Elementary Differential Equations

Linear Equations

Discussion

We solved exact equations by assuming there was a solution and working backwards. But there will be a solution to any equation we can solve, and it is a theorem that all (reasonable) first order equations can be solved. So why aren't all equations exact? Consider the following example. $$ \begin{align} (x^2+y)+(x-\sin(y))\frac{dy}{dx}&=0\qquad\text{is exact} \\ \frac{x^2+y}{x}+\frac{x-\sin(y)}{x}\frac{dy}{dx}&=0\qquad\text{is not exact} \end{align} $$ There is no real difference between the two equations, we have just divided through by $x$ to obtain the second equation from the first. The equations have the same solutions. But when we test for exactness, we find that only the first equation is exact. The reason most first order equations aren't exact is that some key factor has been divided out. One way to solve the equation would be to find the factor, called an integrating factor, and put it back in. Unfortunately, finding integrating factors can be extremely difficult. There is one situation where it is fairly straightforward, when the equation is linear. A first order differential equation is said to be linear if it can be written in the form $$ y' + p(x)y = q(x). $$ In any linear equation, we can find an integrating factor which depends only on the independent variable ($x$ in this case) as follows $$ \mu(x) = e^{\int p(x)\,dx} $$ If we multiply through the linear equation by $\mu(x)$ we obtain $$ \mu(x)y'(x) + \mu(x)p(x)y(x) = \mu(x)q(x) $$ Now $\mu'(x) = \mu(x)p(x)$ from the way we defined $\mu(x)$, so we can now write our equation as $$ \mu(x)y'(x) + \mu'(x)y(x) = \mu(x)q(x) $$ or (applying the product rule in reverse) $$ \frac{d}{dx}(\mu(x)y(x)) = \mu(x)q(x) $$ Now we integrate both sides to obtain $$ \mu(x)y(x) = \int \mu(x)q(x)\,dx $$ so $$ y(x) = \frac{1}{\mu(x)}\int \mu(x)q(x)\,dx $$ WARNING: You can't cancel the $\mu(x)$ inside the integral by the $\mu(x)$ outside the integral. This formula is confusing because we are using $x$ as both the dummy variable of integration and as the independent variable of the resulting function. Because of this it is often preferable to write the solution as $$ y(x)=\frac{1}{\mu(x)}\left(\int_a^x \mu(s)q(s)\,ds + C\right) \tag 1$$ In this form we use $s$ as the variable of integration and use $x$ as the upper limit of integration to show we want to treat the result as a function of $x$. The $a$ as the lower limit of integration is an arbitrary constant as is $C$, the constant of integration. This seems to give us two arbitrary constants, but we really only have one "degree of freedom." That is, for any choice of $a$, we can choose $C$ to give us any solution to the equation. This is discussed in more detail in Example 1 below.

Linear equations won't have singular solutions, so the general solution gives all the solutions. You may be wondering about why I left out the constant of integration when computing $\mu(x)$. I could have included a constant of integration, which would give me an infinite family of integrating factors. But I don't need an infinite family of integrating factors to solve the problem; I only need one integrating factor. So I can ignore that constant. We are now ready to give the paradigm. While it is possible to just memorize the formula for the solution, I prefer to work through the whole process. While that takes longer, I find I am less likely to make mistakes if I go through the details. You are welcome to use the paradigm or the formula (1) depending on which you find works best for you.

Paradigm

Solve $$dy/dx+2y/x=4$$

Step 1: Find the integrating factor $$ \mu(x)=e^{\int 2/x\,dx}=e^{2\log(x)}=x^2 $$ Step 2: Multiply through by the integrating factor $$ x^2\frac{dy}{dx}+2xy=4x^2 $$ Step 3: Recognize the left hand side as the derivative of $\mu y$. $$ \frac{d}{dx}(x^2y)=4x^2 $$ Step 4: Integrate both sides $$ x^2y=\int 4x^2\,dx=(4/3)x^3+C $$ Step 5: Solve for $y$. $$ y(x)=(4/3)x+Cx^{-2} $$

Example 1: Using the formula

$$y'+2y=e^x$$ Plugging this into formula (1) gives the following: $$ \begin{align} \mu(x)&=e^{\int2\,dx}=e^{2x} \\ y(x)&=\frac{\int_a^x e^{2s}e^s\,ds+C}{e^{2x}}=\frac{\int_a^x e^{3s}\,ds+C}{e^{2x}} \\ y(x)&=\frac{e^{3x}/3-e^{3a}/3+C}{e^{2x}} \\ y(x)&=e^x/3+\tilde{C}e^{-2x}\qquad(\tilde{C}=C-e^{3a}/3) \end{align} $$ Note that the choice of $a$ for the lower limit of integration doesn't matter. It all gets sucked into the arbitrary constant in the end. That said, Example 2 will illustrate a situation where choosing the correct value of $a$ can make things easier.

