Differential Equations - First Order: Linear - ii

Example 4 Find the solution to the following IVP.
$$t{y}^{\prime }+2y=t^2-t+1y\left(1\right)=\frac{1}{2}$$

First, divide through by the t to get the differential equation into the correct form.
$$y^{\prime }+\frac{2}{t}y=t-1+\frac{1}{t}$$Now let’s get the integrating factor, $\mu \left(t\right)$.
$$\mu \left(t\right)=e^{\int \frac{2}{t}dt}=e^{2\ln \left|t\right|}$$Now, we need to simplify $\mu \left(t\right)$. However, we can’t use $\text{(11)}$ yet as that requires a coefficient of one in front of the logarithm. So, recall that
$$\ln x^r=r\ln x$$and rewrite the integrating factor in a form that will allow us to simplify it.
$$\mu \left(t\right)=e^{2\ln \left|t\right|}=e^{\ln {\left|t\right|}^2}={\left|t\right|}^2=t^2$$We were able to drop the absolute value bars here because we were squaring the $t$, but often they can’t be dropped so be careful with them and don’t drop them unless you know that you can. Often the absolute value bars must remain.
Now, multiply the rewritten differential equation (remember we can’t use the original differential equation here…) by the integrating factor.
$${\left(t^{2}y\right)}^{\prime }=t^3-t^2+t$$Integrate both sides and solve for the solution.
$$\begin{array}{cc}{t}^{2}y& =\int t^3-t^2+tdt\\ & =\frac{1}{4}{t}^4-\frac{1}{3}{t}^3+\frac{1}{2}{t}^2+c\\ y\left(t\right)& =\frac{1}{4}{t}^2-\frac{1}{3}t+\frac{1}{2}+\frac{c}{t^2}\end{array}$$Finally, apply the initial condition to get the value of $c$.
$$\frac{1}{2}=y\left(1\right)=\frac{1}{4}-\frac{1}{3}+\frac{1}{2}+c\Rightarrow c=\frac{1}{12}$$The solution is then,
$$y\left(t\right)=\frac{1}{4}{t}^2-\frac{1}{3}t+\frac{1}{2}+\frac{1}{12t^2}$$Here is a plot of the solution.

Example 5 Find the solution to the following IVP.
$$t{y}^{\prime }-2y=t^5\sin \left(2t\right)-t^3+4t^{4}y\left(\pi \right)=\frac{3}{2}{\pi }^4$$

First, divide through by $t$ to get the differential equation in the correct form.
$$y^{\prime }-\frac{2}{t}y=t^4\sin \left(2t\right)-t^2+4t^3$$Now that we have done this we can find the integrating factor, $\mu \left(t\right)$.
$$\mu \left(t\right)=e^{\int -\frac{2}{t}dt}=e^{-2\ln \left|t\right|}$$Do not forget that the “-” is part of $p\left(t\right)$. Forgetting this minus sign can take a problem that is very easy to do and turn it into a very difficult, if not impossible problem so be careful!
Now, we just need to simplify this as we did in the previous example.
$$\mu \left(t\right)=e^{-2\ln \left|t\right|}=e^{\ln {\left|t\right|}^{-2}}={\left|t\right|}^{-2}=t^{-2}$$Again, we can drop the absolute value bars since we are squaring the term.
Now multiply the differential equation by the integrating factor (again, make sure it’s the rewritten one and not the original differential equation).
$${\left(t^{-2}y\right)}^{\prime }=t^2\sin \left(2t\right)-1+4t$$Integrate both sides and solve for the solution.
$$\begin{array}{cc}{t}^{-2}y\left(t\right)& =\int t^2\sin \left(2t\right)dt+\int -1+4tdt\\ t^{-2}y\left(t\right)& =-\frac{1}{2}{t}^2\cos \left(2t\right)+\frac{1}{2}t\sin \left(2t\right)+\frac{1}{4}\cos \left(2t\right)-t+2t^2+c\\ y\left(t\right)& =-\frac{1}{2}{t}^4\cos \left(2t\right)+\frac{1}{2}{t}^3\sin \left(2t\right)+\frac{1}{4}{t}^2\cos \left(2t\right)-t^3+2t^4+c{t}^2\end{array}$$Apply the initial condition to find the value of $c$.
$$\begin{array}{cc}\frac{3}{2}{\pi }^4=y\left(\pi \right)& =-\frac{1}{2}{\pi }^4+\frac{1}{4}{\pi }^2-{\pi }^3+2{\pi }^4+c{\pi }^2=\frac{3}{2}{\pi }^4-{\pi }^3+\frac{1}{4}{\pi }^2+c{\pi }^2\\ {\pi }^3-\frac{1}{4}{\pi }^2& =c{\pi }^2\\ c& =\pi -\frac{1}{4}\end{array}$$The solution is then
$$y\left(t\right)=-\frac{1}{2}{t}^4\cos \left(2t\right)+\frac{1}{2}{t}^3\sin \left(2t\right)+\frac{1}{4}{t}^2\cos \left(2t\right)-t^3+2t^4+\left(\pi -\frac{1}{4}\right)t^2$$Below is a plot of the solution.

