## Illustrative Example 01:

Solution:

At first, we apply \mathscr{L} to the differential equation:

\mathscr{L}\{y'-y\}=\mathscr{L}\{1\}

And since \mathscr{L} is linear, then

\mathscr{L}\{y'\}-\mathscr{L}\{y\}=\mathscr{L}\{1\}

If \mathscr{L}\{y(t)\}=Y(s), then from Laplace transform of derivatives, we have:

\mathscr{L}\{y'\}s=Y(s)-y(0)

and

\displaystyle\mathscr{L}\{1\}=\frac{1}{s}

as proven in this post. So, the equation becomes:

\begin{aligned} & sY(s)-\overbrace{y(0)}^{0}-Y(s)=\frac{1}{s} \\ & sY(s)-Y(s)=\frac{1}{s} \end{aligned}

i.e., we now have an algebric equation in Y(s):

\begin{aligned} & (s-1)Y(s)=\frac{1}{s} \\ & Y(s)=\frac{1}{s(s-1)} \end{aligned}

Thus, the solution of the differential equation y(t) is such that its Laplace transform is

\displaystyle Y(s)=\frac{1}{s(s-1)}

However, we realize we are not able to find in the table any function that satisfies it. The idea is to turn Y(s) into a sum/difference of two (or more) functions. To do so, we decompose it into partial fractions.

Since s and s-1 are both linear, there are unique real constants A and B such that:

\displaystyle\frac{1}{s(s-1)}=\frac{A}{s}+\frac{B}{s-1}

which gives us

\displaystyle\frac{1}{s(s-1)}=\frac{A(s-1)+Bs}{s(s-1)}

Once the denominators are identical, and so are the numerators:

\begin{aligned} & 1=A(s-1)+Bs \\ & 1=As-A+Bs \\ & 1=(A+B)s-A \end{aligned}

When we compare the coefficients of powers of s, we get the linear system in A and B:

\left\{\begin{array}{l} A+B=0 \\ -A=1 \end{array}\right.

From the last equation

A=-1

and when we substitute A in the first equation, we have

-1+B=0\Rightarrow B=1

Hence

\displaystyle Y(s)=\frac{1}{s(s-1)}=-\frac{1}{s}+\frac{1}{s-1}

and then

\displaystyle y(t)=\mathscr{L}^{-1}\left\{-\frac{1}{s}+\frac{1}{s-1}\right\}=-\mathscr{L}^{-1}\left\{\frac{1}{s}\right\}+\mathscr{L}^{-1}\left\{\frac{1}{s-1}\right\}

As

\displaystyle\mathscr{L}^{-1}\left\{\frac{1}{s}\right\}=1

and since

\displaystyle\mathscr{L}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}

so, for a=1, we get

\displaystyle\mathscr{L}^{-1}\left\{\frac{1}{s-1}\right\}=e^{1t}

Therefore, the solution of the initial-value problem:

\left\{\begin{array}{l} y'-y=1 \\ y(0)=0 \end{array}\right.

is the function:

y(t)=-1+e^{t}

## Illustrative Example 02:

Solution:

Since

\displaystyle\mathscr{L}\{\cos(at)\}=\frac{s}{s^{2}+a^{2}}

then for a=1, we have

\displaystyle\mathscr{L}\{\cos(t)\}=\frac{s}{s^{2}+1^{2}}=\frac{s}{s^{2}+1}

Now, applying \mathscr{L}:

\begin{aligned} & \mathscr{L}\{y''-2y'+2y\}=\mathscr{L}\{\cos(t)\} \\ & \mathscr{L}\{y''\}-2\mathscr{L}\{y'\}+2\mathscr{L}\{y\}=\mathscr{L}\{\cos(t)\} \\ & s^{2}Y(s)-sy(0)-y'(0)-2(sY(s)-y(0))+2Y(s)=\frac{s}{s^{2}+1} \end{aligned}

From the initial values:

\begin{aligned} & s^{2}Y(s)-s\cdot1-0-2(sY(s)-1)+2Y(s)=\frac{s}{s^{2}+1} \\ & s^{2}Y(s)-s-2sY(s)+2+2Y(s)=\frac{s}{s^{2}+1} \end{aligned}

