Problem 57 - Calculus for Biology and Medicine

Question

In Problems you will need to solve differential equations by separation of variables. In these problems it will not always be possible to solve explicitly for \(y\) in terms of \(t ;\) instead your solution may take the form of an implicit function relating the two variables. $$ \frac{d y}{d t}=\frac{t+1}{y+y^{2}} \text { where } y(0)=1 $$

Step-by-Step Solution

Answer

The implicit solution is \( \frac{1}{2}y^2 + \frac{1}{3}y^3 = \frac{1}{2}t^2 + t + \frac{5}{6} \).

Step 1: Identify the separable form

The given differential equation is \( \frac{d y}{d t} = \frac{t+1}{y+y^2} \). This equation can be separated by placing all terms involving \( y \) on one side and those involving \( t \) on the other.

Step 2: Separate the variables

Rearrange the equation so that all terms involving \( y \) are on one side and \( t \) on the other: \( (y+y^2)\, dy = (t+1)\, dt \).

Step 3: Integrate both sides

Integrate both sides of the separated equation: \( \int (y + y^2) \, dy = \int (t + 1) \, dt \).

Step 4: Perform integration

Calculate the integrals: \( \int (y + y^2) \, dy = \frac{1}{2}y^2 + \frac{1}{3}y^3 + C_1 \) and \( \int (t + 1) \, dt = \frac{1}{2}t^2 + t + C_2 \).

Step 5: Combine the integrals

Equate the expressions obtained from the integration: \( \frac{1}{2}y^2 + \frac{1}{3}y^3 + C_1 = \frac{1}{2}t^2 + t + C_2 \).

Step 6: Apply the initial condition

Use the initial condition \( y(0) = 1 \) to find the constant. Substitute \( y = 1 \) and \( t = 0 \) into the combined integral equation: \( \frac{1}{2}(1)^2 + \frac{1}{3}(1)^3 + C_1 = \frac{1}{2}(0)^2 + 0 + C_2 \).

Step 7: Solve for constants

Calculate \( C_1 - C_2 = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \). Substitute back: \( C_1 = C_2 + \frac{5}{6} \).

Step 8: State the implicit solution

Substitute \( C_1 \) in the combined equation: \( \frac{1}{2}y^2 + \frac{1}{3}y^3 = \frac{1}{2}t^2 + t + C_2 + \frac{5}{6} \). This implicit expression relates \( t \) and \( y \).

Key Concepts

Separation of Variables

Separation of variables is a fundamental technique for solving differential equations. If you face a differential equation like \( \frac{dy}{dt} = \frac{t+1}{y+y^2} \), your task is to rearrange the equation in such a way that all terms involving \( y \) are isolated on one side and those involving \( t \) are on the opposite side. The equation should then take a form like \( (y+y^2)\, dy = (t+1)\, dt \).The strategy is to see the equation as a kind of balance: everything related to \( y \) goes on one side, and everything related to \( t \) goes to the other. This allows us to deal with each side independently. To achieve this separation, carefully manipulate the terms by using algebraic operations like multiplication, division, and factoring, whenever necessary.
It's like dividing up the problem into two simpler parts, where each part solely depends on one variable. This makes it easier to tackle because it converts the problem into the form that can be solved by direct integration later on.

Integration

Once the variables are separated, the next step is to integrate both sides of the differential equation. Given the form \( \int (y + y^2) \, dy = \int (t + 1) \, dt \), you integrate each side independently.When integrating the side with respect to \( y \), you need to remember the power rule for integration:

• The integral of \( y \) is \( \frac{1}{2}y^2 \).
• The integral of \( y^2 \) is \( \frac{1}{3}y^3 \).

So you get: \( \int (y + y^2) \, dy = \frac{1}{2}y^2 + \frac{1}{3}y^3 + C_1 \).For the \( t \)-side, the calculations are similar:

• The integral of \( t \) is \( \frac{1}{2}t^2 \).
• The integral of \( 1 \) is \( t \).

Thus, \( \int (t + 1) \, dt = \frac{1}{2}t^2 + t + C_2 \).Combining both integrals gives us one equation where we can relate \( y \) and \( t \):\[ \frac{1}{2}y^2 + \frac{1}{3}y^3 + C_1 = \frac{1}{2}t^2 + t + C_2 \]. This process transforms a complicated-looking differential equation into an easier problem through straightforward calculus techniques.

Initial Conditions

Applying the initial conditions is a vital final step when solving differential equations. Let's consider our example where the initial condition is given as \( y(0) = 1 \).
In simple terms, initial conditions provide specific values to solve for unknowns like constants that show up in the integration process. They help generate a unique solution from the general form of the equation.To use the initial condition, simply substitute the values given for \( y \) and \( t \) into the integral equation already formed. For example:\( y = 1 \) and \( t = 0 \) are plugged into \[ \frac{1}{2}y^2 + \frac{1}{3}y^3 + C_1 = \frac{1}{2}t^2 + t + C_2 \].This simplifies to: \[ \frac{1}{2}(1)^2 + \frac{1}{3}(1)^3 + C_1 = 0 + C_2 \].Calculate it step by step as:

• \( \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \).

Thus, you find \( C_1 = C_2 + \frac{5}{6} \). Finally, you will substitute back for \( C_1 \) in your integral equation, providing the relationship between \( t \) and \( y \) configured to meet the specific initial condition requirements. This key step ensures that the solution reflects the initial set values, concluding the process.