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Linear Differential Equations

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

To solve it, rewrite it in the differential form: dydt+p(t)y=q(t)\frac{{{d}{y}}}{{{d}{t}}}+{p}{\left({t}\right)}{y}={q}{\left({t}\right)}, or (p(t)y−q(t))dt+dy=0{\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(t,y)=p(t)y−q(t){M}{\left({t},{y}\right)}={p}{\left({t}\right)}{y}-{q}{\left({t}\right)} and N(t,y)=1{N}{\left({t},{y}\right)}={1}, we have that ∂M∂y=p(t)\frac{{\partial{M}}}{{\partial{y}}}={p}{\left({t}\right)} and ∂N∂x=0\frac{{\partial{N}}}{{\partial{x}}}={0}. So, 1N(∂M∂y−∂N∂x)=11(p(t)−0)=p(t)\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(t,y)=e∫p(t)dt{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∫p(t)dt+p(t)e∫p(t)dty=q(t)e∫p(t)dt{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(ye∫p(t)dt){d}{\left({y}{{e}}^{{\int{p}{\left({t}\right)}{d}{t}}}\right)}.

So, the equation can be rewritten as d(ye∫p(t)dt)=q(t)e∫p(t)dt{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 ye∫p(t)dt=∫(q(t)e∫p(t)dt)dt+C{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−∫p(t)dt∫(q(t)e∫p(t)dt)dt+Ce−∫p(t)dt{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′−7y=et{y}'-{7}{y}={{e}}^{{{t}}}.

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

So, the solution is

y=e−∫−7dt∫(et⋅e∫−7dt)dt+Ce−∫−7dt=e7t∫(e−6t)dt+Ce7t=e7t⋅(−16)e−6t+Ce7t=Ce7t−16et{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=Ce7t−16et{y}={C}{{e}}^{{{7}{t}}}-\frac{{1}}{{6}}{{e}}^{{t}}.

Let's work another quick example.

Example 2. Solve t2y′−3y−1=0{{t}}^{{2}}{y}'-{3}{y}-{1}={0}, y(3)=0{y}{\left({3}\right)}={0}.

First, rewrite the equation: y′−3t2y=1t2{y}'-\frac{{3}}{{{t}}^{{2}}}{y}=\frac{{1}}{{{t}}^{{2}}}.

Here, p(t)=−3t2{p}{\left({t}\right)}=-\frac{{3}}{{{t}}^{{2}}}, and q(t)=1t2{q}{\left({t}\right)}=\frac{{1}}{{{t}}^{{2}}}.

So,

y=e−∫−3t2dt∫(1t2e∫−3t2dt)dt+Ce−∫−3t2dt=e−3t∫(1t2e3t)dt+Ce−3t=e−3t⋅(−13)e3t+Ce−3t=Ce−3t−13{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=Ce−3t−13{y}={C}{{e}}^{{-\frac{{3}}{{t}}}}-\frac{{1}}{{3}}.

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

0=Ce−33−13{0}={C}{{e}}^{{-\frac{{3}}{{3}}}}-\frac{{1}}{{3}} or C=e3{C}=\frac{{e}}{{3}}

Finally, y=e3e−3t−13=13(e1−3t−1){y}=\frac{{e}}{{3}}{{e}}^{{-\frac{{3}}{{t}}}}-\frac{{1}}{{3}}=\frac{{1}}{{3}}{\left({{e}}^{{{1}-\frac{{3}}{{t}}}}-{1}\right)}.