Most students find using the formula as in this example shorter than going through the details in the paradigm. So why don't I always use the formula? Because sometimes I misremember part of the formula or get mixed up in what goes where when I have a problem that arises in a situation with multiple variables as in Example 3 below. If I go through the steps of the paradigm, I find it easier to keep everything straight. But either approach is acceptable.

EXAMPLE 2: An integral you can't simplify

Solve the initial value problem $$dy/dx+2xy=1, y(1)=2$$.

FIRST: Find the general solution.

Step 1: $$\displaystyle\mu(x)=e^{\int 2x\,dx}=e^{x^2}$$

Step 2: $$\displaystyle e^{x^2}dy/dx+2xe^{x^2}y=e^{x^2}$$

Step 3: $$\displaystyle \frac{d}{dx}(e^{x^2}y)=e^{x^2}$$

Step 4: $$\displaystyle e^{x^2}y=\int e^{x^2}\,dx=???$$

Unfortunately, I don't know the indefinite integral of $e^{x^2}$. So I leave the integral as a definite integral with $x$ as the upper limit to give me a function of $x$ and $1$ as the lower limit because the initial value is given at $1$ (you'll see why that matters in a second). And I need to be sure not to forget the constant of integration.

Step 4: (Take Two) $$\displaystyle e^{x^2}y=\int_1^x e^{s^2}\,ds+C$$

Step 5: $$\displaystyle y(x)=e^{-x^2}\int_1^x e^{s^2}\,ds+Ce^{-x^2}$$

SECOND: Solve the initial value problem by plugging in. $$ \begin{align} y(1)=e^{-1^2}\int_1^1 e^{s^2}\,ds+Ce^{-1^2} &{\buildrel \text{set}\over =} 2 \\ C&=2e \end{align} $$ While I don't know the indefinite integral of $e^{s^2}$, I do know that the integral of anything from $1$ to $1$ is $0$. That is the advantage of choosing the lower limit to be the same as the place where the initial value is given. So the final answer is $$y(x)=e^{-x^2}\int_1^x e^{s^2}\,ds+2e^{1-x^2}$$

You may be wondering if all this was worth it, since we end up with an integral instead of a nice formula. But if we want specific values, numerical integration routines are much faster and more accurate than approximating a solution to a differential equation numerically. And if what you need is qualitative information, we also have better tools for qualitatively analyzing integrals than differential equations. So getting the answer into integral form is in fact helpful.

Example 3: A Mixing Problem

Suppose we have an ideal two tank CSTR (continuous stirred-tank reactor) in series. Fluid enters the first tank and then flows from the first tank to the second tank and then out the second tank (this is useful for example if you wish to have different conditions at different stages in a sequence of reactions). We suppose the tanks each hold 100 liters and the fluid is flowing at a rate of 2 liters/minute. Suppose a contaminant is introduced to the fluid prior to entering the first tank, at a concentration of 4 mg/liter. How long after the contaminant is first introduced will it take for the level of contamination in each tank to reach 2 mg/liter? (Note: since this is an "ideal" model, we assume perfect mixing and steady flow).

Let $Z_1$ and $Z_2$ be the amounts of contaminant in each tank in milligrams and let $t$ be the time in minutes. In this situation, the volume in the tank is a constant 100 liters, so the concentration is $Z_i/100$. Hence the concentration of 2 mg/l is equivalent to Z=200. Now just as the fluid flows from the first tank to the second, we will first solve for $Z_1$ and then use this to solve for $Z_2$. Contaminant is being added to the first tank at a rate of $4 mg/l \times 2 l/min = 8 mg/min$. Contaminant is leaving the first tank (and flowing into the second tank) at a rate of $Z_1/100 mg/l \times 2 l/min = Z/50 mg/min$. So our differential equation becomes $$\frac{dZ_1}{dt}=8-\frac{Z_1}{50}.$$ This is both a linear equation and also a separable equation. Solving using either technique yields $$Z_1 = 400-C_1\exp(-t/50).$$ Assuming no initial contamination, $Z_1(0)=0$, so $C_1=400$ and the amount of contaminant in tank 1 is $$Z_1(t)=400(1-\exp(-t/50)).$$ Then the contamination will reach 2mg/liter when $Z_1=200$, and solving $200 = 400(1-\exp(-t/50))$ gives us $t=50\log(2)\approx 34.66min$. In 34 minutes, 272 mg of contaminant have flowed into the tank, and since some also flows out, the seems to be a reasonable answer.

