Question

Find the particular solution to the differential equation \(y^{\prime} x^{2}=y\) that passes through \(\left(1, \frac{2}{e}\right)\), given that \(y=C e^{-1 / x}\) is a general solution.

Short answer

The particular solution is \( y = 2 e^{-1/x} \).

Step-by-step solution

Write Down the General Solution

The general solution to the differential equation is given as \( y = C e^{-1/x} \).

Apply the Initial Condition

Use the initial condition provided in the problem, which is \( y(1) = \frac{2}{e} \). Substitute \( x = 1 \) and \( y = \frac{2}{e} \) into the general solution: \[ \frac{2}{e} = C e^{-1/1} = C e^{-1}. \]

Solve for the Constant C

To find \( C \), solve the equation from Step 2: \[ \frac{2}{e} = C e^{-1} \]. Divide both sides by \( e^{-1} \) to isolate \( C \): \[ C = \frac{2}{e} \cdot e = 2. \]

Write the Particular Solution

Substitute \( C = 2 \) back into the general solution \( y = C e^{-1/x} \) to obtain the particular solution: \[ y = 2 e^{-1/x}. \]

Key concepts

Particular Solution

A particular solution to a differential equation is one that satisfies not only the differential equation itself but also a given initial condition. In simpler terms, while a general solution provides a family of possible solutions, a particular solution pinpoints a singular solution by using specific initial values.
In the exercise provided, we identified the particular solution by starting with the general form, which is given as \( y = C e^{-1/x} \).
We then applied the initial condition \( y(1) = \frac{2}{e} \) to find the constant \( C \), resulting in the particular solution \( y = 2 e^{-1/x} \). This solution specifically accommodates the initial condition, aligning perfectly with the constraint set by the given point.

General Solution

The general solution of a differential equation is a solution that includes all possible solutions of that equation. It usually contains one or more constants, such as \( C \), which represent an entire set of solutions or a "family" of curves.
For the differential equation in question \( y' x^2 = y \), the given general solution is \( y = C e^{-1/x} \).
This formula represents a continuous family of solutions where each value of \( C \) generates a different curve, allowing flexibility to meet various initial conditions. The beauty of a general solution lies in its adaptability to different scenarios by tweaking the constant values based on specific constraints.

Initial Condition

An initial condition in the context of differential equations acts as a guideline to help zero in on a specific solution from a family of solutions provided by the general solution. It is a specific point on the solution curve that the particular solution must pass through.
In our problem, the initial condition is \( y(1) = \frac{2}{e} \). This condition tells us that when \( x = 1 \), the value of \( y \) should be \( \frac{2}{e} \).
This piece of information allows us to determine the constant \( C \) in the general solution equation. By substituting the initial condition into the general solution, we found that \( C = 2 \). This transforms the general solution to a particular one which precisely meets the initial condition, offering a specific solution perfectly tailored to this point.