Problem 47 - Calculus for Biology and Medicine

Question

In Problems, solve each differential equation with the given initial condition. $$ \frac{d y}{d x}=\frac{x+1}{y}, \text { with } y(0)=2 $$

Step-by-Step Solution

Answer

The solution to the differential equation is \( y^2 = x^2 + 2x + 4 \).

Step 1: Separate Variables

To solve the differential equation \( \frac{d y}{d x} = \frac{x+1}{y} \), first separate the variables by multiplying both sides by \( y \, dx \). This yields: \( y \, dy = (x+1) \, dx \).

Step 2: Integrate Both Sides

Integrate both sides of the equation separately. The left-hand side becomes \( \int y \, dy \) and the right-hand side becomes \( \int (x+1) \, dx \).

Step 3: Solve Integrals

Evaluate the integrals: \( \int y \, dy = \frac{y^2}{2} + C_1 \) and \( \int (x+1) \, dx = \frac{x^2}{2} + x + C_2 \).

Step 4: Combine & Simplify

Combine the results: \( \frac{y^2}{2} = \frac{x^2}{2} + x + C \) where \( C = C_2 - C_1 \). Multiply the entire equation by 2 to clear the fraction, obtaining: \( y^2 = x^2 + 2x + 2C \).

Step 5: Apply Initial Condition

Apply the initial condition \( y(0) = 2 \) to find \( C \). Substitute \( x = 0 \) and \( y = 2 \) into \( y^2 = x^2 + 2x + 2C \) to obtain: \( 4 = 0^2 + 2(0) + 2C \).

Step 6: Solve for the Constant

From the equation \( 4 = 2C \), solve for \( C \) to get \( C = 2 \).

Step 7: Final Solution

Substitute \( C = 2 \) back into the equation \( y^2 = x^2 + 2x + 2C \), resulting in the final solution: \( y^2 = x^2 + 2x + 4 \).

Key Concepts

Initial Condition

In differential equations, an initial condition helps us determine the specific solution that fits a given scenario. Considering any family of solutions to a differential equation, the initial condition allows us to pinpoint one particular solution among them.
In our exercise, the initial condition is given as \( y(0) = 2 \). This means when \( x = 0 \), \( y \) must equal 2.
To apply the initial condition, we plug the values into the solution after separating and integrating the equation. By substituting \( x = 0 \) and \( y = 2 \) into the developed equation, we find the constant \( C \), which allows us to adjust our solution to fit the initial condition. This step ensures that our solution not only solves the general differential equation but also fits the specific situation given.

Separation of Variables

Separation of variables is a method used to solve differential equations by rearranging terms so that each variable appears on a different side of the equation. This approach is applicable specifically to equations that can be manipulated into a form where one side of the equation involves only one variable and its derivative.
Let's illustrate how this applies to our example: Starting with \( \frac{d y}{d x} = \frac{x+1}{y} \), the strategy lies in isolating \( dy \) with \( y \) and \( dx \) with \( x \). We achieve this by multiplying both sides of the equation by \( y \, dx \), leading to \( y \, dy = (x+1) \, dx \).
Thus,

• Left side solely involves \( y \) and \( dy \)
• Right side solely involves \( x \) and \( dx \)

Successfully applying this technique simplifies the integration step, as the separate integrals can then be handled independently.

Integration

Integration is a fundamental step in resolving the rearranged differential equation. By computing the integral of each side independently, one can decipher the relationship between the variables.
For the separated equation \( y \, dy = (x+1) \, dx \), we integrate both sides:

• On the left, \( \int y \, dy \) integrates to \( \frac{y^2}{2} + C_1 \)
• On the right, \( \int (x+1) \, dx \) integrates to \( \frac{x^2}{2} + x + C_2 \)

Combining these results gives us \( \frac{y^2}{2} = \frac{x^2}{2} + x + C \), where \( C = C_2 - C_1 \). This simplification indicates that integration offers us a general solution.
Upon substituting the initial condition, integration also aids in determining the exact form of this solution by solving for the constant \( C \) and ensuring the equation suits specific initial values. This process underscores the utility of integration in solving differential equations and its role in marrying the theoretical with the practical.