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: ${\displaystyle {\displaystyle \frac{dy}{dx}}=G(ax+by+c)}$$\displaystyle \dfrac{dy}{dx}=G(ax+by+c)$

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

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

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

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

4. The DE is now separable: ${\displaystyle \int \frac{1}{bG(z)+a}dz=\int 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. ${\displaystyle \frac{dy}{dx}=(x+y-4{)}^2}$$\displaystyle \frac{dy}{dx}=(x+y-4)^2$

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

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

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

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