Problem 9 - Technical Mathematics with Calculus

Question

Solve each differential equation, including evaluation of the constant of integration.$$y^{\prime}=\sqrt{x}, \text { passes through }(2,4)$$

Step-by-Step Solution

Answer

The solution of the differential equation, given that the curve passes through (2, 4), is \(y = \frac{2}{3}x^{\frac{3}{2}} + \frac{8}{3}\).

Step 1: Integrate Both Sides

Integrate the differential equation \(y' = \sqrt{x}\) with respect to \(x\) to find \(y\). The integral of \(y'\) with respect to \(x\) is \(y\), and the integral of \(\sqrt{x}\) with respect to \(x\) is \(\frac{2}{3}x^{\frac{3}{2}}\) since \(\int x^{n} dx = \frac{x^{n+1}}{n+1}\) when \(n \eq -1\).

Step 2: Add the Constant of Integration

After integration, include the constant of integration, denoted by \(C\). The equation after integrating becomes \(y = \frac{2}{3}x^{\frac{3}{2}} + C\).

Step 3: Apply the Initial Condition

Substitute the given point (2, 4) into the equation to find the value of \(C\). By setting \(x=2\) and \(y=4\), we solve for the constant \(C\).

Step 4: Solve for the Constant of Integration

Plugging the values into the equation, we get \(4 = \frac{2}{3}(2)^{\frac{3}{2}} + C\). Simplify this to find the value of \(C\).

Step 5: Write the Final Solution

After finding \(C\), substitute it back into the equation to get the final solution for \(y\).

Key Concepts

Integration Techniques

Integration is a fundamental technique in calculus, used to find the function when its derivative is known, effectively reversing differentiation. In the context of differential equations, integrating both sides is a common approach to find a solution. The integration of power functions involves a simple rule: when integrating a function of the form \( x^n \) with respect to \( x \), the result is \( \frac{x^{n+1}}{n+1} \), provided that \( n eq -1 \).

In our exercise, we apply this rule to integrate \( \sqrt{x} \), which is \( x^{1/2} \). Applying the power rule gives us \( \frac{2}{3}x^{3/2} \), since \( \frac{1}{2} + 1 = \frac{3}{2} \) and the reciprocal of \( \frac{3}{2} \) is \( \frac{2}{3} \). It's crucial for students to familiarize themselves with integration techniques, as they form the backbone for solving various types of differential equations.

Constants of Integration

When integrating any function, we must always add a constant of integration, denoted by \( C \). This constant accounts for the fact that the antiderivative of a function is not unique; rather, there is a family of antiderivatives, all differing by a constant.

For instance, in our step-by-step solution, after integrating \( \sqrt{x} \), we expressed the integrated function as \( \frac{2}{3}x^{3/2} + C \). The \( C \) is essential because for any value of \( C \), the derivative of \( \frac{2}{3}x^{3/2} + C \) with respect to \( x \) will give us back \( \sqrt{x} \). Understanding how and why to include the constant of integration is a key concept in calculus that ensures the completeness and accuracy of the integration process.

Initial Value Problems

An initial value problem is a type of differential equation along with a specific condition (initial condition) that the solution must satisfy. This condition allows us to find the specific solution from the family of solutions by determining the constant of integration \( C \).

In our exercise, the initial condition is given by the point (2, 4) through which the curve passes. By substituting these values for \( x \) and \( y \) into the integrated equation, we solve for \( C \) to find the specific curve that not only fits the general differential equation but also goes through the given point. Initial value problems are of great importance in fields such as physics and engineering, where solutions must often satisfy specific starting conditions.