Question

Verify the following general solutions and find the particular solution.Find the particular solution to the differential equation \(y^{\prime}=3 x^{3}\) that passes through \((1,4.75)\), given that \(y=C+\frac{3 x^{4}}{4}\) is \(a\) general solution.

Short answer

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

Step-by-step solution

Understand the General Solution

First, we recognize that the given general solution is \(y = C + \frac{3x^4}{4}\), where \(C\) is an arbitrary constant. This represents the family of solutions for the differential equation \(y' = 3x^3\).

Differentiate the General Solution

Differentiate \(y = C + \frac{3x^4}{4}\) with respect to \(x\) to verify it solves the differential equation. The derivative is \(y' = \frac{d}{dx}\left(C + \frac{3x^4}{4}\right) = 0 + \frac{12x^3}{4} = 3x^3\). This confirms that our general solution is correct since \(y' = 3x^3\).

Substitute the Given Point

To find the particular solution, substitute the point \((1,4.75)\) into the general solution \(y = C + \frac{3x^4}{4}\). This substitution gives:\[ 4.75 = C + \frac{3(1)^4}{4} = C + \frac{3}{4}. \]

Solve for Constant C

Solve for \(C\) by rearranging the equation from Step 3:\[ C = 4.75 - \frac{3}{4} = 4.75 - 0.75 = 4. \]

Write the Particular Solution

Substitute \(C = 4\) back into the general solution equation, giving us the particular solution:\[ y = 4 + \frac{3x^4}{4}. \]

Key concepts

Particular Solution

A particular solution is a unique solution to a differential equation that satisfies a specific condition or passes through a given point. In our example, the given specific condition is the point

• he solution curve passes through he point expressed as "(1,4.75)".

To find the particular solution, substitute this point into the general solution of the differential equation.
This typically involves finding the value of the arbitrary constant, denoted as "C" here, so that the solution coincides with the given point.
For our problem, after substituting and simplifying, the constant is determined to be 4, leading to the particular solution:

• y = 4 + \(\frac{3x^4}{4}\).

Every time you have a specific condition like a point, you're looking for a particular solution.

General Solution

The general solution of a differential equation encompasses an entire family of possible solutions. These solutions are usually expressed in terms of an arbitrary constant, which represents the different solutions depending on specific conditions.
In our case, the general solution is given as

• y = C + \(\frac{3x^4}{4}\).

Here, "C" is an arbitrary constant that can take any value, representing numerous curves each a solution to the differential equation \(y' = 3x^3\).
To verify the general solution, differentiate it with respect to \(x\), and ensure it simplifies back to the original differential equation.
Through differentiation, any valid general solution will return the original form of the differential equation, affirming its correctness.

Derivative

Understanding derivatives is essential in solving differential equations. They represent the rate of change or slope of a curve at any given point.

• In differential equations, these are denoted as \(y'\) or \(dy/dx\).

The derivative plays a central role, as demonstrated in our given problem, where we compute the derivative of the general solution:

• y = C + \(\frac{3x^4}{4}\).

We differentiate with respect to \(x\) to get

• y' = \(\frac{d}{dx}(C + \frac{3x^4}{4}) = 0 + \frac{12x^3}{4} = 3x^3\).

This confirms that our general solution satisfies the given differential equation \(y' = 3x^3\).
Working with derivatives involves using basic rules, such as the power rule, constant rule, etc., making it a fundamental step in understanding and solving differential equations.

Initial Value Problem

An Initial Value Problem (IVP) includes both a differential equation and an initial condition that specifies the value of the unknown function at a given point. This type of problem helps in identifying a particular solution from the general solution.
In this case, the IVP is defined by the differential equation \(y' = 3x^3\) and the initial condition \(y(1) = 4.75\).

• The initial condition implies that at \(x = 1\), the value of \(y\) is 4.75.

This additional information narrows down the general solution to a specific curve that passes through the point (1, 4.75).
Solving an IVP involves substituting the initial condition into the general solution, allowing you to resolve the constant values and outline the particular solution uniquely fitting the criteria given by the problem.