MS-A0111 - Differential and Integral Calculus 1, Lecture, 13.9.2021-27.10.2021

The differential equation $y' = f(x,y)$ can be interpreted geometrically: if the solution curve (i.e. graph of a solution) goes through the point $(x_0, y_0)$, then it holds that $y'(x_0) = f(x_0, y_0)$, i.e. we can find the slopes of the tangents of the curve even if we do not know the solution itself. Direction field or slope field is a vector field $\vec{i} + f(x_k, y_k)\vec{j}$ drawn through the points $(x_k, y_k)$. The direction field provides a fairly accurate image of the behavior of the solution curves.

Linear 1st order ODE

If a differential equation is of the form

$p_n(x)y^{(n)} + p_{n-1}(x)y^{(n-1)} + \cdots + p_1(x)y' + p_0(x)y = r(x),$

then it is called a linear differential equation. The left side of the equation is a linear combination of the derivatives with multipliers $p_k(x)$. Thus a first order linear ODE is of the form

$p_1(x)y' + p_0(x)y = r(x).$

If $r(x) = 0$ for all $x$, then the equation is called homogeneous. Otherwise the equation is nonhomogeneous.

A first order linear ODE can be solved by using an integrating factor method. The idea of the method is to multiply both sides of the equation  $y' + p(x)y = r(x)$ by the integrating factor $\displaystyle e^{\int p(x) dx} =e^{P(x)}$, which allows the equation to be written in the form

 $\displaystyle y'(x)e^{P(x)} + p(x)e^{P(x)}y(x) = r(x)e^{P(x)} \Leftrightarrow \frac{d}{dx}\left( y(x)e^{P(x)}\right) = r(x)e^{P(x)}.$

Integrating both sides of the equation, we get

$\displaystyle y(x)e^{P(x)} = \int r(x)e^{P(x)}\, \mathrm{d}x + C  \Leftrightarrow y(x)= Ce^{-P(x)} + e^{-P(x)}\int r(x) e^{P(x)}\, \mathrm{d}x.$

It is not advisable to try to remember the formula as it is, but rather keep in mind the idea of how the equation should be modified in order to proceed.

Example 1.

Let us solve the differential equation $\displaystyle y'-y = e^x+1.$ The integrating factor is $\displaystyle e^{\int (-1)\, \mathrm{d}x} = e^{-x}$ so we multiply both sides by this expression:

$\displaystyle e^{-x}y'-e^{-x}y = 1+e^{-x}$

$\displaystyle \frac{d}{dx}(y(x)e^{-x}) = 1+e^{-x}$

$\displaystyle y(x)e^{-x} = \int 1+e^{-x}\, \mathrm{d}x + C = x - e^{-x} + C$

$\displaystyle y(x)= e^xx - 1 + Ce^x.$

Example 2.

Let us solve the initial value problem

$\left\{\begin{align}xy' = x^2 + 3y \\ y(0) = 1 \end{align} \right.$

First, we want to express the problem in normal form:

$\displaystyle y' - \frac{3}{x}y = x.$

Now the integrating factor is $\displaystyle e^{ \int \frac{3}{x} dx } =\displaystyle e^{ -3 \ln \vert x \vert } =\displaystyle e^{ \ln x^{-3} } =\displaystyle \frac{1}{x^3},\> x>0.$ Hence, we get

$\displaystyle \frac{y'}{x^3} - \frac{3}{x^4}y = \frac{1}{x^2}$

$\displaystyle \frac{d}{dx}(\frac{y}{x^3}) = \frac{1}{x^2}$

$\displaystyle \frac{y}{x^3} = \int \frac{1}{x^2}\, \mathrm{d}x + C = - \frac{1}{x} + C$

$y = Cx^3 - x^2$

We have found the general solution. Because  $y(0) = C\cdot 0 - 0 = 0,$ that is, the value of the function does not equal the given initial value, the problem does not have a solution. The main reason for this is that the initial condition is given at $x_0=0$, where the normal form of the equation is not defined. Any other choice for $x_0$ will lead to a unique solution.

