Initial Value Problems

Example 1. Solve $$${y}''+{y}'-{2}{y}={0}$$$, $$${y}{\left({0}\right)}={1}$$$, $$${y}'{\left({0}\right)}={2}$$$.

This is a homogeneous differential equation. The characteristic equation is $$${{r}}^{{2}}+{r}-{2}={0}$$$, or $$${\left({r}+{2}\right)}{\left({r}-{1}\right)}={0}$$$, which has the roots $$${r}_{{1}}=-{2},\ {r}_{{2}}={1}$$$.

So, the general solution is $$${y}={c}_{{1}}{{e}}^{{-{2}{x}}}+{c}_{{2}}{{e}}^{{x}}$$$.

Thus, $$${y}'=-{2}{c}_{{1}}{{e}}^{{-{2}{x}}}+{c}_{{2}}{{e}}^{{x}}$$$.

Applying the initial conditions gives

$$${\left\{\begin{array}{c}{y}{\left({0}\right)}={1}={c}_{{1}}{{e}}^{{-{2}\cdot{0}}}+{c}_{{2}}{{e}}^{{0}}\\{y}'{\left({0}\right)}={2}=-{2}{c}_{{1}}{{e}}^{{-{2}\cdot{0}}}+{c}_{{2}}{{e}}^{{0}}\\ \end{array}\right.}$$$,

Or

$$${\left\{\begin{array}{c}{c}_{{1}}+{c}_{{2}}={1}\\-{2}{c}_{{1}}+{c}_{{2}}={2}\\ \end{array}\right.}$$$.

Subtracting the second equation from the first gives $$${3}{c}_{{1}}=-{1}$$$, or $$${c}_{{1}}=-\frac{{1}}{{3}}$$$.

From the first equation, $$${c}_{{2}}={1}-{c}_{{1}}={1}-{\left(-\frac{{1}}{{3}}\right)}=\frac{{4}}{{3}}$$$.

So, the solution is $$${y}=-\frac{{1}}{{3}}{{e}}^{{-{2}{x}}}+\frac{{4}}{{3}}{{e}}^{{x}}$$$.

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

First, solve the corresponding homogeneous equation $$${y}''+{2}{y}'+{y}={0}$$$. The characteristic equation is $$${{r}}^{{2}}+{2}{r}+{1}={0}$$$, or $$${{\left({r}+{1}\right)}}^{{2}}={0}$$$, which has the roots $$${r}_{{1}}=-{1},{r}_{{2}}=-{1}$$$.

There is one root of the multiplicity 2; so, the solution of the homogeneous equation is $$${y}_{{h}}={c}_{{1}}{{e}}^{{-{t}}}+{c}_{{2}}{t}{{e}}^{{-{t}}}$$$.

Once again, we don't apply the initial conditions on this stage. We first have to find the general solution (in this case, $$${y}_{{h}}+{y}_{{p}}$$$) and only then apply the initial conditions.

To find the particular solution, use the method of undetermined coefficients.

Assume that $$${y}_{{p}}={A}{\cos{{\left({t}\right)}}}+{B}{\sin{{\left({t}\right)}}}$$$. Then, $$${y}_{{p}}'=-{A}{\sin{{\left({t}\right)}}}+{B}{\cos{{\left({t}\right)}}}$$$, and $$${y}_{{p}}''=-{A}{\cos{{\left({t}\right)}}}-{B}{\sin{{\left({t}\right)}}}$$$.

Plugging this into the equation gives

$$$-{A}{\cos{{\left({t}\right)}}}-{B}{\sin{{\left({t}\right)}}}+{2}{\left(-{A}{\sin{{\left({t}\right)}}}+{B}{\cos{{\left({t}\right)}}}\right)}+{A}{\cos{{\left({t}\right)}}}+{B}{\sin{{\left({t}\right)}}}={2}{B}{\cos{{\left({t}\right)}}}-{2}{A}{\sin{{\left({t}\right)}}}$$$.

Equating the like terms with $$${\cos{{\left({t}\right)}}}$$$ gives

$$${\left\{\begin{array}{c}{2}{B}={1}\\-{2}{A}={0}\\ \end{array}\right.}$$$,

Which has solution $$${A}={0},{B}=\frac{{1}}{{2}}$$$.

So, $$${y}_{{p}}=\frac{{1}}{{2}}{\sin{{\left({t}\right)}}}$$$.

Thus, the general solution is $$${y}={y}_{{h}}+{y}_{{p}}={c}_{{1}}{{e}}^{{-{t}}}+{c}_{{2}}{t}{{e}}^{{-{t}}}+\frac{{1}}{{2}}{\sin{{\left({t}\right)}}}$$$.

Now, we can use the initial conditions.

For this, find the derivative of the general solution: $$${y}'=-{c}_{{1}}{{e}}^{{-{t}}}+{c}_{{2}}{{e}}^{{-{t}}}-{c}_{{2}}{t}{{e}}^{{-{t}}}+\frac{{1}}{{2}}{\cos{{\left({t}\right)}}}$$$.

So,

$$${\left\{\begin{array}{c}{y}{\left({0}\right)}={3}={c}_{{1}}{{e}}^{{-{0}}}+{c}_{{2}}\cdot{0}\cdot{{e}}^{{-{0}}}+\frac{{1}}{{2}}{\sin{{\left({0}\right)}}}\\{y}'{\left({0}\right)}={0}=-{c}_{{1}}{{e}}^{{-{0}}}+{c}_{{2}}{{e}}^{{-{0}}}-{c}_{{2}}\cdot{0}\cdot{{e}}^{{-{0}}}+\frac{{1}}{{2}}{\cos{{\left({0}\right)}}}\\ \end{array}\right.}$$$,

Or

$$${\left\{\begin{array}{c}{c}_{{1}}={3}\\-{c}_{{1}}+{c}_{{2}}=-\frac{{1}}{{2}}\\ \end{array}\right.}$$$,

Which has the solution $$${c}_{{1}}={3},\ {c}_{{2}}=\frac{{5}}{{2}}$$$.

Finally, the solution is $$${y}={3}{{e}}^{{-{t}}}+\frac{{5}}{{2}}{t}{{e}}^{{-{t}}}+\frac{{1}}{{2}}{\sin{{\left({t}\right)}}}$$$.