# Laplace transforms... How do I go about getting y? y''-2y'-3y=0, y(4)=-3, y'(4)=-17

Sep 9, 2017

I assume $y \left(0\right) = A$ and $y ' \left(0\right) = B$

After taking Laplace transform both sides,

${s}^{2} \cdot Y \left(s\right) - s \cdot y \left(0\right) - y ' \left(0\right) - 2 \cdot \left[s \cdot Y \left(s\right) - y \left(0\right)\right] - 3 Y \left(s\right) = 0$

${s}^{2} \cdot Y \left(s\right) - A s - B - 2 s \cdot Y \left(s\right) - 2 A - 3 Y \left(s\right) = 0$

$\left({s}^{2} - 2 s - 3\right) \cdot Y \left(s\right) = A s + 2 A + B$

$\left(s + 1\right) \cdot \left(s - 3\right) \cdot Y \left(s\right) = A s + 2 A + B$

$Y \left(s\right) = \frac{A s + 2 A + B}{\left(s + 1\right) \cdot \left(s - 3\right)}$

$Y \left(s\right) = \frac{5 A + B}{4} \cdot \frac{1}{s + 1} - \frac{A + B}{4} \cdot \frac{1}{s - 3}$

After taking inverse Laplace transformation both sides,

$y = \frac{5 A + B}{4} \cdot {e}^{- x} - \frac{A + B}{4} \cdot {e}^{3 x}$

Now, I use $y \left(4\right) = - 3$ condition,

$\frac{5 A + B}{4} \cdot {e}^{- 4} - \frac{A + B}{4} \cdot {e}^{12} = - 3$

I differentiate both sides and apply $y ' \left(4\right) = - 17$ condition,

$y ' = - \frac{5 A + B}{4} \cdot {e}^{- x} - \frac{3 A + 3 B}{4} \cdot {e}^{3 x}$

$- \frac{5 A + B}{4} \cdot {e}^{- 4} - \frac{3 A + 3 B}{4} \cdot {e}^{12} = - 17$ or,

$\frac{5 A + B}{4} \cdot {e}^{- 4} + \frac{3 A + 3 B}{4} \cdot {e}^{12} = 17$

From these equations, $A = 2 {e}^{4} - 5 {e}^{- 12}$ and $B = 25 {e}^{- 12} - 2 {e}^{4}$

Thus, $y = 2 {e}^{4} \cdot {e}^{- x} - 5 {e}^{- 12} \cdot {e}^{3 x}$ or,

$y = 2 {e}^{4 - x} - 5 {e}^{3 x - 12}$

#### Explanation:

1) I used Laplace transformation for $y \left(0\right) = A$ and $y ' \left(0\right) = B$ conditions.

2) I solved $Y \left(s\right)$

3) I applied inverse Laplace transformation for finding $y$

4) I used $y \left(4\right) = - 3$ and $y ' \left(4\right) = - 17$ conditions for finding A and B

Sep 9, 2017

$y \left(t\right) = 2 {e}^{4 - t} - 5 {e}^{3 t - 12}$

#### Explanation:

We have:

$y ' ' - 2 y ' - 3 y = 0$ with $y \left(4\right) = - 3 , y ' \left(4\right) = - 17$ ..... [A]

We note that this is a shifted value IVP so we cannot just take Laplace Transforms "as is", and we must perform a substitution to shift the initial conditions to that of an IVP.

We have initial conditions pivoting on $t = 4$, so we need a $- 4$ IVP shift:

Let $\tau = t - 4 \implies t = \tau + 4$
And $u \left(\tau\right) = y \left(\tau + 4\right) \implies y \left(t\right) = u \left(t - 4\right)$

Then substituting into the DE [A] we get:

$u ' ' - 2 u ' - 3 u = 0$ with $u \left(0\right) = - 3 , u ' \left(0\right) = - 17$ ..... [B]

We will need the following standard Laplace transform and inverses:

