• MHB
Gold Member
MHB
Solve the following IVP using Laplace Transforms:
$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t} + 11\,y = 3\,t, \quad y\left( 0 \right) = 5$

Take the Laplace Transform of the equation:

\displaystyle \begin{align*} s\,Y\left( s \right) - y\left( 0 \right) + 11\,Y\left( s \right) &= \frac{3}{s^2} \\ s\,Y\left( s \right) - 5 + 11\,Y\left( s \right) &= \frac{3}{s^2} \\ \left( s + 11 \right) Y\left( s \right) &= \frac{3}{s^2} + 5 \\ Y\left( s \right) &= \frac{3}{s^2 \,\left( s + 11 \right) } + \frac{5}{s + 11} \end{align*}

Apply Partial Fractions:

\displaystyle \begin{align*} \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + 11} &\equiv \frac{3}{s^2 \,\left( s + 11 \right) } \\ A\,s\left( s + 11 \right) + B \left( s + 11 \right) + C\,s^2 &\equiv 3 \end{align*}

Let $\displaystyle s = 0 \implies 11\,B = 3 \implies B = \frac{3}{11}$

Let $\displaystyle s = -11 \implies 121\,C = 3 \implies C = \frac{3}{121}$

Then $\displaystyle A\,s\left( s + 11 \right) + \frac{3}{11} \left( s + 11 \right) + \frac{3}{121}\,s^2 \equiv 3$

Let $\displaystyle s = 1$

\displaystyle \begin{align*} 12\,A + \frac{36}{11} + \frac{3}{121} &= 3 \\ 12\,A + \frac{396}{121} + \frac{3}{121} &= \frac{33}{121} \\ 12\,A &= -\frac{366}{121} \\ A &= -\frac{61}{242} \end{align*}

So

\displaystyle \begin{align*} Y\left( s \right) &= -\frac{61}{242}\left( \frac{1}{s} \right) + \frac{3}{11} \left( \frac{1}{s^2} \right) + \frac{3}{121} \left( \frac{1}{s + 11} \right) + 5 \left( \frac{1}{s + 11 } \right) \\ Y\left( s \right) &= -\frac{61}{242} \left( \frac{1}{s} \right) + \frac{3}{11} \left( \frac{1}{s^2} \right) + \frac{608}{121} \left( \frac{1}{s + 11} \right) \\ \\ y\left( t \right) &= -\frac{61}{242} + \frac{3}{11}\,t + \frac{608}{121}\,\mathrm{e}^{-11\,t} \end{align*}

Homework Helper
Let me, yet again, state my dislike for the "Laplace Transform Method"! And, in fact, Prove It made a slight arithmetic error
(easy to do with something as complicated as "Laplace Transform")
that resulted in an incorrect answer:
if $$y(t)= 61/242+ (3/11)t+ (608/121)e^{-11t}$$ then
$$y'(t)= 3/11- (608/11)e^{-11t}$$
$$11y= 61/22+ 3t+ (608/11)e^{-11t}$$

so $$y'(t)+ 11t= 6/22+ 61/22+ 3t= 67/22+ 3t$$, not "3t".

It is far easier just to recognize that this is a linear differential equation with constant coefficients. Its "characteristic equation" is r+ 11= 0 so r= -11. The general solution to the associated homogeneous equation is $$y(t)= Ce{-11t}$$ where C can be any constant.

Since the "non-homogeous" part is a linear polynomial, 3t, we look for a solution to the entire equation of the form y(t)= At+ B, for constants A and B. Then y'= A so the equation becomes A+ 11(At+ B)= 11At+ A+ 11B= 3t. In order for that to be true for all t, we must have both 11A= 3 and A+ 11B= 0. So A= 3/11 and then 3/11+ 11B= 0. B= -3/121.