And solving the algebraic equation for Y(s), we get:

\begin{aligned}& s^{2}Y(s)-2sY(s)+2Y(s)=\frac{s}{s^{2}+1}+s-2 \\ & (s^{2}-2s+2)Y(s)=\frac{s}{s^{2}+1}+s-2 \\ & Y(s)=\frac{s}{(s^2-2s+2)(s^2+1)}+\frac{s-2}{s^{2}-2s+2} \end{aligned}

We have to decompose Y(s) via partial fractions. As

cannot be factored into binomials, there are then unique A,B,C,D\in\mathbb{R} such that

\displaystyle\frac{s}{(s^2-2s+2)(s^2+1)}=\frac{As+B}{s^{2}-2s+2}+\frac{Cs+D}{s^{2}+1}

which gives us

\begin{aligned} \frac{s}{(s^2-2s+2)(s^2+1)} & = \frac{(As+B)(s^{2}+1)+(Cs+D)(s^{2}-2s+2)}{(s^{2}-2s+2)(s^2+1)} \\ & = \frac{As^{3}+As+Bs^{2}+B}{(s^{2}-2s+2)(s^{2}+1)} \\ & +\frac{Cs^{3}-2Cs^{2}+2Cs+Ds^{2}-2Ds+2D}{(s^{2}-2s+2)(s^{2}+1)} \end{aligned}

As we combine the powers of s, we may rearrange as follows

\begin{aligned} \frac{s}{(s^{2}-2s+2)(s^{2}+1)} &=\frac{(A+C)s^{3}+(B-2C+D)s^{2}}{(s^{2}-2s+2)(s^{2}+1)} \\ &+\frac{(A+2C-2D)s+(B+2D)}{(s^{2}-2s+2)(s^{2}+1)} \end{aligned}

And like we did in the last example, the numerators must be equal. From this, we have the linear system:

\left\{\begin{array}{l} A+C=0 \\ B-2C+D=0 \\ A+2C-2D=1 \\ B+2D=0 \end{array}\right.

In the first equation:

A+C=0\Rightarrow A+C+C=C\Rightarrow A+2C=C

and substituting it in the third equation, we have

\begin{aligned} & A+2C-2D=1 \\ & C-2D=1 \end{aligned}

As we subtract the fourth equation from the second one, we get

-2C-D=0

Thus, now there is a linear system in C and D:

\left\{\begin{array}{l} C-2D=1 \\ -2C-D=0 \end{array}\right.

From the last equation:

D=-2C

and when we substitute it in the other one, we are able to find the value of C:

\displaystyle C+4C=1\Rightarrow C=\frac{1}{5}

So

\displaystyle D=-\frac{2}{5}

In the equality

A+C=0

we may conclude

\displaystyle A=-C\Rightarrow A=-\frac{1}{5}

Also

\displaystyle B=-2D\Rightarrow B=\frac{4}{5}

Thus

\displaystyle \frac{s}{(s^{2}-2s+2)}=\dfrac{-\dfrac{1}{5}s+\dfrac{4}{5}}{s^{2}-2s+2}+\dfrac{\dfrac{1}{5}s-\dfrac{2}{5}}{s^{2}+1}

which may be written as

\begin{aligned} \frac{s}{(s^{2}-2s+2)} & =\frac{1}{5}\left(\frac{-s+4}{s^{2}-2s+2}+\frac{s-2}{s^{2}+1}\right) \\ & =\frac{1}{5}\left(\frac{-s+4}{s^2-2s+2}+\frac{s}{s^{2}+1}-\frac{2}{s^2+1}\right) \end{aligned}

The second fraction in the expression of Y(s) above cannot be decomposed via partial fractions. Then, we have:

\displaystyle Y(s)=\frac{1}{5}\left(\frac{-s+4}{s^2-2s+2}+\frac{s}{s^{2}+1}-\frac{2}{s^2+1}\right)+\frac{s-2}{s^2-2s+2}

i.e.,

\displaystyle y(t)=\mathscr{L}^{-1}\left\{\frac{1}{5}\left(\frac{-s+4}{s^2-2s+2}+\frac{s}{s^{2}+1}-\frac{2}{s^2+1}\right)+\frac{s-2}{s^2-2s+2}\right\}

and from the linearity of the inverse Laplace transform

\begin{aligned} y(t) &=\frac{1}{5}\mathscr{L}^{-1}\left\{\frac{-s+4}{s^2-2s+2}\right\}+\frac{1}{5}\mathscr{L}^{-1}\left\{\frac{s}{s^{2}+1}\right\} \\ &-\frac{1}{5}\mathscr{L}^{-1}\left\{\frac{2}{s^{2}+1}\right\}+\mathscr{L}^{-1}\left\{\frac{s-2}{s^{2}-2s+2}\right\} \end{aligned}

