Question

Find the particular solution to the differential equation \(y^{\prime}=3 x^{2} y\) that passes through \((0,12)\), given that \(y=C e^{x^{3}}\) is a general solution.

Short answer

The particular solution is \(y = 12 e^{x^3}\).

Step-by-step solution

Write the General Solution

We are given that the general solution to the differential equation \(y' = 3x^2 y\) is \(y = C e^{x^3}\). This means that for the general case, the solution depends on the constant \(C\).

Apply Initial Condition

To find the particular solution, we use the initial condition provided: the solution must pass through the point \((0,12)\). This means that when \(x = 0\), \(y\) should equal 12. Substitute these values into the general solution: \(12 = C e^{0^3}\).

Simplify the Equation

Since \(e^{0^3} = e^0 = 1\), the equation simplifies to \(12 = C \cdot 1\), which means \(C = 12\).

Write the Particular Solution

Substitute \(C = 12\) back into the general solution to get the particular solution. Hence, the particular solution to the differential equation is \(y = 12 e^{x^3}\).

Key concepts

Understanding a Particular Solution

In the world of differential equations, finding a particular solution refers to identifying a single, specific solution that satisfies not only the differential equation itself but also one or more additional conditions known as initial conditions. A differential equation can have countless solutions, collectively referred to as the general solution. The general solution contains arbitrary constants. When the arbitrary constants are replaced with specific values derived from initial conditions, we are left with one specific or particular solution.

In the context of our exercise, we start with the general solution of the differential equation, given as \(y = C e^{x^3}\). The presence of the constant \(C\) indicates that there are many solutions, each representing a different situation or initial condition.

• The particular solution is found by substituting the initial conditions into this general formula.
• In our example, the initial condition that the curve passes through the point \((0, 12)\) allows us to determine the specific value of \(C\), yielding a solution unique to this particular problem.

The Role of Initial Conditions

Initial conditions are vital in solving differential equations because they help pin down exactly one solution from the infinite possibilities in a general solution. Typically, an initial condition provides a specific value that the solution must satisfy at a certain point.

For this exercise, the initial condition is that the solution passes through the point \((0, 12)\). This means that when \(x = 0\), the value of \(y\) must be 12. By substituting these values into the general solution \(y = C e^{x^3}\), we can solve for the constant \(C\).

• First, substitute \(x = 0\) and \(y = 12\) into the equation to get \(12 = C e^{0}\).
• Since \(e^0\) equals 1, it simplifies the equation to \(12 = C \).
• This shows that \(C = 12\), which specifies our particular solution.

Initial conditions tell us exactly what particular scenario we are analyzing or what constraint we wish the solution to satisfy.

Exploring Exponential Functions

Exponential functions are a cornerstone in solving various differential equations, especially those with growth and decay models. These functions are generally expressed in the form \(y = Ce^{k x}\), where \(e\) is the base of natural logarithms, approximately equal to 2.718. In our solution, the component \(e^{x^3}\) makes the contribution of the exponential function quite dominant.

Exponential functions grow rapidly or decay based on the exponent. The presence of \(e^{x^3}\) in the general solution \(y = C e^{x^3}\) suggests that the function's rate of increase is influenced by \(x^3\).

• This form takes advantage of the unique properties of exponentials: nonzero derivatives and the fact that their derivatives are proportional to the function itself.
• In our case, the exponential not only shapes the behavior of the solution, but also its application to modeling real-world scenarios such as radioactive decay, population growth, and atmospheric pressure changes where similar rapid changes might be observed.

Understanding exponential functions helps interpret how solutions change over the domain of interest.