Let’s work one final example that looks more at interpreting a solution rather than finding a solution.

Example 6 Find the solution to the following IVP and determine all possible behaviors of the solution as $t\to \infty$. If this behavior depends on the value of $y_0$ give this dependence.$$2y^{\prime }-y=4\sin \left(3t\right)y\left(0\right)=y_0$$

First, divide through by a 2 to get the differential equation in the correct form.
$$y^{\prime }-\frac{1}{2}y=2\sin \left(3t\right)$$Now find $\mu \left(t\right)$.
$$\mu \left(t\right)=e^{\int -\frac{1}{2}dt}=e^{-\frac{t}{2}}$$Multiply $\mu \left(t\right)$through the differential equation and rewrite the left side as a product rule.
$${\left(e^{-\frac{t}{2}}y\right)}^{\prime }=2e^{-\frac{t}{2}}\sin \left(3t\right)$$Integrate both sides (the right side requires integration by parts – you can do that right?) and solve for the solution.
$$\begin{array}{cc}{e}^{-\frac{t}{2}}y& =\int 2e^{-\frac{t}{2}}\sin \left(3t\right)dt+c\\ e^{-\frac{t}{2}}y& =-\frac{24}{37}{e}^{-\frac{t}{2}}\cos \left(3t\right)-\frac{4}{37}{e}^{-\frac{t}{2}}\sin \left(3t\right)+c\\ y\left(t\right)& =-\frac{24}{37}\cos \left(3t\right)-\frac{4}{37}\sin \left(3t\right)+c{e}^{\frac{t}{2}}\end{array}$$Apply the initial condition to find the value of $c$ and note that it will contain $y_0$ as we don’t have a value for that.
$$y_0=y\left(0\right)=-\frac{24}{37}+c\Rightarrow c=y_0+\frac{24}{37}$$So, the solution is
$$y\left(t\right)=-\frac{24}{37}\cos \left(3t\right)-\frac{4}{37}\sin \left(3t\right)+\left(y_0+\frac{24}{37}\right)e^{\frac{t}{2}}$$Now that we have the solution, let’s look at the long term behavior (i.e. $t\to \infty$) of the solution. The first two terms of the solution will remain finite for all values of $t$. It is the last term that will determine the behavior of the solution. The exponential will always go to infinity as $t\to \infty$, however depending on the sign of the coefficient $c$ (yes we’ve already found it, but for ease of this discussion we’ll continue to call it $c$). The following table gives the long term behavior of the solution for all values of $c$.

Range of $c$	Behavior of solution as$t\to \infty$
$c$ < 0	$y\left(t\right)\to -\infty$
$c$ = 0	$y\left(t\right)$ remains finite
$c$ > 0	$y\left(t\right)\to \infty$

This behavior can also be seen in the following graph of several of the solutions.

Now, because we know how $c$ relates to $y_0$ we can relate the behavior of the solution to $y_0$. The following table give the behavior of the solution in terms of $y_0$ instead of $c$.

Range of $y_0$	Behavior of solution as$t\to \infty$
$y_0<-\frac{24}{37}$	$y\left(t\right)\to -\infty$
$y_0=-\frac{24}{37}$	$y\left(t\right)$ remains finite
$y_0>-\frac{24}{37}$	$y\left(t\right)\to \infty$

Note that for $y_0=-\frac{24}{37}$ the solution will remain finite. That will not always happen.

Investigating the long term behavior of solutions is sometimes more important than the solution itself. Suppose that the solution above gave the temperature in a bar of metal. In this case we would want the solution(s) that remains finite in the long term. With this investigation we would now have the value of the initial condition that will give us that solution and more importantly values of the initial condition that we would need to avoid so that we didn’t melt the bar.