Solving IVPs with Laplace Transform

Example 1. Solve $$${y}''+{4}{y}={t}$$$, $$${y}{\left({0}\right)}={0}$$$, $$${y}'{\left({0}\right)}={0}$$$.

The first step is always to take the Laplace transform of both sides.

$$${L}{\left({y}''+{4}{y}\right)}={L}{\left({t}\right)}$$$,

Or

$$${L}{\left({y}''\right)}+{4}{L}{\left({y}\right)}=\frac{{1}}{{{s}}^{{2}}}$$$.

Now, since $$${L}{\left({y}''\right)}={{s}}^{{2}}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}$$$ (see the table of Laplace transforms), we obtain that

$$${{s}}^{{2}}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{4}{Y}{\left({s}\right)}=\frac{{1}}{{{s}}^{{2}}}$$$, or, using the initial conditions,

$$${{s}}^{{2}}{Y}{\left({s}\right)}-{s}\cdot{0}-{0}+{4}{Y}{\left({s}\right)}=\frac{{1}}{{{s}}^{{2}}}$$$, which yields $$${Y}{\left({s}\right)}=\frac{{1}}{{{{s}}^{{2}}{\left({{s}}^{{2}}+{4}\right)}}}$$$.

So, to find $$${y}{\left({t}\right)}$$$, we have to find the inverse Laplace transform.

The partial fraction decomposition is $$$\frac{{A}}{{s}}+\frac{{B}}{{{s}}^{{2}}}+\frac{{{C}{s}+{D}}}{{{{s}}^{{2}}+{4}}}=\frac{{{A}{s}{\left({{s}}^{{2}}+{4}\right)}+{B}{\left({{s}}^{{2}}+{4}\right)}+{\left({C}{s}+{D}\right)}{{s}}^{{2}}}}{{{{s}}^{{2}}{\left({{s}}^{{2}}+{4}\right)}}}=\frac{{{\left({A}+{C}\right)}{{s}}^{{3}}+{\left({B}+{D}\right)}{{s}}^{{2}}+{4}{A}{s}+{4}{B}}}{{{{s}}^{{2}}{\left({{s}}^{{2}}+{4}\right)}}}=\frac{{1}}{{{{s}}^{{2}}{\left({{s}}^{{2}}+{4}\right)}}}$$$

So, the following system is obtained:

$$${\left\{\begin{array}{c}{A}+{C}={0}\\{B}+{D}={0}\\{4}{A}={0}\\{4}{B}={1}\\ \end{array}\right.}$$$, which has the solution $$${A}={0},{B}=\frac{{1}}{{4}},{C}={0},{D}=-\frac{{1}}{{4}}$$$.

Thus,

$$${y}{\left({t}\right)}={{L}}^{{-{{1}}}}{\left({Y}{\left({s}\right)}\right)}={{L}}^{{-{1}}}{\left(\frac{{1}}{{{{s}}^{{2}}{\left({{s}}^{{2}}+{4}\right)}}}\right)}={{L}}^{{-{{1}}}}{\left(\frac{{1}}{{4}}\frac{{1}}{{{{s}}^{{2}}}}-\frac{{1}}{{4}}\frac{{1}}{{{{s}}^{{2}}+{4}}}\right)}=\frac{{1}}{{4}}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}}^{{2}}}\right)}-\frac{{1}}{{4}}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{{s}}^{{2}}+{4}}}\right)}=\frac{{1}}{{4}}{t}-\frac{{1}}{{8}}{{L}}^{{-{{1}}}}{\left(\frac{{2}}{{{{s}}^{{2}}+{4}}}\right)}=\frac{{1}}{{4}}{t}-\frac{{1}}{{8}}{\sin{{\left({2}{t}\right)}}}$$$

Example 2. Solve $$${y}''+{y}'={u}_{{1}}{\left({t}\right)}$$$, $$${y}{\left({0}\right)}={0}$$$, $$${y}'{\left({0}\right)}={0}$$$.

Take the Laplace transform

$$${L}{\left({y}''\right)}+{L}{\left({y}'\right)}={L}{\left({u}_{{1}}{\left({t}\right)}\right)}$$$,

Or

$$${\left({{s}}^{{2}}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\right)}+{\left({s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\right)}=\frac{{{{e}}^{{-{{s}}}}}}{{s}}$$$.

Applying the initial conditions gives:

$$${{s}}^{{2}}{Y}{\left({s}\right)}+{s}{Y}{\left({s}\right)}=\frac{{{e}}^{{-{s}}}}{{s}}$$$ or $$${Y}{\left({s}\right)}={{e}}^{{-{s}}}\frac{{1}}{{{{s}}^{{2}}{\left({s}+{1}\right)}}}$$$.

