Question

Verify the following general solutions and find the particular solution.Find the particular solution to the differential equation \(y^{\prime}=4 x^{2}\) that passes through \((-3,-30)\), given that \(y=C+\frac{4 x^{3}}{3}\) is a general solution.

Short answer

The particular solution is \(y = 6 + \frac{4x^3}{3}\).

Step-by-step solution

Verify the General Solution

First, we need to verify the general solution. The given differential equation is \(y' = 4x^2\). We also have that the general solution is \(y = C + \frac{4x^3}{3}\). To verify, differentiate the general solution with respect to \(x\): \(\frac{d}{dx}\left(C + \frac{4x^3}{3}\right) = 4x^2 \). This matches the original differential equation \(y' = 4x^2\), so the general solution is verified.

Apply Initial Condition

Now we use the given point \((-3, -30)\) to find the particular solution. Substitute \(x = -3\) and \(y = -30\) into the general solution equation: \(-30 = C + \frac{4(-3)^3}{3}\).

Simplify and Solve for C

Calculate \((-3)^3 = -27\), so \(\frac{4(-27)}{3} = -36\). Thus, the equation becomes \(-30 = C - 36\). Solve for \(C\) by adding 36 to both sides: \(-30 + 36 = C\). This gives \(C = 6\).

Write the Particular Solution

Now that we have \(C = 6\), substitute it back into the general solution to get the particular solution: \(y = 6 + \frac{4x^3}{3}\).

Key concepts

General Solution Verification

A general solution of a differential equation is an equation that encompasses all possible solutions for that equation. To verify it, we need to differentiate it and check if it matches the original differential equation. In this problem, we began with the differential equation:

• \( y' = 4x^2 \)
• General solution given as \( y = C + \frac{4x^3}{3} \)

To verify, differentiate the general solution with respect to \( x \):

• \( \frac{d}{dx}\left(C + \frac{4x^3}{3}\right) = 4x^2 \)

This matches the original differential equation \( y' = 4x^2 \), confirming that our general solution is indeed correct. This step ensures that the derivative of the general solution aligns perfectly with the given differential equation.

Particular Solution

Once we have a verified general solution, finding a particular solution involves determining the specific constant \( C \) that satisfies given initial conditions. In this case, we need the solution to pass through the point \( (-3, -30) \). To find \( C \):

• Substitute \( x = -3 \) and \( y = -30 \) into the general solution equation: \(-30 = C + \frac{4(-3)^3}{3}\).
• Calculate \((-3)^3 = -27\), then \( \frac{4(-27)}{3} = -36 \). So the equation becomes \(-30 = C - 36\).
• Solve for \( C \) by adding 36 to both sides: \(-30 + 36 = C\), which gives \( C = 6 \).

Substitute \( C = 6 \) back into the general solution to get the particular solution: \( y = 6 + \frac{4x^3}{3} \). This particular solution is tailored for the point \( (-3, -30) \) and satisfies the differential equation and the initial condition.

Initial Conditions

Initial conditions are key to deriving particular solutions from a general solution. They provide specific values for the function or its derivatives at certain points. In this exercise:

• The initial condition was specified as the point \( (-3, -30) \).
• This condition helps to determine the unique constant \( C \) needed for a particular solution.
• By substituting this point into the general solution, we find the specific \( C \) such that the particular solution passes through the given point.

Hence, initial conditions are crucial for pinpointing one precise solution among the infinite possibilities represented by the general solution.

Integration of Polynomials

Integration of polynomials is fundamental when solving differential equations, especially when deriving general solutions. To integrate a polynomial, apply the power rule: add one to the power and divide by the new power. In our problem:

• Consider the polynomial \( 4x^2 \) from the differential equation \( y' = 4x^2 \).
• Integrate it to find the general solution: \( \int 4x^2 \, dx = \frac{4x^3}{3} + C \).

This integration step yields a term \( \frac{4x^3}{3} \), which is part of the general solution. Remember, integration adds a constant \( C \) which reflects the family of solutions, signifying why it is necessary to use initial conditions to find a particular solution.