Problem 3 - Active Calculus

Question

Find the solution to the differential equation $$ \begin{array}{l} \frac{d y}{d t}=y^{2}(5+t), \\ y=5 \text { when } t=1 . \end{array} $$

Step-by-Step Solution

Answer

The solution is \( y = -\frac{1}{\frac{5}{2} t^2 + 5 t - \frac{11}{2}}\).

Step 1: Identify the Type of Differential Equation

Recognize that the given equation \(\frac{d y}{d t}=y^{2}(5+t)\) is a separable differential equation because the variables \(y\) and \(t\) can be separated to different sides of the equation.

Step 2: Separate the Variables

Rewrite the equation by separating the variables: \(\frac{1}{y^2} \frac{d y}{dt} = 5 + t\). This can be further simplified to \(\frac{1}{y^2} d y = (5 + t) dt\).

Step 3: Integrate Both Sides

Integrate both sides to find \(\frac{-1}{y} = \frac{5}{2} t^2 + 5 t + C\). Integrate \(\frac{1}{y^2} d y\) to get \(-\frac{1}{y}\) and integrate \((5 + t) dt\) to get \(\frac{5}{2} t^2 + 5 t + C\).

Step 4: Solve for the Constant of Integration

Use the initial condition \(y = 5\) when \(t = 1\) to solve for \(C\). Substitute \(y = 5\) and \(t = 1\) into \(\frac{-1}{y} = \frac{5}{2} t^2 + 5 t + C\) to get \(\frac{-1}{5} = \frac{5}{2} (1)^2 + 5(1) + C\). Simplifying this will give \(C = \frac{-11}{2}\).

Step 5: Construct the General Solution

Substitute \(C\) back into the integrated equation to get \(\frac{-1}{y} = \frac{5}{2} t^2 + 5 t - \frac{11}{2}\).

Step 6: Solve for \(y\)

Finally, solve for \(y\) by taking the reciprocal of both sides and simplifying. \(y = -\frac{1}{\frac{5}{2} t^2 + 5 t - \frac{11}{2}}\).

Key Concepts

Separable Differential Equations

In differential equations, a separable differential equation is one that can be written in the form \(\frac{dy}{dt} = g(y) h(t)\). This means that the differential equation can be separated into two parts: one involving only the dependent variable \(y\) and its differential \(dy\), and the other involving only the independent variable \(t\) and its differential \(dt\).

Initial Conditions in Differential Equations

When solving differential equations, initial conditions are very important. They are used to find the exact solution to a differential equation. Initial conditions provide specific values for the variables at a particular point. This helps to determine the constants of integration that appear after integrating the differential equation.

In the provided exercise, the initial condition is given as \(y = 5\) when \(t = 1\). This means that at time \(t = 1\), the value of \(y\) is 5. These conditions help to find the constant \(C\) after integrating the equation.

Integration of Differential Equations

To solve a separable differential equation, we integrate both sides of the separated equation. After separating the variables, integrate both sides to find the solution.

For example, the equation \(\frac{1}{y^2} dy = (5 + t) dt\) was integrated to obtain:

• \( -\frac{1}{y} = \frac{5}{2}t^2 + 5t + C\)

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Here:

• The left side \(-\frac{1}{y}\) is the result of integrating \(\frac{1}{y^2} dy\).
• The right side is the result of integrating \((5 + t) dt\), which gives \( \frac{5}{2} t^2 + 5t + C \).

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Using the initial condition to determine \( C\), the final step is to plug this value back into the equation, which allows us to solve for \(y\). This provides the complete and specific solution to the differential equation.