Example 3.

Let us solve the ODE $xy'-2y=2$ given the initial conditions

1. $y(1)=0$
2. $y(0)=0$.

From the form $y'-(2/x)y=2/x$ we see that the equation in question is a linear ODE. The integrating factor is

Multiplying by the integrating factor, we get

so the general solution to the ODE is $y(x)=x^2 (-1/x^2+C)=Cx^2-1$. From the initial value $y(1)=0$ it follows that $C=1$, but the other condition $y(0)=0$ leads to a contradiction $-1=0$. Therefore, the solution in the (a) part is $y(x)=x^2-1$, but a solution satisfying the initial condition of part b) does not exist: by substituting $x=0$, the equation forces $y(0)=-1$.

A first order differential equation is separable, if it can be written in the form $y' = f(x)g(y),$ where $f$ and $g$ are integrable functions in the domain of interest. Treating formally $y'(x)=dy/dx$ as a fraction, multiplying by $dx$ and dividing by $g(y)$, we obtain $\frac{dy}{g(y)}=f(x)\, dx$. Integrating the left hand side with respect to $y$ and the right hand side with respect to $x$, we get

$\displaystyle \int \frac{\mathrm{d}y}{g(y)} = \int f(x)\, \mathrm{d}x + C$

This method gives the solution to the differential equation in implicit form, which we may further be able to solve explicitly for $y =y(x)$. The justification for this formal treatment can be made by using change of variables in integrals.

Example 4.

Let us solve the differential equation $\displaystyle y'+\frac{2}{5}x = 0$ by separating the variables. (We could also solve the equation by using the method of integrating factors.)

 $\displaystyle y'+\frac{2}{5}y = 0$

 $\displaystyle \frac{dy}{dx} = -\frac{2}{5}y$  

 $\displaystyle \int \frac{1}{y}\, \mathrm{d}y = -\frac{2}{5} \int \, \mathrm{d}x$  

$\displaystyle \ln |y| = -\frac{2}{5}x + C_1$

$\displaystyle y =\pm e^{-\frac{2}{5}x+C_1} = \pm e^{-\frac{2}{5}x}e^{C_1} = Ce^{-\frac{2}{5}x}, \: C\neq 0.$

In the last step, we wrote $C =\pm e^{C_1}$ for simplicity. The case $C=0$ is also allowed, since it leads to the trivial solution $y(x)\equiv 0$, see below.

Example 5.

Let us solve the initial value problem

$\left\{\begin{align}y' = \frac{x}{y} \\ y(0) = 1 \end{align} \right.$

Because the general solution is not required, we may take a little shortcut by applying integrals in the following way:

$\displaystyle \frac{dy}{dx} = \frac{x}{y}$

$\displaystyle \int_1^y y \, \mathrm{d}y =\int_0^x x \, \mathrm{d}x$

$\displaystyle \frac{1}{2}y^2 - \frac{1}{2} =\frac{1}{2}x^2$

The solution is $y=y(x)=\sqrt{x^2+1}$.

General solution achieved by applying the method for separable functions typically lacks information about solutions related to the zeros of the function $g(y)$. The reason for this is that in the separation method we need to assume that $g(y(x)) \neq 0$ in order to be able to divide the expression by $g(y(x))$. We notice that for each zero $\alpha$ of the function $g$ there exists a corresponding constant solution $y(x)\equiv \alpha$ of the ODE $y'=f(x)g(y)$, since $y'(x)\equiv 0=g(\alpha)\equiv g(y(x))$. These solutions are called trivial solutions (in contrast to the general solution).

If the conditions of the following theorem hold, then all the solutions to a separable differential equation can be derived from either the general solution or the trivial solutions.

The proof of the theorem is based on a technique known as Picard-Lindelöf iteration, which was invented by Emile Picard and further developed by the Finnish mathematician Ernst Lindelöf (1870-1946), and others.

