Since second-term (integral) calculus, we have been solving differential equations, we just didn't know it.
Integrals as Solutions
Consider the differential equation $$y'=f(x).$$ The general solution to this equation $y(x)$ is nothing more than the family of antiderivatives of $f(x).$
That is, $$ y=\int f(x)\, dx $$
Example
Find the general solution to the differential equation $$ \frac{dy}{dx}=\sin(5x) $$ and then satisfy the initial condition $y(0)=2.$
The general solution is
$$
y=\int \sin(5x)\, dx=-\frac{1}{5}\cos(5x)+C
$$
Satisfying the initial condition,
$$
\begin{array}{lll}
&\displaystyle y(0)=2&\mbox{}\\
\implies &\displaystyle -\frac{1}{5}\cos(5\cdot 0)+C=2&\mbox{}\\
\implies &\displaystyle -\frac{1}{5}+C=2&\mbox{}\\
\implies &\displaystyle C=\frac{11}{5}.&\mbox{}\\
\end{array}
$$
Thus, the solution which satisfies the initial condition is
$$
y=-\frac{1}{5}\cos(5x)+\frac{11}{5}
$$
Expressing Solutions as Definite Integrals
As we have likely seen, some functions do not have elementary antiderivatives.
A classic example is $f(x)=e^{-x^2}.$
Question: How do we write the solution of the equation below? $$ \frac{dy}{dx}=e^{-x^2} $$
Expressing Solutions as Definite Integrals
One answer is that we could simply write $$ y=\int e^{-x^2}\,dx $$ Another Question: How do we work with this notation when satisfying an initial condition?
For example, try satisfying the initial condition $y(0)=1.$
Expressing Solutions as Definite Integrals
When solving differential equations, a better notation for an antiderivative of $f(x)=e^{-x^2}$ is $$ y(x)=\int_{x_0}^{x} e^{-t^2}\,dt+C $$ The above is justified by Fundamental Theorem of calculus which tells us that $$ y'(x)=\frac{d}{dx}\left(\int_{x_0}^{x} e^{-t^2}\,dt+C\right)=e^{-x^2} $$
Expressing Solutions as Definite Integrals
Expressing the general solution as a definite integral $$ y(x)=\int_{x_0}^{x} e^{-t^2}\,dt+C $$ allows us to easily satisfy an initial condition, say $y(x_0)=y_0.$ $$ \begin{array}{lll} &\displaystyle y(x_0)=1&\mbox{}\\ \implies &\displaystyle \int_{x_0}^{x_0} e^{-t^2}\,dt+y_0=1&\mbox{}\\ \implies &\displaystyle y_0=1&\mbox{}\\ \end{array} $$
Expressing Solutions as Definite Integrals
Thus $$ y(x)=\int_{0}^{x} e^{-t^2}\,dt+1 $$ satisfies $$ \frac{dy}{dx}=e^{-x^2} $$ with the initial condition $y(0)=1.$
Expressing Solutions as Definite Integrals
Another advantage of expressing the solution to the above as a definite integral $$ y(x)=\int_{0}^{x} e^{-t^2}\,dt+1 $$ is that we may use approximation techniques like Simpson's Method for evaluating $\displaystyle \int_{0}^{x} e^{-t^2}\,dt.$
In practice, the above situation occurs quite often.
To Summarize
In general, we may express the solution to the differential equation $$ y'=f(x) $$ with initial condition $y(x_0)=y_0$ as $$ y(x)=\int_{x_0}^{x} f(t)\,dt+y_0 $$
Example
Solve the initial value problem $$ \frac{dy}{dx}=y^3,\,y(0)=1. $$
We first note that $y=0$ is a solution. We now find a general solution assuming that $y$ is not identically $0.$
Treating $x$ as a function of $y,$ we may use the Inverse Function Theorem from first-term calculus and write $$ \frac{dx}{dy}=\frac{1}{y^3}=y^{-3} $$ so that $$ x=\int \frac{1}{y^3}\, dy=-\frac{1}{2}y^{-2}+C=-\frac{1}{2y^2}+C $$ We may now solve for $y.$ $$ \begin{array}{lll} &\displaystyle x=-\frac{1}{2y^2}+C&\mbox{}\\ \implies &\displaystyle \frac{1}{2y^2}=C-x&\mbox{}\\ \implies &\displaystyle \frac{1}{2y^2}=C-x&\mbox{}\\ \implies &\displaystyle 2y^2=\frac{1}{C-x}&\mbox{}\\ \implies &\displaystyle y^2=\frac{1}{2(C-x)}&\mbox{}\\ \implies &\displaystyle y^2=\frac{1}{2C-2x}&\mbox{}\\ \end{array} $$ Since $2C$ is just a constant, we shall relabel the constant again as simply $C.$ That is, $$ y^2=\frac{1}{C-2x} $$ so that our general solution is $$ y=\pm\sqrt{\frac{1}{C-2x}}=\pm\frac{1}{\sqrt{C-2x}} $$ We now satisfy the initial condition.
