# Differential Equation Using Laplace Transforms

• Apr 30th 2009, 04:49 PM
louboutinlover
i have stalled in another problem.

$\displaystyle 3\frac{dy}{dx} + 4y = 4 + 6x + 4x^2$

Given that $\displaystyle y=1$ when$\displaystyle x=0$

$\displaystyle 3(sL(y)-1) + 4L(y) = \frac{4}{s} + \frac{6}{s^2} + \frac{8}{s^3}$

$\displaystyle L(y) = \frac{4}{(s)(3s+4)} + \frac{6}{(s^2)(3s+4)} + \frac{8}{(s^3)(3s+4)} + \frac{3}{(3s+4)}$

Solving using partial fractions:

$\displaystyle A(3s+4) + B(s) = 4$
$\displaystyle A=1 B=-3$

$\displaystyle C(3s+4) + D(s^2) = 6$
$\displaystyle C = \frac{3}{2} D = \frac{27}{8}$

$\displaystyle E(3s+4) + F(s^3)$
$\displaystyle E = 2 F=\frac{-27}{8}$

Therefore:

$\displaystyle L(y) = \frac{1}{s} - \frac{3}{3s+4} + \frac{3}{2s^2} + \frac{27}{8(3s+4)} + \frac{2}{s^3} - \frac{27}{8(3s+4)} + \frac{3}{3s+4}$

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

$\displaystyle y = 1 + x^2$ is the solution. however if i take the inverse of $\displaystyle L(y)$ that does not match the solution.
can you spot where i have gone wrong ?

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I seem to be making the same mistake again with this question:

$\displaystyle \frac{d^2x}{dt^2} + 9x = 2sin2t$

Given that:

$\displaystyle s=0$ and $\displaystyle \frac{ds}{dt} = 1$ when $\displaystyle t=0$

Taking Laplace Transforms:

$\displaystyle L(x) = \frac{s^2+8}{(s^2+4)(s^2+9)}$

Now solving using partial fractions:

$\displaystyle \frac{A}{s^2+4} + \frac{B}{s^2+9} = s^2+8$

$\displaystyle A(s^2+9) + B(s^2+4) = s^2+8$

$\displaystyle As^2 + 9A + Bs^2 +4B = s^2+8$

From which:

$\displaystyle A+B=1$ and $\displaystyle 9A+4B=8$

$\displaystyle A=\frac{4}{5}$ $\displaystyle B=\frac{1}{5}$

Hence:

$\displaystyle L(x) = \frac{4}{5(s^2+4)} + \frac{1}{5(s^2+9)}$

The solution gives $\displaystyle x=\frac{1}{15}sin3t + \frac{2}{5}sin2t$

What step am i missing out or where have i gone wrong ?
• May 6th 2009, 06:24 AM
Calculus26
For

http://www.mathhelpforum.com/math-he...8ee2743f-1.gif

you don't need to use PFD write s^2 + 8 as s^2 + 4 + 4 and separate

Write as [(s^2+4)/[(s^2+4)(s^2+9)] + 4/[(s^2+4)(s^2+9)]

L(s) = 1/(s^2+9) + 4/(s^2+4)(s^2+9) both of which are standard transforms

y(t) = 1/3sin(3t) + 4[-1/5*(1/3sin(3t)-1/2sin(2t)]

= 1/3sin(3t)-4/15sin(3t) + 2/5sin(2t)

= 1/15sin(3t) + 2/5sin(2t)
• May 6th 2009, 06:40 AM
Calculus26
in general you are using the incorrect PFD

For an irreducible quadratic 1/(s^2+a) the term in the decomposition

is (As+B)/(s^2+a) not A/(s^2+a)

It is for linear factors1/(s+a) you have A/(s+a)

For linear factor1/[(s-a)^n] include A1/(s-a) +A2/(s-a)^2 + ..........+ An/(s-a)^n

For example for 1/s^2 you would have a A/s and a B/s^2 in the decomposition