First-Order Linear Equations

Definition

A first order linear equation is of the form:

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

Where $a$ and $b$ are continuous functions.

Homogeneous First Order Linear Equation

If $b(t)=0$, we will say that the equation is homogeneous.

$$\frac{dy}{dt}+a(t)y=0$$

Initial Condition (Initial Value Problems - IVPs)

Sometimes, there will be an extra equation that we will call the initial condition. Typically of the form:

$$y(t_0)=y_0$$

Where $y(t)$ is the solution function. The subscript ${\text{var}}_0$ just signifies the initial condition or state.

Solving First-Order Linear Equations

Given $a(t)$, we define a new function called the integrating factor. We have:

$\mu (t)=\exp \left[\int a(t)dt\right]$

Where $\int a(t)dE$ is any antiderivative of $a(t)$

Remark: The exponential function is $\exp (Z)=e^Z$, thus $\mu (t)=\exp \left[\int a(t)dt\right]=e^{\int a(t)dt}$

Remark: If we have an initial condition $y(t_0)=y_0$, it is convenient to use

a(s)ds$$ *** If $u=\int a(t)dt$ then $\mu(t)=e^u$ and $\frac{d\mu}{dt}=\frac{d}{dt}e^u=e^u \frac{du}{dt}$ We conclude $$ \frac{du}{dt=\mu(t_{1}a(t))}

If $y(t)$ satisfies

$$\frac{d}{dt}(\mu (t)y(t))=\frac{d\mu }{dt}y(t)+\mu (t)\frac{dy}{dt}$$

After substitution, we find

$$\begin{array}{r}\frac{d}{dt}(\mu (t)y(t))=\mu (t)a(t)y(t)+\mu (t)[-a(t)y(t)+b(t)]\\ \\ =\end{array}$$

Useful Antiderivatives

$$\begin{array}{r}\int e^{at}\cos (bt)dt=\frac{e^{at}}{a^2+b^2}(a\cos (bt)+b(\sin (bt)))\\ \\ \int e^{at}\sin (bt)dt=\frac{e^{at}}{a^2+b^2}(a\sin (bt)-b(\cos (bt)))\\ \\ \text{where a and b are constants}\end{array}$$

Example

Find the general solution of

$$\frac{dy}{dt}+2y=10\sin (t)$$

There is no initial value, so the solution will include a constant.

We use $a(t)=2$ and $b(t)=10\sin (t)$.

The integrating factor is

$$\mu (t)=\exp \left[\int a(t)dt\right]=\exp [2t]$$ $$\frac{d}{dt}(e^{2t}y(t))=10e^{2t}\sin (t)$$

This implies that

$$\begin{array}{r}{e}^{2t}y(t)=\int 10e^{2t}\sin (t)dt\\ =10\int e^{2t}\sin (t)dt\\ =10\frac{e^{2t}}{4+1}(2\sin (t)-\cos (t))\\ \\ =2e^{2t}(2\sin (t)-\cos (t))\\ \\ 4e^{2t}\sin (t)-2e^{2t}\cos (t)+C\end{array}$$

Example with DE given in correct form

Solve

$$\frac{dy}{dt}+2y=5e^{3t}\text{ with }y(0)=-10$$

In this case, $a(t)=2$, $b(t)=5e^{3t}$.

We have an initial condition $y(t_0{)}_{=}{y}_0$ where $t_0=0$ and $y_0=-10$.

The integrating factor is:

$$\begin{array}{r}\mu (t)=\exp \left[\int a(t)dt\right]=\exp \left[{\int }_{2}dt\right]\\ =\exp [2t]=e^{2t}\end{array}$$

We use the formula

$$\frac{d}{dt}(\mu (t)y(t))=\mu (t)b(t)$$

In our case,

$$\frac{d}{dt}(e^{2t}y(t))=e^{2t}5e^{3t}=5e^{5t}$$

Find antiderivatives:

$$e^{2t}y(t)=\int 5e^{5t}dt=e^{5t}+C$$

We have that $e^{2t}y(t)=e^{5t}+C$. We need to find $C$, so we can use the initial condition from earlier $y(0)=-10$. Substitute $t=0$ into the equation we have.

Thus:

$$\begin{array}{r}y(0)=1+C=-10\\ C=-11\end{array}$$

We can now substitute $C$ into the equation, and get $e^{2t}y(t)=e^{5t}-11$. Solving for $y(t)$, we divide by $e^{2t}$ and get:

$$y(t)=e^{3t}-11e^{-2t}$$

This is the solution to the differential equation.

We can verify the solution by computing the left-hand side (LHS) and compare with the right-hand side (RHS) of the equation.

We have:

$$\begin{array}{r}\frac{dy}{dt}+2y=3e^{3t}+22e^{-2t}+2(e^{3t}-11e^{-2t})\\ =3e^{3t}+2e^{3t}=5e^{3t}\end{array}$$

In addition, $y(0)=1-11(1)=-10$ which satisfies the initial condition.

Example with DE given in incorrect form

Solve

$$t\frac{dy}{dt}-3y=10t^7\text{ with }y(1)=6$$

This is not in the form of our expected pattern, so we need to divide by $t$ to get that.

$$\frac{dy}{dt}-\frac{3y}{t}=10t^6$$

In this case, $a(t)=-\frac{3}{t}$, $b(t)=10t^6$

The integrating factor is $\mu (t)=\exp \left[\int a(t)dt\right]=\exp \left[-3\int \frac{1}{t}dt\right]$

We have the formula:

$$\frac{d}{dt}(\mu (t)y(t))=\mu (t)b(t)$$

In this case, $\frac{d}{dt}(t^{-3}y(t))=t^{-3}10t^6=10t^3$. Now we can find the antiderivatives:

$$\begin{array}{r}{t}^{-3}y(t)=\int 10t^{3}dt=\frac{5}{2}{t}^4+C\end{array}$$

Now find the constant $C$ using the initial condition $y(1)=6$:

$$\begin{array}{r}y(1)=\frac{5}{2}+C=6\\ C=6-\frac{5}{2}=\frac{7}{2}\end{array}$$

Substitute this value back into the solution, and we get:

$$\begin{array}{r}{t}^{-3}y(t)=\frac{5}{2}{t}^4+\frac{7}{2}\end{array}$$

Solving for $y(t)$,

$$y(t)=\frac{5}{2}{t}^7+\frac{7}{2}{t}^3$$