Since $y(0)=1\gt 0,$ we choose the positive root of the above so that $$ y=\frac{1}{\sqrt{C-2x}} $$ Then $$ \begin{array}{lll} &\displaystyle y(0)=1&\mbox{}\\ \implies &\displaystyle \frac{1}{\sqrt{C-2 \cdot 0}}=1&\mbox{}\\ \implies &\displaystyle \frac{1}{\sqrt{C}}=1&\mbox{}\\ \implies &\displaystyle \frac{1}{C}=1&\mbox{}\\ \implies &\displaystyle C=1&\mbox{}\\ \end{array} $$ Thus, our solution to the initial value problem is $$ y=\frac{1}{\sqrt{1-2x}} $$ Check
We now verify that the above is a solution to the original equation.
We see that $$ \begin{array}{lll} \displaystyle y'&\displaystyle=\frac{d}{dx}\frac{1}{\sqrt{1-2x}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{2}\left(1-2x\right)^{-3/2}\cdot(-2) &\mbox{}\\ \displaystyle &\displaystyle=\left(1-2x\right)^{-3/2} &\mbox{}\\ \displaystyle &\displaystyle=\left(\left(1-2x\right)^{-1/2}\right)^3 &\mbox{}\\ \displaystyle &\displaystyle=\left(\frac{1}{\sqrt{1-2x}}\right)^3 &\mbox{}\\ \displaystyle &\displaystyle=y^3 &\mbox{}\\ \end{array} $$
Treating $x$ as a function of $y,$ we may use the Inverse Function Theorem from first-term calculus and write $$ \frac{dx}{dy}=\frac{1}{y^3}=y^{-3} $$ so that $$ x=\int \frac{1}{y^3}\, dy=-\frac{1}{2}y^{-2}+C=-\frac{1}{2y^2}+C $$ We may now solve for $y.$ $$ \begin{array}{lll} &\displaystyle x=-\frac{1}{2y^2}+C&\mbox{}\\ \implies &\displaystyle \frac{1}{2y^2}=C-x&\mbox{}\\ \implies &\displaystyle \frac{1}{2y^2}=C-x&\mbox{}\\ \implies &\displaystyle 2y^2=\frac{1}{C-x}&\mbox{}\\ \implies &\displaystyle y^2=\frac{1}{2(C-x)}&\mbox{}\\ \implies &\displaystyle y^2=\frac{1}{2C-2x}&\mbox{}\\ \end{array} $$ Since $2C$ is just a constant, we shall relabel the constant again as simply $C.$ That is, $$ y^2=\frac{1}{C-2x} $$ so that our general solution is $$ y=\pm\sqrt{\frac{1}{C-2x}}=\pm\frac{1}{\sqrt{C-2x}} $$ We now satisfy the initial condition.
Since $y(0)=1\gt 0,$ we choose the positive root of the above so that $$ y=\frac{1}{\sqrt{C-2x}} $$ Then $$ \begin{array}{lll} &\displaystyle y(0)=1&\mbox{}\\ \implies &\displaystyle \frac{1}{\sqrt{C-2 \cdot 0}}=1&\mbox{}\\ \implies &\displaystyle \frac{1}{\sqrt{C}}=1&\mbox{}\\ \implies &\displaystyle \frac{1}{C}=1&\mbox{}\\ \implies &\displaystyle C=1&\mbox{}\\ \end{array} $$ Thus, our solution to the initial value problem is $$ y=\frac{1}{\sqrt{1-2x}} $$ Check
We now verify that the above is a solution to the original equation.
We see that $$ \begin{array}{lll} \displaystyle y'&\displaystyle=\frac{d}{dx}\frac{1}{\sqrt{1-2x}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{2}\left(1-2x\right)^{-3/2}\cdot(-2) &\mbox{}\\ \displaystyle &\displaystyle=\left(1-2x\right)^{-3/2} &\mbox{}\\ \displaystyle &\displaystyle=\left(\left(1-2x\right)^{-1/2}\right)^3 &\mbox{}\\ \displaystyle &\displaystyle=\left(\frac{1}{\sqrt{1-2x}}\right)^3 &\mbox{}\\ \displaystyle &\displaystyle=y^3 &\mbox{}\\ \end{array} $$
Position, Velocity, and Acceleration
Recall that velocity $v(t)$ is the derivative of position $s(t),$ and that acceleration $a(t)$ is the derivative of velocity $v(t).$
In several cases we were given a position function $s(t)$ and found the velocity function as $v(t)=s'(t)$ by taking a derivative, of the acceleration function $a(t)=v'(t)=s''(t)$ as the derivative of velocity or the second derivative of position.
What if we were given the velocity function or the acceleration function but not position?
How would we recover the position from the velocity or acceleration functions?
Answer: By solving an initial value problem! :D
Example
According to nacto.org, "if a street surface is dry, the average driver can safely decelerate an automobile or light truck with reasonably good tires at the rate of about $15$ feet per second."