Now we use our formula for $Z_1$ to determine how $Z_2$ increases. The amount of contaminant flowing into tank 2 is $Z_1(t)/100 mg/l \times 2 l/min = 8(1-\exp(-t/50)) mg/min$. The amount of contaminant flowing out of tank 2 is $Z_2/100 mg/l \times 2 l/min = Z_2/50 mg/min$. So our equation for $Z_2$ becomes $$\begin{align} \frac{dZ_2}{dt}= 8(1-\exp(-t/50)) - Z_2/50, \text{ or } \\ \frac{dZ_2}{dt}+\frac{1}{50}Z_2 = 8-8\exp(-t/50). \end{align}$$ This is a linear equation (and this time is not separable). We solve it as follows:

Step 1: $$\mu(t) = \exp\left(\int \frac{dt}{50}\right) = \exp(t/50)$$ Step 2: $$\exp(t/50)\frac{dZ_2}{dt}+\frac{exp(t/50)}{50}Z_2 = 8\exp(t/50)-8$$ Step 3: $$\frac{d(\exp(t/50)Z_2}{dt}=8\exp(t/50)-8$$ Step 4: $$\begin{align} \exp(t/50)Z_2 &= \int 8\exp(t/50) - 8 dt \\ &=400\exp(t/50) - 8t + C_2 \end{align}$$ Step 5: $$Z_2=400 - (8t+C_2)\exp(-t/50)$$ Assuming no initial contamination, $Z_2(0)=0$ implies $C_2=400$ and we get the amount of contaminant in tank 2 is $$Z_2=400\left(1-(1+t/50)\exp(-t/50)\right).$$ This is actually a somewhat messy equation to solve for $Z_2(t)=200$ because of the $t\exp(-t/50)$ term. But numerically we find $t \approx 83.9 min$. So it takes more than twice as long for the contamination to reach that level in the second tank as in the first tank.

Example 4: Linear For $x$ As A Function Of $y$

Find the general solution to $\displaystyle\frac{dy}{dx}=\frac{1}{x+y}$.
This is not a linear equation if we think of $y$ in terms of $x$. But in differential equations, we often think of $x$ and $y$ each as their own variable and not just of $y$ as a function of $x$ (this attitude is also quite useful in Physics and Chemistry when you are dealing with formulas like $PV=nRT$ where each variable depends on the others). Look what happens if we think of $x$ as a function of $y$ in the equation above. $$\begin{align} \frac{dy}{dx}&=\frac{1}{x+y} \\ (x+y)dy &= dx \\ x+y &= \frac{dx}{dy} \\ y = \frac{dx}{dy} - x. \end{align}$$ Now we do have a linear equation. While this may seem like more of the sleight of hand we used in dealing with separable equations, it can be fully justified, either by using the inverse function theorem from Calculus 3 or just by looking for an integrating factor which depends only on $y$ for $(x+y)dy - dx = 0$. Now that we have our linear equation in $x$ as a function of $y$, we use the paradigm above, just being careful which letters we use in each step.

Step 1: $$\mu(y)=\exp\left(\int -1\,dy\right) = e^{-y}$$ Step 2: $$e^{-y}\frac{dx}{dy} - xe^{-y} = ye^{-y}$$ Step 3: $$\frac{d}{dy}\left(xe^{-y}\right) = ye^{-y}$$ Step 4: $$xe^{-y} = \int ye^{-y} = -ye^{-y} - e{-y} + C$$ Step 5: $$x = -y-1 + Ce^y\text{, or } x+y+1=Ce^y$$

Randomly Generated Practice Problems

A randomly generated practice problem is below. Once you submit an answer, you will have a link to see the detailed solution for the problem, as well as the option to generate a new problem. Note that you can just quickly enter a "dummy" answer like $y=x$ to quickly get to the detailed solution if that is all you want to see.

Kansas State University

Mathematics Department