# Linear Differential Equations

A first-order linear differential equation has form y'+p(t)y=q(t) .

To solve it, rewrite it in the differential form: (dy)/(dt)+p(t)y=q(t) or (p(t)y-q(t))dt+dy=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) and N(t,y)=1 then (partial M)/(partial y)=p(t) and (partial N)/(partial x)=0 . So, 1/N((partial M)/(partial y)-(partial N)/(partial x))=1/1 (p(t)-0)=p(t) . Thus, integrating factor I(t,y)=e^(int p(t) dt) .

y'e^(int p(t) dt)+p(t)e^(int p(t)dt)y=q(t)e^(int p(t) dt) .

Closely looking at the left side we note that it equals d(ye^(int p(t) dt))

So, equation can be rewritten as d(ye^(int p(t) dt))=q(t)e^(int p(t) dt) .

Integrating both sides gives ye^(int p(t) dt)=int (q(t)e^(int p(t)dt))dt+C

Finally, y=e^(-int p(t) dt) int (q(t)e^(int p(t)dt))dt+Ce^(-int p(t) dt)

This formula allows to find solution directly.

Example 1. Solve y'-7y=e^(t)

Here p(t)=-7 and q(t)=e^t .

So, solution is

 y=e^(-int -7 dt) int (e^t*e^(int -7dt))dt+Ce^(-int -7 dt)=e^(7t) int (e^(-6t))dt+Ce^(7t)=e^(7t)*(-1/6)e^(-6t)+Ce^(7t)=Ce^(7t)-1/6 e^t

Thus, y=Ce^(7t)-1/6 e^t.

Example 2. Solve t^2y'-3y-1=0 , y(3)=0

First rewrite equation: y'-3/t^2 y=1/t^2 .

Here p(t)=-3/t^2 and q(t)=1/t^2 .

So,

y=e^(-int -3/t^2 dt) int (1/t^2 e^(int -3/t^2 dt))dt+Ce^(-int -3/t^2 dt)=e^(-3/t) int (1/t^2 e^(3/t))dt+Ce^(-3/t)=e^(-3/t) *(-1/3)e^(3/t)+Ce^(-3/t)=Ce^(-3/t)-1/3

y=Ce^(-3/t)-1/3 .

To find particular solution, plug in initial condition:

0=Ce^(-3/3)-1/3 or C=e/3 .

Finally, y=e/3 e^(-3/t)-1/3=1/3 (e^(1-3/t)-1)