3.5 Ordinary Differential Equations

In this section, we cover four typical differential equation patterns that are commonly seen in interviews.

Separable differential equations

A separable differential equation has the form $\frac{dy}{dx}=g(x)h(y)$. Since it is separable, we can express the original equation as $\frac{dy}{h(y)}=g(x)dx$. Integrating both sides, we have the solution $\int\frac{dy}{h(y)}=\int g(x)dx$.

A. Solve ordinary differential equation $y'+6xy=0,\ y(0)=1$

Solution: Let $g(x)=-6x$ and $h(y)=y$, we have $\frac{dy}{y}=-6xdx$. Integrate both sides of the equation:

where $c$ is a constant. Plugging in the initial condition $y(0)=1$, we have $c=0$ and $y=e^{-3x^2}$.

B. Solve ordinary differential equation $y'=\frac{x-y}{x+y}$.⁷

Solution: Unlike the last example, this equation is not separable in its current form. But we can use a change of variable to turn it into a separable differential equation. Let $z=x+y$, then the original differential equation is converted to

First-order linear differential equations

A first-order differential linear equation has the form $\frac{dy}{dx}+P(x)y=Q(x)$. The standard approach to solving a first-order differential equation is to identify a suitable function $I(x)$, called an integrating factor, such that $I(x)(y'+P(x)y)=I(x)y'+I(x)P(x)y$

$=(I(x)y)'$; Then we have $(I(x)y)'=I(x)Q(x)$ and we can integrate both sides to solve for $y$:

The integrating factor, $I(x)$, must satisfy $\frac{dI(x)}{dx}=I(x)P(x)$, which means $I(x)$ is a separable differential equation with general solution $I(x)=e^{\int P(x)dx}$.⁸

Solve ordinary different equation $y'+\frac{y}{x}=\frac{1}{x^2},\ y(1)=1$, where $x>0$.

Solution: This is a typical example of first-order linear equations with $P(x)=\frac{1}{x}$ and $Q(x)=\frac{1}{x^2}$. So $I(x)=e^{\int P(x)dx}=e^{\int(1/x)dx}=e^{\ln x}=x$ and we have $I(x)Q(x)=\frac{1}{x}$.

Taking integration on both sides, $xy=\int(1/x)dx=\ln x+c\Rightarrow y=\frac{\ln x+c}{x}$.

Plugging in $y(1)=1$, we get $c=1$ and $y=\frac{\ln x+1}{x}$.

Homogeneous linear equations

A homogenous linear equation is a second-order differential equation with the form $a(x)\frac{d^2y}{dx^2}+b(x)\frac{dy}{dx}+c(x)=0$.

It is easy to show that, if $y_1$ and $y_2$ are linearly independent solutions to the homogeneous linear equation, then any $y(x)=c_1y_1(x)+c_2y_2(x)$, where $c_1$ and $c_2$ are arbitrary constants, is a solution to the homogeneous linear equation as well.

When $a,b$ and $c$ ($a\ne0$) are constants instead of functions of $x$, the homogenous linear equation has closed form solutions:

Let $r_1$ and $r_2$ be the roots of the characteristic equation $ar^2+br+c=0$,⁹

1. If $r_1$ and $r_2$ are real and $r_1\ne r_2$, then the general solution is $y=c_1e^{r_1x}+c_2e^{r_2x}$;
2. If $r_1$ and $r_2$ are real and $r_1=r_2=r$, then the general solution is $y=c_1e^{rx}+c_2xe^{rx}$;
3. If $r_1$ and $r_2$ are complex numbers $\alpha\pm i\beta$, then the general solution is $y=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)$.

It is easy to verify that the general solutions indeed satisfy the homogeneous linear solutions by taking the first and secondary derivatives of the general solutions.

What is the solution of ordinary differential equation $y''+y'+y=0$?

Solution: In this specific case, we have $a=b=c=1$ and $b^2-4ac=-3<0$, so we have complex roots $r=-1/2\pm\sqrt{3}/2i$ ($\alpha=-1/2,\ \beta=\sqrt{3}/2$), and the general solution to the differential equation is therefore

Nonhomogeneous linear equations

Unlike a homogenous linear equation $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+c=0$, a nonhomogeneous linear equation $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+c=d(x)$ has no closed-form solution. But if we can find a particular solution $y_p(x)$ for $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+c=d(x)$, then $y=y_p(x)+y_g(x)$, where $y_g(x)$ is the general solution of the homogeneous equation $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+c=0$, is a general solution of the nonhomogeneous equation $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+c=d(x)$.

Although it may be difficult to identify a particular solution $y_p(x)$ in general, in the special case when $d(x)$ is a simple polynomial, the particular solution is often a polynomial of the same degree.

What is the solution of ODEs $y''+y'+y=1$ and $y''+y'+y=x$?

Solution: In these ODEs, we again have $a=b=c=1$ and $b^2-4ac=-3<0$, so we have complex solutions $r=-1/2\pm\sqrt{3}/2i$ ($\alpha=-1/2,\ \beta=\sqrt{3}/2$) and the general solution is $y=e^{-1/2x}\left(c_1\cos(\sqrt{3}/2x)+c_2\sin(\sqrt{3}/2x)\right)$.

What is a particular solution for $y''+y'+y=1$? Clearly $y=1$ is. So the solution to $y''+y'+y=1$ is

To find a particular solution for $y''+y'+y=x$, Let $y_p(x)=mx+n$, then we have $y''+y'+y=0+m+(mx+n)=x\Rightarrow m=1,n=-1$. So the particular solution is $x-1$ and the solution to $y''+y'+y=x$ is