91Ó°ÊÓ

In solving differential equations, initial conditions are crucial, as they help to determine the particular solution among a family of solutions. In the given problem, the initial condition is specified as \(y(1) = 5\). This means that when \(x = 1\), the function \(y\) should equal 5. This pinpoints a precise solution on the curve that fits both the differential equation and the specific starting point.Without initial conditions, we can only find a general solution. Initial conditions let us fold in additional real-world constraints or specific requirements. Imagine initial conditions as signposts that guide us to the right path on a maze of possible solutions. In our problem, applying the initial condition means taking the solution after integration and ensuring any constants (unknowns) are adjusted so that \(y(1) = 5\) holds true.

When dealing with first-order separable differential equations, integrating both sides of the equation is often required to solve the problem. In this problem, you've got to integrate \( \frac{1}{x^2} \) with respect to \(x\). This involves using the power rule for integration.The Power Rule for Integration
The power rule states that the integral of \( x^{-n} \) is \( \frac{x^{-n+1}}{-n+1} + C \). In our case, \( n = 2 \), so the integral becomes \( -\frac{1}{x} + C \). Here, you simplify the expression \( x^{-2} \) by increasing the exponent by one and dividing by the new exponent, making sure to tack on the constant of integration \( C \) at the end.These integration techniques allow us to transform a differential equation into a direct expression for \( y \), making it manageable to apply the initial conditions and solve for constants.

Every time you integrate, a constant of integration \( C \) is added to your solution. In the context of differential equations, this constant reflects the family of solutions generated by indefinite integration.Why is \( C \) Important?

• When you integrate, you’re finding an antiderivative that fits a wide array of possible functions.
• The constant \( C \) represents variations in those functions consistent with the differential equation.

To pin down the exact value of \( C \), we use the initial condition, such as \( y(1) = 5 \). With \( y = -\frac{1}{x} + C \), substituting \( x = 1 \) and \( y = 5 \) helps us isolate \( C \) to find \( C = 6 \).The constant, therefore, personalizes the solution to fit the specific scenario laid out by the initial conditions, allowing us to shift from a general solution to one that precisely meets the problem's requirements.