# Substitutions

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 2.6 – Substitutions and Transformations

## Expected Educational Results

• Objective 4–1: I can identify the appropriate substitution needed for a given ODE.

• Objective 4–2: I can apply the appropriate substitution to solve an ODE.

## Equations of the form: $\frac{dy}{dx}=G\left(ax+by+c\right)$$\displaystyle \dfrac{dy}{dx}=G(ax+by+c)$

### Method for Solving Equations in the Form $\frac{dy}{dx}=G\left(ax+by+c\right)$$\displaystyle \dfrac{dy}{dx}=G(ax+by+c)$

1. Use the substitution $z=ax+by+c$$\displaystyle z=ax+by+c$.

2. Differentiate using product rule: $\frac{dz}{dx}=a+b\frac{dy}{dx}$$\displaystyle\dfrac{dz}{dx}=a + b\dfrac{dy}{dx}$ This is the second substitution: $\frac{dy}{dx}=\frac{\frac{dz}{dx}-a}{b}$$\displaystyle\dfrac{dy}{dx}=\dfrac{\dfrac{dz}{dx}-a}{b}$.

3. Rewrite the DE as $\frac{\frac{dz}{dx}-a}{b}=G\left(z\right)$$\displaystyle\dfrac{\dfrac{dz}{dx}-a}{b}=G(z)$ or $\frac{dz}{dx}=bG\left(z\right)+a$$\displaystyle\dfrac{dz}{dx}=bG(z)+a$.

4. The DE is now separable: $\int \frac{1}{bG\left(z\right)+a}\phantom{\rule{0.167em}{0ex}}dz=\int \phantom{\rule{0.167em}{0ex}}dx$$\displaystyle \int{\frac{1}{bG(z)+a}\,dz}=\int{\,dx}$.

5. Rewrite the solution in terms of $ax+by+c$$ax+by+c$.

6. Identify missing solutions or solutions that are not solutions of the original DE.

#### Question 02

Solve the following DEs:

1. $\frac{dy}{dx}=\left(x+y-4{\right)}^{2}$$\displaystyle \frac{dy}{dx}=(x+y-4)^2$

2. $\frac{dy}{dx}=\left(-16x+y+3{\right)}^{4}$$\displaystyle \frac{dy}{dx}=(-16x+y+3)^4$

3. $\frac{dy}{dx}={\mathrm{tan}}^{2}\left(x+y\right)$$\displaystyle \frac{dy}{dx}=\tan^2{(x+y)}$

4. $\frac{dy}{dx}=\left(x-y-5{\right)}^{2}$$\displaystyle \frac{dy}{dx}=(x-y-5)^2$, $y\left(1\right)=1$$y(1)=1$