Find out the answer to "Please solve the 2nd order differential equation by..."

Caelan

Answered question

2021-01-27

Please solve the 2nd order differential equation by (PLEASE FOLLOW GIVEN METHOD) LAPLACE TRANSFORMATION
ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
$L (y) = [\frac{w}{(s^{2} + a^{2}) (s^{2} + w^{2})}] \cdot b$
Problem 2 Solve the differential equation
$\frac{d^{2} y}{d t^{2}} + a^{2} y = b \sin (ω t)$ where $y (0) = 0$
and $y^{'} (0) = 0$

Answer & Explanation

Daphne Broadhurst

Skilled2021-01-28Added 109 answers

Step 1
Given a differential equation,

y " + a^{2} y = b \sin (ω t)

, where

y (0) = 0, y^{'} (0) = 0

Taking the Laplace transform of both sides of the given differential equation,

L (y ") + a^{2} L (y) = b L (\sin (ω t))

s^{2} L (y) - s y (0) - y^{'} (0) + a^{2} L (y) = \frac{b ω}{s^{2} + ω^{2}}

(s^{2} + a^{2}) L (y) = \frac{b ω}{s^{2} + ω^{2}}

∴, L (y) = \frac{b ω}{(s^{2} + ω^{2})} (s^{2} + ω^{2})

Step 2
Applying partial fraction,

\frac{b ω}{(s^{2} + ω^{2}) (s^{2} + a^{2})} = \frac{A}{s^{2} + ω^{2}} + \frac{B}{s^{2} + ω^{2}} \dots (1)

A (s^{2} + a^{2}) + B (s^{2} + ω^{2}) = b ω

A s^{2} + A a^{2} + B s^{2} + B ω^{2} = b ω

(A + B) s^{2} + A a^{2} + B ω^{2} = b ω

Comparing coefficients,

A + B = 0 \Rightarrow A = - B

A a^{2} + B ω^{2} = b ω

Substitute A in the above equation,

- B a^{2} + B ω^{2} = b ω

B = \frac{b ω}{ω^{2} - a^{2}}

Therefore,

A = (\frac{b ω}{a})^{2} - ω^{2}

Step 3
Substitute A and B in equation (1),

\frac{b ω}{(s^{2} + ω^{2}) (s^{2} + a^{2})} = \frac{b ω}{(a^{2} - ω^{2}) (s^{2} + ω^{2})} - \frac{b ω}{(a^{2} - w^{2}) (s^{2} + a^{2})}

= \frac{b ω}{(a^{2} - w^{2}) \frac{1}{s^{2} + ω^{2}}} - \frac{1}{(s^{2} + a^{2})}

Substitute in L(y),

L (y) = \frac{b ω}{a^{2} - ω^{2}} (\frac{1}{s^{2} + ω^{2}} - \frac{1}{s^{2} + a^{2}})

Taking inverse Laplace transform,

y (t) = \frac{b ω}{a^{2} - ω^{2}} L^{- 1} (\frac{1}{s^{2} + ω^{2}} - \frac{1}{s^{2} + a^{2}})

y (t) = \frac{b ω}{a^{2} - ω^{2}} [L^{- 1} (\frac{1}{s^{2} + ω^{2}}) - L^{- 1} (\frac{1}{s^{2} + a^{2}})]

= \frac{b ω}{a^{2} - ω^{2}} [\frac{1}{ω} \sin (ω t) - \frac{1}{a} \sin (a t)]

Therefore, the required solution is,

This is helpful 15 Flag

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Didn't find what you were looking for?