Linear Differential Equations

A first-order linear differential equation has the form $$${y}'+{p}{\left({t}\right)}{y}={q}{\left({t}\right)}$$$.

To solve it, rewrite it in the differential form: $$$\frac{{{d}{y}}}{{{d}{t}}}+{p}{\left({t}\right)}{y}={q}{\left({t}\right)}$$$, or $$${\left({p}{\left({t}\right)}{y}-{q}{\left({t}\right)}\right)}{d}{t}+{d}{y}={0}$$$.

Now, using the facts about exact equations, we see that this equation is not exact. However, since $$${M}{\left({t},{y}\right)}={p}{\left({t}\right)}{y}-{q}{\left({t}\right)}$$$ and $$${N}{\left({t},{y}\right)}={1}$$$, we have that $$$\frac{{\partial{M}}}{{\partial{y}}}={p}{\left({t}\right)}$$$ and $$$\frac{{\partial{N}}}{{\partial{x}}}={0}$$$. So, $$$\frac{{1}}{{N}}{\left(\frac{{\partial{M}}}{{\partial{y}}}-\frac{{\partial{N}}}{{\partial{x}}}\right)}=\frac{{1}}{{1}}{\left({p}{\left({t}\right)}-{0}\right)}={p}{\left({t}\right)}$$$. Thus, the integrating factor is $$${I}{\left({t},{y}\right)}={{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}$$$.

Now, return to the initial equation and multiply it by the integrating factor:

$$${y}'{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}+{p}{\left({t}\right)}{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}{y}={q}{\left({t}\right)}{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}$$$.

Closely looking at the left side, we note that it equals $$${d}{\left({y}{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}\right)}$$$.

So, the equation can be rewritten as $$${d}{\left({y}{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}\right)}={q}{\left({t}\right)}{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}$$$.

Integrating both sides gives $$${y}{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}=\int{\left({q}{\left({t}\right)}{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}\right)}{d}{t}+{C}$$$.

Finally, $$${y}={{e}}^{{-\int{p}{\left({t}\right)}{d}{t}}}\int{\left({q}{\left({t}\right)}{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}\right)}{d}{t}+{C}{{e}}^{{-\int{p}{\left({t}\right)}{d}{t}}}$$$.

This formula allows to find the solution directly.

Example 1. Solve $$${y}'-{7}{y}={{e}}^{{{t}}}$$$.

Here, $$${p}{\left({t}\right)}=-{7}$$$, and $$${q}{\left({t}\right)}={{e}}^{{t}}$$$.

So, the solution is

$$${y}={{e}}^{{-\int-{7}{d}{t}}}\int{\left({{e}}^{{t}}\cdot{{e}}^{{\int-{7}{d}{t}}}\right)}{d}{t}+{C}{{e}}^{{-\int-{7}{d}{t}}}={{e}}^{{{7}{t}}}\int{\left({{e}}^{{-{6}{t}}}\right)}{d}{t}+{C}{{e}}^{{{7}{t}}}={{e}}^{{{7}{t}}}\cdot{\left(-\frac{{1}}{{6}}\right)}{{e}}^{{-{6}{t}}}+{C}{{e}}^{{{7}{t}}}={C}{{e}}^{{{7}{t}}}-\frac{{1}}{{6}}{{e}}^{{t}}$$$

Thus, $$${y}={C}{{e}}^{{{7}{t}}}-\frac{{1}}{{6}}{{e}}^{{t}}$$$.

Let's work another quick example.

Example 2. Solve $$${{t}}^{{2}}{y}'-{3}{y}-{1}={0}$$$, $$${y}{\left({3}\right)}={0}$$$.

First, rewrite the equation: $$${y}'-\frac{{3}}{{{t}}^{{2}}}{y}=\frac{{1}}{{{t}}^{{2}}}$$$.

Here, $$${p}{\left({t}\right)}=-\frac{{3}}{{{t}}^{{2}}}$$$, and $$${q}{\left({t}\right)}=\frac{{1}}{{{t}}^{{2}}}$$$.

So,

$$${y}={{e}}^{{-\int-\frac{{3}}{{{t}}^{{2}}}{d}{t}}}\int{\left(\frac{{1}}{{{t}}^{{2}}}{{e}}^{{\int-\frac{{3}}{{{t}}^{{2}}}{d}{t}}}\right)}{d}{t}+{C}{{e}}^{{-\int-\frac{{3}}{{{t}}^{{2}}}{d}{t}}}={{e}}^{{-\frac{{3}}{{t}}}}\int{\left(\frac{{1}}{{{t}}^{{2}}}{{e}}^{{\frac{{3}}{{t}}}}\right)}{d}{t}+{C}{{e}}^{{-\frac{{3}}{{t}}}}={{e}}^{{-\frac{{3}}{{t}}}}\cdot{\left(-\frac{{1}}{{3}}\right)}{{e}}^{{\frac{{3}}{{t}}}}+{C}{{e}}^{{-\frac{{3}}{{t}}}}={C}{{e}}^{{-\frac{{3}}{{t}}}}-\frac{{1}}{{3}}$$$

$$${y}={C}{{e}}^{{-\frac{{3}}{{t}}}}-\frac{{1}}{{3}}$$$.

To find the particular solution, plug in the initial condition:

$$${0}={C}{{e}}^{{-\frac{{3}}{{3}}}}-\frac{{1}}{{3}}$$$ or $$${C}=\frac{{e}}{{3}}$$$

Finally, $$${y}=\frac{{e}}{{3}}{{e}}^{{-\frac{{3}}{{t}}}}-\frac{{1}}{{3}}=\frac{{1}}{{3}}{\left({{e}}^{{{1}-\frac{{3}}{{t}}}}-{1}\right)}$$$.