Applying the previous theorem, we can formulate the following result for separable equations.

The solution curves at each point $(x_0,y_0)$ of the domain of the equation are always unique. In particular, the curves cannot intersect and it is not possible for a single curve to split into two or several parts.

∴ The other solution curves of the ODE cannot intersect the curves $y=\alpha$ corresponding to the trivial solutions. That is, for all the other solutions the condition $g(y(x))\neq 0$ automatically holds!

Example 6.

Let us solve the linear homogeneous differential equation $y'+p(x)y=0$ by applying the method of separation.

The equation has the trivial solution $y_0(x)\equiv 0$. Since the other solutions do not get the value 0 anywhere, it holds that

$\begin{aligned} \frac{dy}{dx} &= y'= -p(x)y \\ &\Leftrightarrow \int\frac{\mathrm{d}y}{y} = -\int p(x)\, \mathrm{d}x +C_1 \\ &\Leftrightarrow \ln|y| = -P(x)+C_1 \\ &\Leftrightarrow |y| =e^{C_1-P(x)} \\ &\Leftrightarrow y=y(x)=\pm e^{C_1} e^{-P(x)} =Ce^{-P(x)}.\end{aligned}$

Here, the expression $\pm e^{C_1}$ has been replaced by a simpler constant $C\in\mathbb{R}$.

Some differential equations can made separable by using a suitable substitution.

i) ODEs of the form $y'(x)= f\Big(\frac{y(x)}{x}\Big).$

Example 7.

Let us solve the differential equation $y'= \frac{x+y}{x-y}.$ The equation is not separable in this form, but we can make if separable by substituting $u = \frac{y}{x},$ resulting to $y' = u + xu'.$ We get

 $u + xu'= \displaystyle \frac{1+u}{1-u}.$ 

Separating the variables and integrating both sides, we get

 $\displaystyle \int \frac{1-u}{1+u^2} \, \mathrm{d}u= \int \frac{1}{x} \, \mathrm{d}x$

  $\arctan{u} - \displaystyle \frac{1}{2} \ln(u^2 +1)= \ln{x} + C.$

Substituting  $u = \frac{y}{x}$ and simplifying yields

  $\displaystyle \arctan{\frac{y}{x}} = \ln{C\sqrt{x^2 + y^2}}.$

Here, it is not possible to derive an expression for y so we have to make do with just the implicit solution. The solutions can be visualized graphically: 

As we can see, the solutions are spirals expanding in the positive direction that are suitably cut for demonstration purposes. This is clear from the solutions' polar coordinate representation which we obtain by using the substitution

$\theta = \displaystyle \arctan{\frac{y}{x}}, r = \sqrt{x^2 + y^2}.$

Hence, the solution is

$\theta = \ln(Cr) \Leftrightarrow r = Ce^{\theta}.$

Another type of differential equation that can be made separable are equations of the form 

$\displaystyle y' = f(ax+by+c).$

To rewrite the equation as separable, we use the substitution  $\displaystyle u = ax+by+c.$

Example 8.

Let us find the solution to the differential equation

$\displaystyle y' =(x-y)^2 +1.$

Here, a natural substitution is $\displaystyle u = x-y \Leftrightarrow y = x-u \Rightarrow y' = 1-u'.$ Substitution yields

$\displaystyle 1-u' =u^2 +1$

$\displaystyle \int -\frac{1}{u^2} \, \mathrm{d}u = \int \, \mathrm{d}x$

$\displaystyle \frac{1}{u} = x +C$

$\displaystyle y = x -  \frac{1}{x +C}.$

$\star$ Euler's method

In practice, it is usually not feasible to find analytical solutions to differential equations. In these cases, the only choice for us is to resort to numerical methods. A prominent example of this kind of technique is called Euler's method. The idea behind the method is the observation made earlier with direction fields: even if we do not know the solution itself, we are still able to determine the tangents of the solution curve. In other words, we are seeking solutions for the initial value problem

