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UPDATED WTH ANSWER:

Please help me solve the following differential equation using Laplace transform: $$y''+y'+y=0;y(0)=4,y'(0)=-3$$

My answer so far: $$\mathcal{L}\{y''\} + \mathcal{L}\{y'\} + \mathcal{L}\{y\}=0$$ We know that $\mathcal{L}\{y'\} = s\mathcal{L}\{y\}-y(o)$ and $\mathcal{L}\{y''\}=s\mathcal{L}\{y'\}-y'(0)$. Upon substituting this to the equation, it becomes $$s\mathcal{L}\{y'\}-y'(0)+s\mathcal{L}\{y\}-y(0)+\mathcal{L}\{y\}=0$$ $$s(s\mathcal{L}\{y\}-y(0))-y'(0)+s\mathcal{L}\{y\}-y(0)+\mathcal{L}\{y\}=0$$ $$s^2\mathcal{L}\{y\}-sy(0)-y'(0)+s\mathcal{L}\{y\}-y(0)+\mathcal{L}\{y\}=0$$ Upon substituting the values of initial conditions the equation becomes $$s^2\mathcal{L}\{y\}-4s+3+s\mathcal{L}\{y\}-4+\mathcal{L}\{y\}=0$$ Solving for $\mathcal{L}\{y\}$: $$\mathcal{L}\{y\}(s^2+s+1)-4s+3-4=0$$ $$\mathcal{L}\{y\}(s^2+s+1)=4s+1$$ $$\mathcal{L}\{y\}=\frac{4s+1}{s^2+s+1}$$ Modifying the numerator and denominator to obtain the form similar to one of the them in the Laplace transform table. First applying the method of completing the square to the denominator: $$\frac{4s+1}{s^2+s+1} = \frac{4s+1}{(s^2+s+\frac{1}{4})+(1-\frac{1}{4})}=\frac{4s+1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$$ Then modifying the numerator: $$\frac{4s+1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\frac{4s+2-1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\frac{4(s+\frac{1}{2})-1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$$ Divide the fractions: $$\mathcal{L}\{y\}=\frac{4(s+\frac{1}{2})}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}-\frac{1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$$

Taking the inverse of Laplace transform of $\mathcal{L}\{y\}$ and let it be $y$. Then the equation becomes $$y=4\mathcal{L}^{-1}[\frac{(s+\frac{1}{2})}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2)}] - \mathcal{L}^{-1}[\frac{1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}]$$

The inverse of the first fraction is: $$4e^{\frac{-1x}{2}}\cos(\frac{\sqrt3x}{2})$$

To match the second fraction to one of the forms from the Laplace Transform table, we need to multiply the numerator and denominator by $\frac{\sqrt3}{2}$. Then it becomes: $$\frac{2}{\sqrt3}\mathcal{L}^{-1}[\frac{\frac{\sqrt3}{2}}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}]$$ Therefore the inverse of the second fraction is: $$\frac{2}{\sqrt3}e^{\frac{-1x}{2}}\sin(\frac{\sqrt3x}{2})$$ FINAL ANSWER: $$y=4e^{\frac{-1x}{2}}\cos(\frac{\sqrt3x}{2})+\frac{2}{\sqrt3}e^{\frac{-1x}{2}}\sin(\frac{\sqrt3x}{2})$$ //

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$$\mathcal{L}(s)=\frac{4\left(s+\frac{1}{2}\right)-1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$$

therefore

$$ f(x)=e^{-\frac{1}{2}x}\left(4\cos \left(\frac{\sqrt{3}}{2}x\right)-\frac{2\sqrt{3}}{3}\sin \left(\frac{\sqrt{3}}{2}x\right)\right)$$

In case there remains some mystery concerning how to know to replace the $4s+1$ with $4\left(s+\frac{1}{2}\right)-1$, since after completing the square in the denominator we see that it contains $\left(s+\frac{1}{2}\right)^2$ we must then replace the $4s$ with $4\left(s+\frac{1}{2}\right)$. But seeing that this changes the numerator to $4s+2$ we subtract the $1$ to keep the numerator the same after the adjustment.

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  • $\begingroup$ If you factor out 4, then the numerator of the first part becomes 4(s-1/8) $\endgroup$
    – socrates
    May 7, 2017 at 23:51
  • $\begingroup$ @socrates Is not $4\left(s+\frac{1}{2}\right)-1=4s+1$? $\endgroup$ May 7, 2017 at 23:53
  • $\begingroup$ right........ ... $\endgroup$
    – socrates
    May 7, 2017 at 23:55
  • $\begingroup$ how did you get 2sqrt(3)/3 in front of the second part? $\endgroup$
    – socrates
    May 7, 2017 at 23:56
  • $\begingroup$ Still have not checked my answer, however. Doing that now. $\endgroup$ May 7, 2017 at 23:56

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