How to solve differential equations by substitution

If the same combination of x and y keeps showing up in a differential equation, we can try giving that combination its own name. The new variable can turn the equation into one we already know how to solve. Let’s see how to make the substitution and get back to y.

Use the expression inside the function

For an equation of the form $y' = F (a x + b y)$ , set $u = a x + b y$ . Differentiating gives $u' = a + b y'$ , so the equation becomes $u' = a + b F (u)$ . Now the right side depends only on $u$ . When $b \neq 0$ , we can solve for $y$ after finding $u$ .

Work through a first example

For $y' = 2 x + y$ , set $u = 2 x + y$ . Then $u' = 2 + y' = u + 2$ . The shifted variable $u + 2$ has a derivative equal to itself, so it equals $C e^{x}$ . This includes the constant solution $u = - 2$ when $C = 0$ .

\begin{array}{l} 2 x + y + 2 = C e^{x} \\ y = C e^{x} - 2 x - 2 \end{array}

Use substitution when the expression is squared

For $y' = (4 x - y - 1)^{2}$ , set $u = 4 x - y$ . This time differentiating gives $u' = 4 - y'$ , so

\begin{array}{l} u' = 4 - (u - 1)^{2} \\ u' = (u + 1) (3 - u) \end{array}

The constant values $u = - 1$ and $u = 3$ give the solutions $y = 4 x + 1$ and $y = 4 x - 3$ . On intervals where $u$ is between or outside those values, separate and integrate.

\begin{array}{l} \frac{1}{4} \ln | \frac{u + 1}{3 - u} | = x + C \\ \frac{u + 1}{3 - u} = K e^{4 x} \\ y = 4 x + \frac{1 - 3 K e^{4 x}}{1 + K e^{4 x}} \end{array}

The last expression is defined on intervals where $1 + K e^{4 x} \neq 0$ .

Learn mathKrista KingMarch 17, 2021math, learn online, online course, online math, equilibrium solutions

How to solve differential equations by substitution

Use the expression inside the function

Work through a first example

Use substitution when the expression is squared

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