$\left\{\begin{align}y' = f(x,y) \\ y(x_0) = y_0. \end{align} \right.$

In Euler's method, we begin the solving process by choosing the step length $h$ and using the iteration formula

$\displaystyle y_{k+1} = y_k +  hf(x_k, y_k).$

The iteration starts from the index $\displaystyle k=0$ by substituting the given initial value to the right side of the iteration formula. Since $f(x_k, y_k) = y'(x_k)$ is the slope of the tangent of the solution at $x_k$, on each step we move the distance expressed by the step length in the direction of the tangent. Because of this, an error occurs, which grows as the step length is increased.

Example 9.

Use the gadget on the right to examine the solution to the initial value problem

$\left\{\begin{align}y' = \sin(xy) \\ y(x_{0}) = y_{0} \end{align} \right.$

obtained by using Euler's method and compare the result to the precise solution.

For second order linear equations, there is no easy way to find a general solution. We begin by examining a homogeneous equation

$y’’ + p(x)y’ + q(x)y = 0,$

where $p$ and $q$ are continuous functions on their domains. Then, it holds that

1) the equation has linearly independent solutions $y_1$ and $y_2$, called fundamental solutions. Roughly speaking, linear independence means that the ratio $y_2(x)/y_1(x)$ is not constant, so that the solutions are essentially different from each other.

2) the general solution can expressed by means of any linearly independent pair of solutions in the form $y(x) = C_1y_1(x) + C_2y_2(x)$, where $C_1$ and $C_2$ are constants.

3) if the initial values $y(x_0) = a, y'(x_0) = b$ are fixed, then the solution is unique.

A general method for finding explicitly the fundamental solutions $y_1(x)$ and $y_2(x)$ does not exist. To find the solution, a typical approach is to try to make an educated guess about the solution's form and check the details by substituting this into the equation.

The above results can be generalized to higher order homogeneous equations as well, but then the number of required fundamental solutions and initial conditions increases with respect to the order of the equation.

Example 1.

The equation $y’’-y= 0$ has solutions $y = e^x$ and $y = e^{-x}.$ These solutions are linearly independent, so the general solution is of the form $y(x) = C_1e^x + C_2e^{-x}.$

Equations with constant coefficients

As a relatively simple special case, let us consider the 2nd order equation

$y’’ + py’ + qy = 0.$

In order to solve the equation, we use the guess $y(x) = e^{\lambda x}$, where $\lambda$ is an unknown constant. Substituting the guess into the equation yields

$\lambda^2 e^{\lambda x} + p\lambda e^{\lambda x} + qe^{\lambda x} = 0.$

$\lambda^2 + p\lambda + q = 0.$

The last equation is called the characteristic equation of the ODE. Solving the characteristic equation allows us to find the solutions for the actual ODE. The roots of the characteristic equation can be divided into three cases:

1) The characteristic equation has two distinct real roots. Then, the ODE has the solutions $y_1(x) = e^{\lambda_1x}$ and $y_2(x) = e^{\lambda_2x}.$

2) The characteristic equation has a double root. Then, the ODE has the solutions $y_1(x) = e^{\lambda x}$ and $y_2(x) = xe^{\lambda x}.$

3) The roots of the characteristic equation are of the form $\lambda = a \pm bi.$ Then, the ODE has the solutions $y_1(x) = e^{ax}\cos(bx)$ and $y_2(x) = e^{ax}\sin(bx).$

The second case can be justified by substitution into the original ODE, and the third case by using the Euler formula $e^{ix}=\cos x+i\sin x$. With minor changes, these results can also be generalized to higher order differential equations.

Since the characteristic equation has exactly the same coefficients as the original ODE, it is not necessary to derive it again in concrete examples: just write it down by looking at the ODE!

Example 2.

