Problem 52 - Calculus for Biology and Medicine

Question

Find the general solution of the differential equation. $$ \frac{d y}{d t}=t^{2}\left(1+t^{2}\right), t \geq 0 $$

Step-by-Step Solution

Answer

The general solution is \( y = \frac{t^3}{3} + \frac{t^5}{5} + C \).

Step 1: Identify the Type of Differential Equation

This is a first-order ordinary differential equation. It can be solved by integration since the equation is separable.

Step 2: Separate the Variables

Write the equation in a form where all terms involving just \( y \) and \( dy \) are on one side and all terms involving just \( t \) and \( dt \) are on the other. The given equation is \(\frac{d y}{d t} = t^2(1 + t^2)\). This can be rewritten as \( dy = t^2(1 + t^2) dt \).

Step 3: Integrate Both Sides

Integrate both sides of the equation separately. The left-hand side is \( \int dy = y + C_1 \). The right-hand side is \( \int t^2(1 + t^2) dt \).

Step 4: Simplify and Evaluate the Integral

First expand the integrand on the right: \( t^2(1 + t^2) = t^2 + t^4 \). Now, integrate term-by-term: \[ \int t^2 dt + \int t^4 dt = \frac{t^3}{3} + \frac{t^5}{5} + C_2 \].

Step 5: Combine the Results

Combine the expressions from Step 3 to get the general solution. We equate \( y + C_1 = \frac{t^3}{3} + \frac{t^5}{5} + C_2 \). For a single constant, let \( C = C_2 - C_1 \), then the general solution is \( y = \frac{t^3}{3} + \frac{t^5}{5} + C \).

Key Concepts

First-order Differential Equations

First-order differential equations involve the first derivative of a function and are generally written with the derivative \( \frac{dy}{dt} \) or similar notations. These equations are called 'first-order' because the highest derivative in the equation is of the first order.
The equation provided, \( \frac{dy}{dt} = t^2(1 + t^2) \), is an example of a first-order differential equation. This means the change rate of the function \( y \) with respect to time \( t \) is described by this equation.
First-order differential equations arise in many natural processes, such as population growth and radioactive decay. Solving these equations helps predict how a quantity of interest changes over time.
There are various methods to solve first-order differential equations, and the one chosen often depends on the form and complexity of the equation. In this case, we can use 'separable equations' as the equation structure supports this method.

Separable Equations

Separable differential equations are a special class where you can split the variables, allowing separation of terms involving \( y \) from those involving \( t \). This is crucial because it allows the use of basic integration techniques to find a solution.
In our example, the equation \( \frac{dy}{dt} = t^2(1 + t^2) \) is separable. We separated variables by rewriting it as \( dy = t^2(1 + t^2) dt \). This rearrangement enables the integration of each side independently.

• Place all terms with \( y \) on one side of the equation (in this case, \( dy \)).
• And place all terms with \( t \) (here, \( t^2(1 + t^2) dt \)) on the other side.

Once separated, each side of the equation can be integrated independently, facilitating the solving process.

Integration Methods

After separating the variables, the next essential step in solving a separable equation is to integrate both sides to derive the function. Integration is a mathematical process used to find functions given their derivatives.
For the left-hand side of our example \( \int dy \), integration is straightforward, resulting in \( y + C_1 \), where \( C_1 \) is a constant of integration. On the right-hand side, \( \int t^2(1 + t^2) dt \), we simplify by expanding the expression: \( t^2 + t^4 \).
To solve \( \int t^2 dt + \int t^4 dt \), we employ basic integration rules:

• For \( \int t^n dt \), use \( \frac{t^{n+1}}{n+1} \).

So we get:\[ \int t^2 dt = \frac{t^3}{3} \] \[ \int t^4 dt = \frac{t^5}{5} \]
Combining these results gives us \( \frac{t^3}{3} + \frac{t^5}{5} + C_2 \).
The general solution to our differential equation is then formed by combining these integrated results and the constants: \( y = \frac{t^3}{3} + \frac{t^5}{5} + C \), where \( C \) incorporates all constant terms from integration.