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 ${\displaystyle \frac{dy}{dx}=f(x,y)}$$\displaystyle\frac{dy}{dx}=f(x,y)$ can be written as a function of the ratio ${\displaystyle \frac{y}{x}}$$\displaystyle\frac{y}{x}$ alone, i.e., ${\displaystyle \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(tx,ty)=f(x,y)$$f(tx,ty)=f(x,y)$.

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

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

   • Rewrite as $y=\nu x$$y=\nu x$;

   • Differentiate using product rule: ${\displaystyle \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 ${\displaystyle \nu +x\frac{d\nu }{dx}=G(\nu )}$$\displaystyle\nu + x\frac{d\nu}{dx}=G(\nu)$.

5. The DE is now separable: ${\displaystyle \int \frac{1}{G(\nu )-\nu }d\nu =\int \frac{1}{x}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. ${\displaystyle \frac{dy}{dx}=\frac{x^2+y^2}{xy}}$$\displaystyle \frac{dy}{dx}=\frac{x^2+y^2}{xy}$

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

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

4. ${\displaystyle (x-y)dx+xdy=0}$$\displaystyle (x-y)dx+xdy=0$

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

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

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Last Modified: Sunday, 6 September 2020 14:15 EDT