Let us solve the initial value problem

$\left\{\begin{align}y'' -y' +2y=0 \\ y(0) = 1, y(1)=0 \end{align} \right.$

The characteristic equation is $\lambda^2 -\lambda -2 = 0,$ which has the roots $\lambda_1 = 2$ and $\lambda_2 = -1.$ Thus, the general solution is $y(x) = C_1e^{2x} + C_2e^{-x}.$ The constants can be determined by using the initial conditions:

$\left\{\begin{align}C_1 + C_2=1 \\ e^2C_1 + e^{-1}C_2 = 0 \end{align} \right.$

$\left\{\begin{align}C_1 = -\frac{1}{e^3-1} \\ C_2 = \frac{e^3}{e^3-1} \end{align} \right.$

Hence, the general solution is $y(x) = \frac{1}{e^3-1} (-e^{2x} + e^{3-x}).$

Example 3.

Let us have a look at how the above results hold in higher order equations by solving

$y^{(4)} - 4y''' +14y'' -20y' +25y = 0.$

Now, the characteristic equation is $\lambda^4 - 4\lambda^3 +14\lambda^2 -20\lambda +25 = 0,$ which has the roots $\lambda_1 = \lambda_2 = 1 + 2i$ and $\lambda_3 = \lambda_4 = 1 - 2i.$ Thus, the fundamental solutions to the ODE are $e^x\sin(2x)$, $e^x\cos(2x)$, $xe^x\sin(2x)$ and $xe^x\cos(2x)$. The general solution is

$y = C_1e^x\sin(2x) + C_2e^x\cos(2x) + C_3xe^x\sin(2x) + C_4xe^x\cos(2x).$

Example 4.

Let $\omega >0$ be a constant. The characteristic equation of the ODE $y''+\omega^2y=0$ is $\lambda^2+\omega^2=0$ with roots $\lambda=\pm i \omega$. So $\alpha =0$ and $\beta =\omega$ in Case 3). Since this ODE is a model for harmonic oscillation, we use time $t$ as variable, and obtain the general solution $y(t)=A\cos (\omega t) +B\sin (\omega t),$ with $A,B$ constants. They will be uniquely determined if we know the initial location $y(0)$ and the initial velocity $y'(0)$. All solutions are periodic and their period is $T=2\pi/\omega$. In the animation to the right we have $y'(0)=0$ and you can choose $\omega$ and the initial displacement $y(0)=y_0$.

Another relatively common type of 2nd order differential equation is Euler's differential equation

$x^2y'' + axy' + by = 0,$

where $a$ and $b$ are constants. An equation of this form is solved by using the guess $y(x)= x^r$. Substituting the guess in the equation yields

$r^2 + (a-1)r + b = 0.$

Using the roots of this equation, we obtain the solutions for the ODE in the following way:

1) If the roots are distinct and real, then $y_1(x)= |x|^{r_1}$ and $y_2(x)= |x|^{r_2}$.

2) If the equation has a double root, then $y_1(x)= |x|^{r}$ and $y_2(x)= |x|^{r}\ln |x|$.

3) If the equation has roots of the form $r = a \pm bi$, then $y_1(x)= |x|^{a}\cos(b\ln |x|)$ and $y_2(x)= |x|^{a}\sin(b\ln |x|)$.

Example 5.

Let us solve the equation $x^2y'' - 3xy' + y = 0.$ Noticing that the equation is Euler's differential equation, we proceed by using the guess $y= x^r.$ Substituting the guess into the equation, we get $r(r-1)x^r - 3rx^r + x^r = 0 \Rightarrow r^2 - 4r + 1 = 0,$ which yields $r = 2 \pm \sqrt{3}.$ Therefore, the general solution to the ODE is

$y = C_1 x^{2+\sqrt{3}} + C_2x^{2-\sqrt{3}}$.

The general solution to a nonhomogeneous equation

$y'' + p(x)y' + q(x)y = r(x)$

is the general solution to the corresponding homogeneous equation $+$ particular solution to the nonhomogeneous equation, i.e.

$y(x) = C_1y_1(x) + C_2y_2(x) + y_0(x)$.

