Question

using laplace transform solve y'+5y'+6y=2 given y(0)=0 y'(0)=0​

Answer 1

Step-by-step explanation:

*Given:y"+5y'+6y=2; y(0)=0,y'(0)=0

To find: Solve using Laplace transform.

Solution:

We know that using Laplace transform we can easily solve these types of equation.

Step 1: Take Laplace transform both sides

$L(y''+5y'+6y)=L(2)\\\\L(y'')+L(5y')+L(6y)=L(2)\\\\s^2Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)=2$

Step 2: Put initial conditions

$s^2Y(s)-s.(0)-0+5sY(s)-5.(0)+6Y(s)=2$

$s^2Y(s)+5sY(s)+6Y(s)=2$

Step 3: Take Y(s) common

$Y(s)(s^2+5s+6)=2$

or

$Y(s)= \frac{2}{(s^2+5s+6)}$

or

$Y(s)= \frac{2}{(s^2+3s + 2s+6)} \\\\Y(s)= \frac{2}{(s+3)(s + 2)}$

Step 4: Do partial fraction

to split the RHS,partial fraction should be done; so that inverse Laplace transform can be done

$\frac{2}{(s+3)(s + 2)} = \frac{A}{s + 3} + \frac{B}{s + 2} \\ \\ \frac{2}{(s+3)(s + 2)} = \frac{A(s + 2) + B(s + 3)}{(s + 3)(s + 2)} \\ \\ 2 = s(A + B) + 2A + 3B$

compare the terms

$A + B = 0...eq1 \\ 2A + 3B = 2...eq2$

solve these two eqs to find value of a and b,put value of B from eq1 in eq2

$2A - 3A = 2 \\ \\ A = - 2 \\ \\ B = 2$

Put the value of A and B

$Y(s)= \frac{ - 2}{(s+3)} + \frac{2}{s + 2}$

Step 5: Take inverse Laplace transform

${L}^{ - 1} Y(s)= {L}^{ - 1}\left(\frac{ - 2}{(s+3)} \right) + {L}^{ - 1}\left(\frac{2}{s + 2} \right)\\ \\ y(t) = - 2 {e}^{ - 3t} + 2 {e}^{ - 2t} \\ \\ y(t) = 2( {e}^{ - 2t} - {e}^{ - 3t} )$

Final answer:

$\bold{\red{y(t) = 2( {e}^{ - 2t} - {e}^{ - 3t} )}}$

or in terms of function of x

$\bold{\green{y(x) = 2( {e}^{ - 2x} - {e}^{ - 3x} )}}$

Hope it helps you.

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