sin{(t/2)}+\cos{(t/2)}))=-1\dfrac{d}{ds}\bigg(\dfrac{\dfrac{1}{2}+s}{s^2+\dfrac{1}{4}}\bigg)=$

$=-\dfrac{s^2+\dfrac{1}{4}-s-2s^2}{(s^2+\dfrac{1}{4})^2}=$

$=\dfrac{s^2+s-\dfrac{1}{4}}{(s^2+\dfrac{1}{4})^2}=\dfrac{4(4s^2+4s-1)}{(4s^2+1)^2}$

$L(t\sqrt{1+\sin{t}})=\dfrac{4(4s^2+4s-1)}{(4s^2+1)^2}$