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#### Prove It

##### Well-known member
MHB Math Helper
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*}

#### HallsofIvy

##### Well-known member
MHB Math 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 $$11y(t)= (602/11)e^{-11t}+ 3t- 3/11$$

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

Last edited:

#### Prove It

##### Well-known member
MHB Math 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 $$11y(t)= (602/11)e^{-11t}+ 3t- 3/11$$

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

#### Prove It

##### Well-known member
MHB Math Helper
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.

#### HallsofIvy

##### Well-known member
MHB Math Helper
Argh! Arithmetic! I never was any good at that!