Suppose a car is being driven at a rate of $45$ mph when the brakes are applied. The car decelerates at a constant rate of $15$ ft/sec2.
How long does it take for the car to stop?
We shall model the situation with a differential equation.
Since acceleration is the derivative of the velocity function $v(t),$ a constant deceleration of $15$ feet per second is modelled by $$\displaystyle \frac{d v}{dt}=-15$$ Since the velocity at the moment the brakes we applied (when the clock starts) is $45$ miles per hour, we have an initial condition.
However, we must state the velocity in feet per second: $$ \frac{45 \mbox{ mi}}{1 \mbox{ hr}}\cdot \frac{ 5280\mbox{ ft}}{ 1 \mbox{ mi}}\cdot \frac{ 1 \mbox{ hr}}{ 3600 \mbox{ sec}}=\frac{66 \mbox{ ft}}{1 \mbox{ sec}} $$ Thus, our initial condition is $v(0)=66.$
We will now solve the above initial value problem. First we find a general solution to the differential equation by finding an antiderivative: $$v(t)=\int (-15) \,dt =-15t+C.$$ Imposing the initial condition we have $$ \begin{array}{lll} &\displaystyle v(0)=66 & \\ \implies & \displaystyle-15\cdot 0+C =66 & \mbox{}\\ \implies & \displaystyle C =66 & \mbox{}\\ \end{array} $$ Thus, our velocity function is $v(t)=-15t+66.$
The car stops when the velocity is $0.$ Thus, we set $v(t)=0$ which gives $-15t+66=0.$
So, the car stops after $t=4.4$ seconds.
Since acceleration is the derivative of the velocity function $v(t),$ a constant deceleration of $15$ feet per second is modelled by $$\displaystyle \frac{d v}{dt}=-15$$ Since the velocity at the moment the brakes we applied (when the clock starts) is $45$ miles per hour, we have an initial condition.
However, we must state the velocity in feet per second: $$ \frac{45 \mbox{ mi}}{1 \mbox{ hr}}\cdot \frac{ 5280\mbox{ ft}}{ 1 \mbox{ mi}}\cdot \frac{ 1 \mbox{ hr}}{ 3600 \mbox{ sec}}=\frac{66 \mbox{ ft}}{1 \mbox{ sec}} $$ Thus, our initial condition is $v(0)=66.$
We will now solve the above initial value problem. First we find a general solution to the differential equation by finding an antiderivative: $$v(t)=\int (-15) \,dt =-15t+C.$$ Imposing the initial condition we have $$ \begin{array}{lll} &\displaystyle v(0)=66 & \\ \implies & \displaystyle-15\cdot 0+C =66 & \mbox{}\\ \implies & \displaystyle C =66 & \mbox{}\\ \end{array} $$ Thus, our velocity function is $v(t)=-15t+66.$
The car stops when the velocity is $0.$ Thus, we set $v(t)=0$ which gives $-15t+66=0.$
So, the car stops after $t=4.4$ seconds.
How far will the car travel from the time the brakes are applied to the time the vehicle stops?
To answer the question, we now need to find the position function
$s(t).$ Since velocity is the derivative of position, we may use the above solution to write another differential equation:
$$\frac{ds}{dt}=-15t+66.$$
We will say the $s(t)$ is the distance the car travels after the brakes are applied so that our initial condition is
$s(0)=0.$
Again, we have a differential equation and an initial condition, that is, an initial value problem.
And again, we find a general solution to the differential equation by finding an anti derivative: $$s(t)=\int (-15t+66) \,dt =-\frac{15}{2}t^2 +66 t + C.$$ Imposing the initial condition we have $$ \begin{array}{lll} &\displaystyle s(0)=0 & \\ \implies & \displaystyle -\frac{15}{2}0^2 +66 \cdot 0 + C=0 & \mbox{}\\ \implies & \displaystyle C =0 & \mbox{}\\ \end{array} $$ Thus, our position function is $\displaystyle s(t)=-\frac{15}{2}t^2 +66 t.$ Since the car will travel $t=4.4$ seconds before stopping, the car will go $\displaystyle s(4.4)=-\frac{15}{2}(4.4)^2 +66 \cdot 4.4= 145.2$ feet before stopping.
Again, we have a differential equation and an initial condition, that is, an initial value problem.
And again, we find a general solution to the differential equation by finding an anti derivative: $$s(t)=\int (-15t+66) \,dt =-\frac{15}{2}t^2 +66 t + C.$$ Imposing the initial condition we have $$ \begin{array}{lll} &\displaystyle s(0)=0 & \\ \implies & \displaystyle -\frac{15}{2}0^2 +66 \cdot 0 + C=0 & \mbox{}\\ \implies & \displaystyle C =0 & \mbox{}\\ \end{array} $$ Thus, our position function is $\displaystyle s(t)=-\frac{15}{2}t^2 +66 t.$ Since the car will travel $t=4.4$ seconds before stopping, the car will go $\displaystyle s(4.4)=-\frac{15}{2}(4.4)^2 +66 \cdot 4.4= 145.2$ feet before stopping.