Answer to Question #209334 in Differential Equations for Mumu

Question

Use Laplace transform to solve initial value problem.

Y^2+Y={t,0 less than or equal to t less than 1

{0,t greater or equal to 1

y(0)=0,y'(0)=0

Answer 1

Given Y'' + Y =  $\begin{cases} t, \text{ if } 0 \leq t < 1 \\ 0, \text{ if } t \geq 1 \end{cases}$

let f(t) = $\begin{cases} t, \text{ if } 0 \leq t < 1 \\ 0, \text{ if } t \geq 1 \end{cases}$

we can express f(t) using unit step function U(t - a) = $\begin{cases} 0, \text{ if } t < a \\ 1, \text{ if } t \geq a \end{cases}$

hence we can express f(t) as f(t) = $t$ $-$ $t$U(t-1)

now given differential equation becomes, Y'' + Y = $t$ $-$ $t$U(t-1)

applying Laplace transform both side, ( $\mathcal{L}$(f(t)) = F(s) )

$\mathcal{L}$(Y'' + Y) = $\mathcal{L}$$(t$ $-$ $t$U(t-1))

$\mathcal{L}$(Y'') + $\mathcal{L}$ (Y) = $\mathcal{L}$( $t$ ) $-$ $\mathcal{L}$($t$U(t-1))

s²Y(s) - sY(0) - Y'(0) + Y(s) = $\frac{1}{s^2}$ - e^-s$\mathcal{L}$(t+1)

s²Y(s) - sY(0) - Y'(0) + Y(s) = $\frac{1}{s^2}$ - e^-s$\frac{s+1}{s^2}$

since Y(0) = 0 and Y'(0) = 0 (as given)

s²Y(s) + Y(s) = $\frac{1}{s^2}$ - e^-s$\frac{s+1}{s^2}$

Y(s) = $\frac{1}{s^2 (s^2+1)}$ - e^-s$\frac{s+1}{s^2(s^2+1)}$

now applying the inverse Laplace transform we get,

$\mathcal{L}^{-1}$ Y(s) = $\mathcal{L}^{-1}$ ( $\frac{1}{s^2 (s^2+1)}$ - e^-s$\frac{s+1}{s^2(s^2+1)}$ )

Y(t) = $\mathcal{L}^{-1}$ ( $\frac{1}{s^2 (s^2+1)}$) -$\mathcal{L}^{-1}$ (e^-s$\frac{s+1}{s^2(s^2+1)}$ )

Y(t) = $\mathcal{L}^{-1}$ ( $\frac{1}{s^2 } - \frac{1}{(s^2+1)}$) -$\mathcal{L}^{-1}$ (e^-s$\frac{s+1}{s^2(s^2+1)}$ )

Y(t) = $\mathcal{L}^{-1}$ ( $\frac{1}{s^2 } - \frac{1}{(s^2+1)}$) -$\mathcal{L}^{-1}$ (e^-s$\frac{s+1}{s^2(s^2+1)}$ )

Y(t) = $\mathcal{L}^{-1}$ $(\frac{1}{s^2 }) - \mathcal{L}^{-1}(\frac{1}{(s^2+1)})$ $-\mathcal{L}^{-1}$ (e^-s$\frac{1}{s^2 (s^2+1)}$+ e^-s $\frac{s}{s^2 (s^2+1)}$)

Y(t) = $\mathcal{L}^{-1}$ $(\frac{1}{s^2 }) - \mathcal{L}^{-1}(\frac{1}{(s^2+1)})$$+\mathcal{L}^{-1}$ e^-s$(\frac{1}{s^2 })$ $-$ $\mathcal{L}^{-1}$ e^-s $(\frac{1}{(s^2+1)})$ +$\mathcal{L}^{-1}$ e^-s$(\frac{1}{s })$ $-$ $\mathcal{L}^{-1}$ e^-s $(\frac{s}{(s^2+1)})$

Y(t) = t - sin(t) + tU(t-1) -sin(t-1)U(t-1) - cos(t-1)U(t-1) +U(t-1)

Y(t) = t - sin(t) + (t -sin(t-1) - cos(t-1) +1)U(t-1) where U(t-1) is the unit step function.