As

\displaystyle \mathscr{L}^{-1}\left\{\frac{s-a}{(s-a)^{2}+b^{2}}\right\}=e^{at}\cos(bt)

we have to rewrite the first and the fourth fractions such that it appears a square of sum/difference of two terms. Note that

s^{2}-2s+2=s^{2}-2\cdot s\cdot1+2

and since 2=1+1 and clearly, 1=1^{2}, we get

s^{2}-2s+2=s^{2}-2\cdot s\cdot1+1^{2}+1=(s-1)^{2}+1^{2}

Comparing the last result with the formula of Laplace inverse above, we have

Hence, we may rewrite the fractions as follows

\begin{aligned} \frac{-s+4}{s^2-2s+2} & =-\frac{s-4}{s^2-2s+2} \\ & =-\frac{s-1-3}{(s-1)^{2}+1^{2}} \\ & =-\frac{s-1}{(s-1)^{2}+1^{2}}-\frac{-3}{(s-1)^{2}+1^{2}} \\ & =-\frac{s-1}{(s-1)^{2}+1^{2}}+\frac{3}{(s-1)^{2}+1^{2}} \end{aligned}

and

\begin{aligned} \frac{s-2}{s^2-2s+2} &=\frac{s-1-1}{s^2-2s+2} \\ &=\frac{s-1}{(s-1)^{2}+1^{2}}-\frac{1}{(s-1)^{2}+1^{2}} \end{aligned}

We also know

\displaystyle \mathscr{L}^{-1}\left\{\frac{b}{(s-a)^{2}+b^{2}}\right\}=e^{at}\sin(bt)

then

\begin{aligned} \mathscr{L}^{-1}\left\{\frac{-s+4}{s^2-2s+2}\right\} & =-\mathscr{L}^{-1}\left\{\frac{s-1}{(s-1)^{2}+1^{2}}\right\}+3\cdot\mathscr{L}^{-1}\left\{\frac{1}{(s-1)^{2}+1^{2}}\right\} \\ & =-e^{1t}\cos(1t)+3e^{1t}\sin(1t) \end{aligned}

and

\begin{aligned} \mathscr{L}^{-1}\left\{\frac{s-2}{s^2-2s+2}\right\} & =\mathscr{L}^{-1}\left\{\frac{s-1}{(s-1)^{2}+1^{2}}\right\}-\mathscr{L}^{-1}\left\{\frac{1}{(s-1)^{2}+1^{2}}\right\} \\ & =e^{1t}\cos(1t)-e^{1t}\sin(1t) \end{aligned}

From the table of Laplace transforms:

Therefore, for b=1

\displaystyle \mathscr{L}^{-1}\left\{\frac{s}{s^2+1}\right\}=\cos(1t)

and also

\begin{aligned} \mathscr{L}^{-1}\left\{\frac{2}{s^2+1}\right\} & =2\cdot\mathscr{L}^{-1}\left\{\frac{1}{s^2+1^2}\right\} \\ & =2\cdot\sin(1t) \end{aligned}

Thus

\begin{aligned} & y(t)=\frac{1}{5}(-e^{t}\cos(t)+3e^{t}\sin(t))+\frac{1}{5}\cos(t)-\frac{1}{5}\cdot2\sin(t)+e^{t}\cos(t)-e^{t}\sin(t) \\ & y(t)=-\frac{e^{t}\cos(t)}{5}+\frac{3e^{t}\sin(t)}{5}+\frac{\cos(t)}{5}-\frac{2\sin(t)}{5}+e^{t}\cos(t)-e^{t}\sin(t) \end{aligned}

Hence, the solution of the 2nd order differential equation:

\left\{\begin{array}{l} y''-2y'+2y=\cos(t), \\ y(0)=1, \\ y'(0)=0 \end{array}\right.

is given by:

\displaystyle y(t)=\frac{4e^{t}\cos(t)}{5}-\frac{2e^{t}\sin(t)}{5}+\frac{\cos(t)}{5}-\frac{2\sin(t)}{5}