# Substitutions and Transformations

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.

## Homogeneous Equations

### Definition: Homogeneous Equations

If the right-hand side of the equation $\frac{dy}{dx}=f\left(x,y\right)$$\displaystyle\frac{dy}{dx}=f(x,y)$ can be written as a function of the ratio $\frac{y}{x}$$\displaystyle\frac{y}{x}$ alone, i.e., $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$$\displaystyle\frac{dy}{dx}=f\left(\frac{y}{x}\right)$, then the equation is homogeneous.

### Method for Solving Homogeneous Equations}

1. Identify if the DE is homogeneous, i.e., if and only if $f\left(tx,ty\right)=f\left(x,y\right)$$f(tx,ty)=f(x,y)$.

2. Rewrite the DE as $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$$\displaystyle\frac{dy}{dx}=f\left(\frac{y}{x}\right)$.

3. Use the substitution $\nu =\frac{y}{x}$$\displaystyle\nu=\frac{y}{x}$:

• Rewrite as $y=\nu x$$y=\nu x$;

• Differentiate using product rule: $\frac{dy}{dx}=\nu +x\frac{d\nu }{dx}$$\displaystyle\frac{dy}{dx}=\nu + x\frac{d\nu}{dx}$ which is the second substitution.

4. Rewrite the DE as $\nu +x\frac{d\nu }{dx}=G\left(\nu \right)$$\displaystyle\nu + x\frac{d\nu}{dx}=G(\nu)$.

5. The DE is now separable: $\int \frac{1}{G\left(\nu \right)-\nu }\phantom{\rule{0.167em}{0ex}}d\nu =\int \frac{1}{x}\phantom{\rule{0.167em}{0ex}}dx$$\displaystyle \int{\frac{1}{G(\nu)-\nu}\,d\nu}=\int{\frac{1}{x}\,dx}$.

6. Rewrite the solution in terms of $x$$x$ and $y$$y$.

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

#### Question 01

Solve the following DEs:

1. $\frac{dy}{dx}=\frac{{x}^{2}+{y}^{2}}{xy}$$\displaystyle \frac{dy}{dx}=\frac{x^2+y^2}{xy}$

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

3. $\frac{dy}{dx}=\frac{x-y}{x+y}$$\displaystyle \frac{dy}{dx}=\frac{x-y}{x+y}$

4. $\left(x-y\right)dx+xdy=0$$\displaystyle (x-y)dx+xdy=0$

5. $\left({x}^{2}-{y}^{2}\right)dx+2xydy=0$$\displaystyle (x^2-y^2)dx+2xydy=0$, $y\left(1\right)=1$$y(1)=1$

6. $\frac{dy}{dx}=\frac{{x}^{2}-xy+{y}^{2}}{xy}$$\displaystyle \frac{dy}{dx}=\frac{x^2-xy+y^2}{xy}$, $y\left(1\right)=0$$y(1)=0$