First Order Linear Differential Equation

A Linear Differential Equation of Order 1. We are assuming it to be a Ordinary Differential Equation.

$$\frac{dy}{dt}+a(t)y=b(t)$$

Techniques

Integrating Factor

If and only if The given equation is a First Order Linear Differential Equation and the coefficient of the first order derivative is 1, i.e. in the given form

$$\frac{dy}{dt}+a(t)y=b(t)$$

Then it is possible to simplify the process of solving the equation by multiplying both sides of the equation by some function $M(t)$

$$M(t)\frac{dy}{dt}+M(t)a(t)y=M(t)b(t)$$

Then let $M^{\prime }(t)=M(t)a(t)$, and you now have a product rule that when integrated and after some algebra gives an explicit solution for $y$ so long as you can solve for whatever $M(t)$ needs to be.

Example

$$y^{\prime }-\frac{3}{4}y=6\exp \left(-\frac{5}{4}t\right)$$ $$M(t)y^{\prime }-M(t)\frac{3}{4}y=M(t)6\exp \left(-\frac{5}{4}t\right)$$ $$\text{Let }{M}^{\prime }(t)=\left(-M(t)\frac{3}{4}\right)⟹(M(t)y{)}^{\prime }=M(t)y^{\prime }-M^{\prime }(t)y=M(t)6\exp \left(-\frac{5}{4}t\right)$$ $$M(t)y=\int M(t)6\exp \left(-\frac{5}{4}t\right)dt$$ $$\frac{M^{\prime }}{M}=-\frac{3}{4}$$ $$\int \frac{M^{\prime }}{M}dt=-\int \frac{3}{4}dt$$ $$\int \frac{dM}{M}=-\int \frac{3}{4}dt$$ $$\ln (M)=-\frac{3}{4}t+C$$ $$M(t)=\exp \left(-\frac{3}{4}t+C\right)$$ $$\boxed{M(t)=D\exp \left(-\frac{3}{4}t\right)}$$ $$\left(D\exp \left(-\frac{3}{4}t\right)\right)y=\int \left(D\exp \left(-\frac{3}{4}t\right)\right)6\exp \left(-\frac{5}{4}t\right)dt$$ $$\exp \left(-\frac{3}{4}t\right)y=6\int \exp \left(-2t\right)dt$$ $$\exp \left(-\frac{3}{4}t\right)y=-3\exp \left(-2t\right)+K$$ $$\exp \left(-\frac{3}{4}t\right)y=-3\exp \left(-2t\right)+K$$ $$\boxed{y(t)=-3\exp \left(-\frac{5}{4}t\right)+K\exp \left(\frac{3}{4}t\right)}$$