 {: (ul(f(t)=ℒ^(-1){F(s)}), ul(F(s)=ℒ{f(t)}), ul("Notes")), (f(t), F(s),), (f'(t), sF(s)-f(0),), (f''(t), s^2F(s)-s f(0)-s f'(0),), (e^(at), 1/(s-a),) :}

We can now use this table to apply Laplace Transforms to [B]:

 ℒ \ {u''} - ℒ \ {2u'} -ℒ \ { 3u} = 0

$\therefore {s}^{2} U - s u ' \left(0\right) - u \left(0\right) - 2 \left\{s U - s \left(0\right)\right\} - 3 U = 0$ where U(s)= ℒ \ {u(t)}

$\therefore \left({s}^{2} U + 3 s + 17\right) - 2 \left(s U + 3\right) - 3 U = 0$

$\therefore {s}^{2} U + 3 s + 17 - 2 s U - 6 - 3 U = 0$

$\therefore \left({s}^{2} - 2 s - 3\right) U + 3 s + 11 = 0$

$\therefore \left(s - 3\right) \left(s + 1\right) U = - 11 - 3 s$
$\therefore U = \frac{- 11 - 3 s}{\left(s - 3\right) \left(s + 1\right)}$

Now we decompose into partial fractions:

$\frac{- 11 - 3 s}{\left(s - 3\right) \left(s + 1\right)} \equiv \frac{a}{s - 3} + \frac{b}{s + 1}$
$\text{ } \equiv \frac{a \left(s + 1\right) + b \left(s - 3\right)}{\left(s - 3\right) \left(s + 1\right)}$
$\therefore - 11 - 3 s = a \left(s + 1\right) + b \left(s - 3\right)$

Put:

$s = 3 \implies - 20 = 4 a \implies a = - 5$
$s = - 1 \implies - 8 = - 4 b \implies b = 2$

Thus:

$U \left(s\right) = - \frac{5}{s - 3} + \frac{2}{s + 1}$

And again using the above table we can now take inverse Laplace Transforms:

$u \left(\tau\right) = - 5 {e}^{3 \tau} + 2 {e}^{- \tau}$
$\setminus \setminus \setminus \setminus \setminus \setminus = 2 {e}^{- \tau} - 5 {e}^{3 \tau}$

And so, the solution we seek, after restoring the substitution is:

$y \left(t\right) = u \left(t - 4\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = 2 {e}^{- \left(t - 4\right)} - 5 {e}^{3 \left(t - 4\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus = 2 {e}^{4 - t} - 5 {e}^{3 t - 12}$

Side Note

Although this demonstrates how to deal with a shifted data IVPs, it is in fact much easier to solve this DE using traditional methods rather than Laplace Transforms, viz:

The associated Auxiliary equation is:

${m}^{2} - 2 m - 3 = 0 \implies \left(m - 3\right) \left(m + 1\right) = 0 \implies m = - 1 , 3$

Yielding a general solution of the homogeneous equation of the form:

$y = A {e}^{- t} + B {e}^{3 t} \implies y ' = - A {e}^{- t} + 3 B {e}^{3 t}$

And using the initial condition $y \left(4\right) = - 3 , y ' \left(4\right) = - 17$ we get:

$- 3 = A {e}^{- 4} + B {e}^{12}$ ..... [C]
$- 17 = - A {e}^{- 4} + 3 B {e}^{12}$ ..... [D]

Solving Simultaneously, we get:

Eg [C]+ Eq [D]:

$- 20 = 4 B {e}^{12} \implies B = - 5 {e}^{- 12}$

Subs $B$ into [C]:

$- 3 = A {e}^{- 4} - 5 {e}^{12} {e}^{- 12}$
$\therefore A {e}^{- 4} = 5 - 3$
$\therefore A {e}^{- 4} = 2$
$\therefore A = 2 {e}^{4}$

So the Particular Solution is:

$y = 2 {e}^{4} {e}^{- t} - 5 {e}^{- 12 t} {e}^{3 t}$
$\setminus \setminus \setminus = 2 {e}^{4 - t} - 5 {e}^{3 t - 12}$, as above