Question

3. Solve the IVP using the Laplace transform y' - 2y =4, given that y(0) = 1 *​

Answer 1

Given: y' - 2y = 4,

           y(0) = 1 (this is initial value)

To find: y(t)

solution:

IVP means initial value Problem

step 1: take Laplace transform for given equation:

           (sY(s) - y(0)) -2Y(s) = 4................................equation -1

     as,  y(0) = 1 (given)

so, put y(0) = 1 in equation -1, we get

           (sY(s) - 1) -2Y(s) = 4

           sY(s) -1- 2Y(s) = 4

            Y(s) (s - 2) - 1 = 4

            Y(s) (s - 2) = 4+1

            Y(s) (s - 2) = 5

            Y(s) = $\frac{5}{(s-2)}$....................................................................equation -2

step 2: take Laplace inverse of equation -2, we get;

             y(t) = 5 $e^{2t}$

so the final answer is y(t) = 5 $e^{2t}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

WORKING PROCEDURE

The procedure stated below is applied to solve a linear differential equation with constant coefficients by transform method :

1. First the Laplace Transform of both sides of the given linear differential equation

2. Use the given initial conditions

3. Express $\overline{y}$ in terms of s

4. Express this function of s into partial fractions

5. Take the inverse transform in both sides

6. The obtained value of y as a function of t is the desired Solution satisfying the given initial conditions

FORMULA TO BE IMPLEMENTED

1.

The LAPLACE TRANSFORMS of f(t), denoted by L{f(t)} and defined as :

$L\{f(t)\} =\displaystyle \int\limits_{0}^{\infty} e^{-st} f(t)\, dt$

2.If f '(t) & its first (n-1) derivatives be continuous then

$L \{ \: {f}^{n}(t) \} = {s}^{n} \: \overline{f} \: (s) - {s}^{n - 1} f(0) - ....... - {f}^{n - 1} (0)$

GIVEN

$y' - 2y =4 \: \: \: given \: that \: \: y(0) = 1$

TO DETERMINE

To find the value of y

CALCULATION

It is given that

$y' - 2y =4$

Taking the Laplace Transform of both sides we get

$\displaystyle \: [s \: \overline{y} - y(0)] - \: 2 \: \overline{y} = \frac{4}{s}$

$\implies \: \displaystyle \: s \: \overline{y} - 1- \: 2 \: \overline{y} = \frac{4}{s}$

$\implies \: \displaystyle \: s \: \overline{y} - \: 2 \: \overline{y} = \frac{4}{s} + 1$

$\implies \: \displaystyle \:( s \: - 2 \: ) \overline{y} = \frac{s + 4}{s}$

$\implies \: \displaystyle \overline{y} = \frac{s + 4}{s(s - 2)} \: \: .........(1)$

Now we are proceeding to the fraction

$\displaystyle \frac{s + 4}{s(s - 2)} \:$

into partial fractions

Let

$\displaystyle \frac{s + 4}{s(s - 2)} \: \: = \frac{a}{s} + \frac{b}{s - 2}$

$\implies \: \displaystyle \frac{s + 4}{s(s - 2)} \: \: = \frac{a(s - 2) + bs}{s}$

$\implies \: \displaystyle \frac{s + 4}{s(s - 2)} \: \: = \frac{(a + b)s - 2a}{s}$

Comparing both sides

$a + b = 1 \: \: and \: \: - 2a = 4$

Now

$- 2a = 4 \: \: gives \: \: a \: = - 2$

From

$a + b = 1 \: \: we \: get$

$</em></strong><strong><em>b \: = 1 - a = 1+ 2 = 3$

$</em></strong><strong><em> </em></strong><strong><em>\</em></strong><strong><em>implies</em></strong><strong><em> </em></strong><strong><em>b \: = 1+ 2 = 3$

So

$\displaystyle \frac{s + 4}{s(s - 2)} \: \: = \frac{ - 2}{s} + \frac{3}{s - 2}$

Hence from Equation (1) we get

$\displaystyle \overline{y} = \frac{ - 2}{s} + \frac{3}{s - 2}$

On Inversion

$\displaystyle y = {L}^{ - 1} \: ( \frac{ - 2}{s} \: ) + {L}^{ - 1} ( \frac{3}{s - 2} \: )$

$\implies \: \displaystyle y = - 2 \: {L}^{ - 1} \: ( \frac{ 1}{s} \: ) + 3 \: {L}^{ - 1} ( \frac{1}{s - 2} \: )$

$\implies \: \displaystyle y = -2 \times 1 + 3 \times {e}^{2t}$

$\therefore \: \: \displaystyle y = -2 + 3 {e}^{2t}$

Which is the desired result

Hence the required solution is

$\therefore \: \: \displaystyle y = -2 + 3 {e}^{2t}$