# Equations with Linear Coefficients

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 with Linear Coefficients

NOTE: You will not be assessed on ODEs with linear coefficients.

### Definition: Equations with Linear Coefficients

Equations with linear coefficients is an ODE in the form:

$\left({a}_{1}x+{b}_{1}y+{c}_{1}\right)\phantom{\rule{0.167em}{0ex}}dx+\left({a}_{2}x+{b}_{2}y+{c}_{2}\right)\phantom{\rule{0.167em}{0ex}}dy=0$$\displaystyle (a_1x+b_1y+c_1)\,dx+(a_2x+b_2y+c_2)\,dy=0$

### Method for Solving Equations with Linear Coefficients

##### Case 1: ${a}_{1}{b}_{2}={a}_{2}{b}_{1}$$a_1b_2 = a_2b_1$
1. Substitute $v=\frac{{a}_{1}x+{b}_{1}y}{{a}_{1}}$$\displaystyle v=\dfrac{a_1x+b_1y}{a_1}$ which implies ${a}_{1}\left(v-x\right)={b}_{1}y$$\displaystyle a_1(v-x)=b_1y$ and $\frac{{a}_{2}{b}_{1}}{{b}_{2}}\left(v-x\right)={b}_{1}y$$\displaystyle \dfrac{a_2b_1}{b_2}(v-x)=b_1y$ since ${a}_{1}{b}_{2}={a}_{2}{b}_{1}$$\displaystyle a_1b_2 = a_2b_1$. $\frac{{a}_{2}{b}_{1}}{{b}_{2}}\left(v-x\right)={b}_{1}y$$\displaystyle \dfrac{a_2b_1}{b_2}(v-x)=b_1y$ further simplifies to ${a}_{2}\left(v-x\right)={b}_{2}y$$\displaystyle a_2(v-x)=b_2y$.

2. Solve ${a}_{2}\left(v-x\right)={b}_{2}y$$\displaystyle a_2(v-x)=b_2y$ for $v$$v$: $v=\frac{{a}_{2}x+{b}_{2}y}{{a}_{2}}$$\displaystyle v=\dfrac{a_2x+b_2y}{a_2}$.

3. Differentiate the substitution expression: $\frac{dv}{dx}=1+\frac{{b}_{1}}{{a}_{1}}\frac{dy}{dx}$$\displaystyle \dfrac{dv}{dx}=1+\dfrac{b_1}{a_1}\dfrac{dy}{dx}$.

4. Rewrite the derivative above into: $\frac{dy}{dx}=\frac{{a}_{1}}{{b}_{1}}\left(\frac{dv}{dx}-1\right)$$\displaystyle \dfrac{dy}{dx}=\dfrac{a_1}{b_1}\left(\dfrac{dv}{dx}-1\right)$.

5. Substitute the formulas for $v$$v$ from steps $1$$1$ and $2$$2$ and $\frac{dy}{dx}$$\displaystyle \dfrac{dy}{dx}$ from step $3$$3$ into $\frac{dy}{dx}=\frac{{a}_{1}x+{b}_{1}y+{c}_{1}}{{a}_{2}x+{b}_{2}y+{c}_{2}}$$\displaystyle \dfrac{dy}{dx}=\dfrac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}$ to obtain: $\frac{{a}_{1}}{{b}_{1}}\left(\frac{dv}{dx}-1\right)=\frac{{a}_{1}v+{c}_{1}}{{a}_{2}v+{c}_{2}}$$\dfrac{a_1}{b_1}\left(\dfrac{dv}{dx}-1\right)=\dfrac{a_1v+c_1}{a_2v+c_2}$

6. Solve for $\frac{dv}{dx}$$\displaystyle \dfrac{dv}{dx}$: $\frac{dv}{dx}=1+\frac{{b}_{1}}{{a}_{1}}\left(\frac{{a}_{1}v+{c}_{1}}{{a}_{2}v+{c}_{2}}\right)$$\displaystyle \dfrac{dv}{dx}=1+\dfrac{b_1}{a_1}\left(\dfrac{a_1v+c_1}{a_2v+c_2}\right)$.

7. The DE is now separable into: $\frac{dv}{1+\frac{{b}_{1}}{{a}_{1}}\left(\frac{{a}_{1}v+{c}_{1}}{{a}_{2}v+{c}_{2}}\right)}=dx$$\displaystyle \dfrac{dv}{1+\dfrac{b_1}{a_1}\left(\dfrac{a_1v+c_1}{a_2v+c_2}\right)}=dx$.

##### Case 2: ${a}_{1}{b}_{2}\ne {a}_{2}{b}_{1}$$a_1b_2\ne a_2b_1$ and ${c}_{1}={c}_{2}=0$$c_1=c_2=0$
1. Rewrite $\left({a}_{1}x+{b}_{1}y\right)dx+\left({a}_{2}x+{b}_{2}y\right)dy=0$$\displaystyle (a_1x+b_1y)dx+(a_2x+b_2y)dy=0$ into $\frac{dy}{dx}=-\frac{{a}_{1}x+{b}_{1}y}{{a}_{2}x+{b}_{2}y}=-\frac{{a}_{1}+{b}_{1}\frac{y}{x}}{{a}_{2}+{b}_{2}\frac{y}{x}}$$\displaystyle\dfrac{dy}{dx}=-\dfrac{a_1x+b_1y}{a_2x+b_2y}=-\dfrac{a_1+b_1\dfrac{y}{x}}{a_2+b_2\dfrac{y}{x}}$.

2. The DE is now homogeneous.

##### Case 3: ${a}_{1}{b}_{2}\ne {a}_{2}{b}_{1}$$a_1b_2\ne a_2b_1$ and both ${c}_{1}\ne 0$$c_1\ne 0$ and ${c}_{2}\ne 0$$c_2\ne 0$
1. Substitute $x=u+h$$x=u+h$ and $y=v+k$$y=v+k$, where $h$$h$ and $k$$k$ are constants.

2. Solve the system: ${a}_{1}h+{b}_{1}k+{c}_{1}=0$$a_1h+b_1k+c_1=0$, ${a}_{2}h+{b}_{2}k+{c}_{2}=0$$a_2h+b_2k+c_2=0$ for $h$$h$ and $k$$k$.

3. Rewrite the DE as $\frac{dv}{du}=-\frac{{a}_{1}u+{b}_{1}v}{{a}_{2}u+{b}_{2}v}=-\frac{{a}_{1}+{b}_{1}\frac{v}{u}}{{a}_{2}+{b}_{2}\frac{v}{u}}$$\displaystyle\dfrac{dv}{du}=-\dfrac{a_1u+b_1v}{a_2u+b_2v}=-\dfrac{a_1+b_1\dfrac{v}{u}}{a_2+b_2\dfrac{v}{u}}$.

4. The DE is now homogeneous.

#### Question 04

Solve the following DEs:

1. $\frac{dy}{dx}=\frac{x-2y-2}{2x+y+6}$$\displaystyle \frac{dy}{dx}=\frac{x-2y-2}{2x+y+6}$

2. $\frac{dy}{dx}=\frac{2x+y}{-y-x}$$\displaystyle \frac{dy}{dx}=\frac{2x+y}{-y-x}$

3. $\frac{dy}{dx}=\frac{x+3y+4}{4y-3x+1}$$\displaystyle \frac{dy}{dx}=\frac{x+3y+4}{4y-3x+1}$