The general solution to the entire equation is $$y(t)= Ce^{-11t}+ (3/11)t- 3/121$$. Since we want y(0)= 5, we must have $$5= C+ 3/121$$ so $$C= 5- 3/121= 605/121- 3/121= 602/121. The solution is [tex]y(t)= (602/121)e^{-11t}+ (3/11)t- 3/121$$.

Check: $$y'(t)= -(602/11)e^{-11t}+ 3/11$$ while $$11y(t)= (602/11)e^{-11t}+ 3t- 3/11$$

$$y'(t)+ 11t= 3t$$.

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Gold Member
MHB
Let me, yet again, state my dislike for the "Laplace Transform Method"! And, in fact, Prove It made a slight arithmetic error
(easy to do with something as complicated as "Laplace Transform")
that resulted in an incorrect answer:
if $$y(t)= 61/242+ (3/11)t+ (608/121)e^{-11t}$$ then
$$y'(t)= 3/11- (608/11)e^{-11t}$$
$$11y= 61/22+ 3t+ (608/11)e^{-11t}$$

so $$y'(t)+ 11t= 6/22+ 61/22+ 3t= 67/22+ 3t$$, not "3t".

It is far easier just to recognize that this is a linear differential equation with constant coefficients. Its "characteristic equation" is r+ 11= 0 so r= -11. The general solution to the associated homogeneous equation is $$y(t)= Ce{-11t}$$ where C can be any constant.

Since the "non-homogeous" part is a linear polynomial, 3t, we look for a solution to the entire equation of the form y(t)= At+ B, for constants A and B. Then y'= A so the equation becomes A+ 11(At+ B)= 11At+ A+ 11B= 3t. In order for that to be true for all t, we must have both 11A= 3 and A+ 11B= 0. So A= 3/11 and then 3/11+ 11B= 0. B= -3/121.

The general solution to the entire equation is $$y(t)= Ce^{-11t}+ (3/11)t- 3/121$$. Since we want y(0)= 5, we must have $$5= C+ 3/121$$ so $$C= 5- 3/121= 605/121- 3/121= 602/121. The solution is [tex]y(t)= (602/121)e^{-11t}+ (3/11)t- 3/121$$.

Check: $$y'(t)= -(602/11)e^{-11t}+ 3/11$$ while $$11y(t)= (602/11)e^{-11t}+ 3t- 3/11$$

$$y'(t)+ 11t= 3t$$.

Thanks for pointing out my error Hallsofivy. Adam is one of my students, and the topic they are learning is Laplace Transforms, so he will have to use that method.

Gold Member
MHB
The general solution to the entire equation is $$y(t)= Ce^{-11t}+ (3/11)t- 3/121$$. Since we want y(0)= 5, we must have $$5= C+ 3/121$$ so [tex]C= 5- 3/121= 605/121- 3/121= 602/121.

You also have an arithmetic error. When $\displaystyle t = 5$ you end up with $\displaystyle 5 = C - \frac{3}{121} \implies C = 5 + \frac{3}{121} = \frac{608}{121}$.

I also see where my mistake was in the initial Laplace Transform. The easiest way to evaluate A is to look at the coefficient of $\displaystyle s^2$, which gives

$\displaystyle A + \frac{3}{121} = 0 \implies A = -\frac{3}{121}$.

Thus

\displaystyle \begin{align*} Y\left( s \right) &= -\frac{3}{121}\left( \frac{1}{s} \right) + \frac{3}{11}\left( \frac{1}{s^2} \right) + \frac{3}{121} \left( \frac{1}{s + 11} \right) + 5 \left( \frac{1}{s + 11 } \right) \\ &= -\frac{3}{121} \left( \frac{1}{s} \right) + \frac{3}{11} \left( \frac{1}{s^2} \right) + \frac{608}{121} \left( \frac{1}{s + 11 } \right) \\ \\ y\left( t \right) &= -\frac{3}{121} + \frac{3}{11}\,t + \frac{608}{121}\,\mathrm{e}^{-11\,t} \end{align*}

and this is definitely correct.