The particular solution $y_0$ is usually found by using a guess that is of the same form as $r(x)$ with general coefficients. Substituting the guess into the ODE, we can solve these coefficients, but only if the guess is of the correct form.

In the table below, we have created a list of possible guesses for second order differential equations with constant coefficients. The form of the guess depends on what kind of elementary functions $r(x)$ consists of. If $r(x)$ is a combination of several different elementary functions, then we need to include corresponding elements for all of these functions in our guess. The characteristic equation of the corresponding homogeneous differential equation is $P(\lambda)=\lambda^2+p\lambda+q=0$.

Note. For roots of a second degree polynomial we have to keep in mind that

• $P(k)=0$ and $P'(k)\neq 0$ $\Leftrightarrow$ $k\in\mathbb{R}$ is a simple root of $P$.

• $P(k)=P'(k)= 0$ $\Leftrightarrow$ $k\in\mathbb{R}$ is a double root of $P$.

• $P(ik)\neq 0$ $\Leftrightarrow$ $ik\in\mathbb{C}$ is not a root of $P$; i.e. $\sin kx$ and $\cos kx$ are not solutions to the homogeneous equation.

Example 6.

Let us find the general solution to the ODE $y''+y'-6y=r(x)$, when

a) $r(x)=12e^{-x}$

b) $r(x)=20e^{2x}$.

The solutions are of the form $y(x)=C_1e^{-3x}+C_2e^{2x}+y_0(x)$.

a) Substituting the guess $y_0(x)=Ae^{-x},$ we get $(A -A -6A)e^{-x} =12e^{-x}$, which solves for $A=-2$.

b) In this case a guess of the form $Be^{2x}$ is useless, as it is part of the general solution to the corresponding homogeneous equation and yields just zero when substituted to the left side of the ODE. Here, a right guess is of the form $y_0(x)=Bxe^{2x}$. Substitution yields

which solves for $B=4$.

Using these values for $A$ and $B$, we can write the general solutions to the given differential equations.

Example 7.

Let us find the solution to the ODE $y''+y'-6y=12e^{-x}$ with the initial conditions $y(0)=0$, $y'(0)=6$.

Based on the previous example, the general solution is of the form $y(x)=C_1e^{-3x}+C_2e^{2x}-2e^{-x}$. Differentiation yields $y'(x)=-3C_1e^{-3x}+2C_2e^{2x}+2e^{-x}$. From the initial conditions, we get the following pair of equations:

which solves for $C_1=0$ and $C_2=2$. Therefore, the solution to the initial value problem is $y(x)=2e^{2x}-2e^{-x}$.

Example 8.

A typical application of a second order nonhomogeneous ODE is an RLC circuit containing a resistor (with resistance $R$), an inductor (with inductance $L$), a capacitor (with capacitance $C$), and a time-dependent electromotive force $E(t)$. The electric current $y(t)$ in the circuit satisfies the ODE $Ly''+Ry'+\frac{1}{C}y=E'(t).$ Let us solve this ODE with artificially chosen numerical values in the form $y''+10y'+61y=370\sin t.$

The homogeneous part has characteristic equation of the form $\lambda^2+10\lambda +61=0$ with solutions $\lambda = -5\pm 6i$. This gives the solutions $y_1(t)=e^{-5t}\cos(6t)$ and $y_2(t)=e^{-5t}\sin(6t)$ for the homogeneous equation. For a particular solution we try $y_0(t)=A\cos t +B\sin t$. Substituting this into the nonhomogeneous ODE and collecting similar terms yields to $(60A+10B)\cos t +(60B-10A)\sin t = 370\sin t.$ This equation will be satisfied for all $t$ (only) if

which solves for $A=-1$ and $B=6$. Therefore, the general solution is $y(t)=e^{-5t}(C_1\cos(6t)+C_2\sin(6t)) -\cos t+6\sin t .$ Note. The exponential terms go to zero very fast and eventually, the current oscillates in the form $y(t)\approx -\cos t+6\sin t.$