11-10-2025 Separable Differential Equations

• $\frac {dy}{dx} = \frac {fx}{gy}$    with    $g(y) \ne 0$

Many proteins that are needed for basic cell function are produced at a constant rate (called constitutive expression). In the cell, all proteins are actively degraded by proteolytic enzymes. The rate of degradation of each protein is proportional to the abundance of that protein in the cell.

Protein abundance p(t) converges exponentially to its equilibrium value.

Model of protein abundance:

•  $\frac d{dt} p(t) = a -kp(t)$

  • $\frac d{dt} p(t)$ is the rate of change of protein abundance

  • $a$ is the rate of production (constant)

  • $kp(t)$ is the rate of degradation (proportional to $p(t)$)

For this model, the equilibrium protein abundance is:

• $A)$ $k/a$

• $B)$ $a$

• $C)$ $a/k$

• $D)$ $-kp$

• $E)$ What?

Answer is $a/k$ since $p = a/k$ give rate of change of zero

General Solution Example

$\frac {dy}{dx} = \frac {x²}{y²}$

• We want one side to have $x$ variables and one side to have $y$ variables

Step 1) Cross multiply

• $\frac {dy}{dx} = \frac {x²}{y²} \Rightarrow y²dy=x²dx$

Step 2) Now that we have separated the variables, we can integrate both sides

• $y²dy=x²dx \Rightarrow \frac {y³}{3} = \frac {x³}{3} + C$

Step 3) Put y by itself

• $\frac {y³}{3} = \frac {x³}{3} + C \Rightarrow y=\sqrt[3]{x³+3C}$

  • Since C is a constant, we can just say C

Particular Solution Example

$y^\prime = xy$        $y(0) = 5$    

$y^\prime = \frac {dy}{dx}$

Step 1) Separate both variables

• $\frac {dy}{dx} = xy \Rightarrow \frac {dy}y = x{dx}$

Step 2) Integrate

• $\frac 1y dy = xdx \Rightarrow \ln|y| = \frac 12 x² + C$

Step 3) Get y by itself

• $e^{\ln|y|}=e^{\frac 12 x² + C}$

• $y = e^{\frac 12 x² + C}$

Step 4) Get C to somewhere manageable

• $e^{\frac 12 x² + C}\Rightarrow e^{\frac 12 x²} \cdot e^C$

  • $e^C$ is a constant, so we can replace it with $C$

• $e^{\frac 12 x²} \cdot C$

Step 5) Find the particulate solution

• $y(0) = 5$

• $y=Ce^{\frac 12 x²}\Rightarrow 5 = Ce^{\frac 12 (0)²}$

• $C = \frac 5{1}$

• $C = 5$

Step 6) Construct the final equation

• $y=5e^{\frac 12 x²}$

Particular Solution Example

$y^\prime = xy$        $y(0) = 3$    

$y^\prime = \frac {dy}{dx}$

Step 1) Separate both variables

• $\frac {dy}{dx} = xy \Rightarrow \frac {dy}y = x{dx}$

Step 2) Integrate

• $\frac 1y dy = xdx \Rightarrow \ln|y| = \frac 12 x² + C$

Step 3) Get y by itself

• $e^{\ln|y|}=e^{\frac 12 x² + C}$

• $y = e^{\frac 12 x² + C}$

Step 4) Get C to somewhere manageable

• $e^{\frac 12 x² + C}\Rightarrow e^{\frac 12 x²} \cdot e^C$

  • $e^C$ is a constant, so we can replace it with $C$

• $e^{\frac 12 x²} \cdot C$

Step 5) Find the particulate solution

• $y(0) = 3$

• $y=Ce^{\frac 12 x²}\Rightarrow 3 = Ce^{\frac 12 (0)²}$

• $C = \frac 3{1}$

• $C = 3$

Step 6) Construct the final equation

• $y=3e^{\frac 12 x²}$