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:

${\displaystyle (a_{1}x+b_{1}y+c_1)dx+(a_{2}x+b_{2}y+c_2)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 ${\displaystyle v={\displaystyle \frac{a_{1}x+b_{1}y}{a_1}}}$$\displaystyle v=\dfrac{a_1x+b_1y}{a_1}$ which implies ${\displaystyle a_1(v-x)=b_{1}y}$$\displaystyle a_1(v-x)=b_1y$ and ${\displaystyle {\displaystyle \frac{a_2{b}_1}{b_2}}(v-x)=b_{1}y}$$\displaystyle \dfrac{a_2b_1}{b_2}(v-x)=b_1y$ since ${\displaystyle a_1{b}_2=a_2{b}_1}$$\displaystyle a_1b_2 = a_2b_1$. ${\displaystyle {\displaystyle \frac{a_2{b}_1}{b_2}}(v-x)=b_{1}y}$$\displaystyle \dfrac{a_2b_1}{b_2}(v-x)=b_1y$ further simplifies to ${\displaystyle a_2(v-x)=b_{2}y}$$\displaystyle a_2(v-x)=b_2y$.

2. Solve ${\displaystyle a_2(v-x)=b_{2}y}$$\displaystyle a_2(v-x)=b_2y$ for $v$$v$: ${\displaystyle v={\displaystyle \frac{a_{2}x+b_{2}y}{a_2}}}$$\displaystyle v=\dfrac{a_2x+b_2y}{a_2}$.

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

4. Rewrite the derivative above into: ${\displaystyle {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{a_1}{b_1}}({\displaystyle \frac{dv}{dx}}-1)}$$\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$$2$ and ${\displaystyle {\displaystyle \frac{dy}{dx}}}$$\displaystyle \dfrac{dy}{dx}$ from step $3$$3$ into ${\displaystyle {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{a_{1}x+b_{1}y+c_1}{a_{2}x+b_{2}y+c_2}}}$$\displaystyle \dfrac{dy}{dx}=\dfrac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}$ to obtain: ${\displaystyle \frac{a_1}{b_1}}({\displaystyle \frac{dv}{dx}}-1)={\displaystyle \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 ${\displaystyle {\displaystyle \frac{dv}{dx}}}$$\displaystyle \dfrac{dv}{dx}$: ${\displaystyle {\displaystyle \frac{dv}{dx}}=1+{\displaystyle \frac{b_1}{a_1}}\left({\displaystyle \frac{a_{1}v+c_1}{a_{2}v+c_2}}\right)}$$\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: ${\displaystyle {\displaystyle \frac{dv}{1+{\displaystyle \frac{b_1}{a_1}}\left({\displaystyle \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 ${\displaystyle (a_{1}x+b_{1}y)dx+(a_{2}x+b_{2}y)dy=0}$$\displaystyle (a_1x+b_1y)dx+(a_2x+b_2y)dy=0$ into ${\displaystyle {\displaystyle \frac{dy}{dx}}=-{\displaystyle \frac{a_{1}x+b_{1}y}{a_{2}x+b_{2}y}}=-{\displaystyle \frac{a_1+b_1{\displaystyle \frac{y}{x}}}{a_2+b_2{\displaystyle \frac{y}{x}}}}}$$\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 ${\displaystyle {\displaystyle \frac{dv}{du}}=-{\displaystyle \frac{a_{1}u+b_{1}v}{a_{2}u+b_{2}v}}=-{\displaystyle \frac{a_1+b_1{\displaystyle \frac{v}{u}}}{a_2+b_2{\displaystyle \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. ${\displaystyle \frac{dy}{dx}=\frac{x-2y-2}{2x+y+6}}$$\displaystyle \frac{dy}{dx}=\frac{x-2y-2}{2x+y+6}$

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

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

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