The partical fraction decomposition for $$$\frac{{1}}{{{{s}}^{{2}}{\left({s}+{1}\right)}}}$$$ is $$$\frac{{1}}{{{s}}^{{2}}}-\frac{{1}}{{s}}+\frac{{1}}{{{s}+{1}}}$$$.

So,

$$${y}{\left({t}\right)}={{L}}^{{-{{1}}}}{\left({{e}}^{{-{s}}}\frac{{1}}{{{s}}^{{2}}}\right)}-{{L}}^{{-{{1}}}}{\left({{e}}^{{-{s}}}\frac{{1}}{{s}}\right)}+{{L}}^{{-{{1}}}}{\left({{e}}^{{-{s}}}\frac{{1}}{{{s}+{1}}}\right)}={u}_{{1}}{\left({t}\right)}{\left({t}-{1}\right)}-{u}_{{1}}{\left({t}\right)}{1}+{u}_{{1}}{\left({t}\right)}{{e}}^{{-{\left({t}-{1}\right)}}}={u}_{{1}}{\left({t}\right)}{\left({t}-{2}+{{e}}^{{{1}-{t}}}\right)}$$$

Of course, we could rewrite the Heaviside function by definition and solve the differential equation on each interval separately, but this would require a greater amount of work.

Example 3. Calculate $$${y}''+{y}'={t}$$$, $$${y}{\left({2}\right)}={0}$$$, $$${y}'{\left({2}\right)}={0}$$$.

Notice that the initial conditions are not at $$${0}$$$. In order to use the Laplace transform, we need the initial conditions to be at $$${0}$$$. For this, use the change of variable: let $$${a}={t}-{2}$$$; then:

$$${y}''{\left({a}+{2}\right)}+{y}'{\left({a}+{2}\right)}={a}+{2}$$$, $$${y}{\left({0}\right)}={0}$$$, $$${y}'{\left({0}\right)}={0}$$$.

To simplify the expression, let $$${y}{\left({a}+{2}\right)}={u}{\left({a}\right)}$$$; then, $$${y}'{\left({a}+{2}\right)}={u}'{\left({a}\right)}$$$, and $$${y}''{\left({a}+{2}\right)}={u}''{\left({a}\right)}$$$; so, the equation can be rewritten as:

$$${u}''+{u}'={a}+{2}$$$, $$${u}{\left({0}\right)}={0}$$$, $$${u}'{\left({0}\right)}={0}$$$.

Now, we can use the Laplace transform:

$$${L}{\left({u}''\right)}+{L}{\left({u}'\right)}={L}{\left({a}\right)}+{L}{\left({2}\right)}$$$

$$${{s}}^{{2}}{U}{\left({s}\right)}-{s}{u}{\left({0}\right)}-{u}'{\left({0}\right)}+{s}{U}{\left({s}\right)}-{u}{\left({0}\right)}=\frac{{1}}{{{s}}^{{2}}}+\frac{{2}}{{s}}$$$

Applying the initial conditions gives:

$$${U}{\left({s}\right)}=\frac{{\frac{{1}}{{{s}}^{{2}}}+\frac{{2}}{{s}}}}{{{{s}}^{{2}}+{s}}}=\frac{{{2}{s}+{1}}}{{{{s}}^{{3}}{\left({s}+{1}\right)}}}$$$

The partial fraction decomposition is (verify!): $$$\frac{{1}}{{{s}}^{{3}}}+\frac{{1}}{{{s}}^{{2}}}+\frac{{1}}{{{s}+{1}}}-\frac{{1}}{{s}}$$$.

So, $$${u}{\left({a}\right)}={{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}}^{{3}}}+\frac{{1}}{{{s}}^{{2}}}+\frac{{1}}{{{s}+{1}}}-\frac{{1}}{{s}}\right)}={{L}}^{{-{{1}}}}{\left(\frac{{1}}{{2}}\frac{{2}}{{{s}}^{{3}}}+\frac{{1}}{{{s}}^{{2}}}+\frac{{1}}{{{s}+{1}}}-\frac{{1}}{{s}}\right)}=\frac{{1}}{{2}}{{a}}^{{2}}+{a}+{{e}}^{{-{a}}}-{1}$$$.

However, we need $$${y}{\left({t}\right)}$$$, not $$${u}{\left({a}\right)}$$$. Recall that $$${y}{\left({t}\right)}={y}{\left({a}+{2}\right)}={u}{\left({a}\right)}={u}{\left({t}-{2}\right)}$$$; so,

$$${y}{\left({t}\right)}=\frac{{1}}{{2}}{{\left({t}-{2}\right)}}^{{2}}+{t}-{2}+{{e}}^{{-{\left({t}-{2}\right)}}}-{1}=\frac{{1}}{{2}}{{t}}^{{2}}-{t}-{1}+{{e}}^{{{2}-{t}}}$$$.