Question

Solve the initial value problem. $$ y^{\prime}-4 y=\frac{48 x}{y^{2}}, \quad y(0)=1 $$

Short answer

Based on the step by step solution provided above, answer the following question: **Question:** Find the solution to the initial value problem given by the differential equation $$y^{\prime} - 4y = \frac{48x}{y^2}, \quad y(0) = 1$$. **Answer:** The solution to the initial value problem is $$y(x) = e^{4x}$$.

Step-by-step solution

Identify the linear part of the differential equation

The given differential equation is of the form: $$ y^{\prime} - 4y = \frac{48x}{y^2} $$ The linear part of the equation is: $$ y^{\prime} - 4y = 0 $$

Find the integrating factor

For the linear part, \(y^{\prime} - 4y = 0\), we will find the integrating factor. The integrating factor is given by $$ \mu(x) = e^{\int{-4}dx} = e^{-4x} $$

Multiply the linear part with the integrating factor

Now, we will multiply the integrating factor with the linear part of the equation: $$ e^{-4x}(y^{\prime} - 4y) = 0 $$ This gives us $$ (e^{-4x}y)^{\prime} = 0 $$

Integrate both sides

Now, we will integrate both sides with respect to x: $$ \int(e^{-4x}y)^{\prime} dx = \int 0 dx $$ Which gives us $$ e^{-4x}y = C $$ Where C is the integration constant.

Solve for y

Now we solve for y: $$ y(x) = Ce^{4x} $$ Now, to find the particular solution, we will use the non-linear part of the equation: $$ y^{\prime} - 4y = \frac{48x}{y^2} $$ Replace y with the general solution we found: $$ (Ce^{4x})^{\prime} - 4(Ce^{4x}) = \frac{48x}{(Ce^{4x})^2} $$

Solve for C using the initial condition

To find the constant C, we will use the initial condition: $$ y(0) = 1 $$ So, replacing x with 0 and y(0) with 1, we get: $$ 1 = Ce^{4(0)} $$ Which gives us: $$ C = 1 $$

Write down the particular solution

Now we can write down the particular solution with the value of C: $$ y(x) = e^{4x} $$ So, the solution to the initial value problem is: $$ y(x) = e^{4x} $$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. These equations help describe various physical phenomena such as heat, motion, and fluid dynamics. In the problem given, we have a first-order differential equation, which means it involves the first derivative of the unknown function, which in this case is \( y' \).

• The general form of a first-order linear differential equation is \( y' + p(x)y = g(x) \), where \( p(x) \) and \( g(x) \) are continuous functions.
• In our exercise, the equation is slightly more complex due to the \( \frac{48x}{y^2} \) term, but it still features the crucial linear part \( y' - 4y = 0 \).

This equation guides us toward finding a solution that satisfies both the differential equation and the given initial condition. Differential equations are employed in numerous fields such as engineering, physics, and economics to model and solve real-world problems.

Integrating Factor

An integrating factor is a function used to solve linear differential equations. It simplifies the equation, making it easier to integrate. For a differential equation of the form \( y' + p(x)y = g(x) \), the integrating factor \( \mu(x) \) is defined by the exponential function \( \mu(x) = e^{\int p(x) dx} \).

• In our case, the differential equation includes \( y' - 4y = 0 \), giving us \( p(x) = -4 \).
• The integrating factor then becomes \( \mu(x) = e^{-4x} \).
• By multiplying both sides of our differential equation by this integrating factor, we obtain a simplified form that can be readily integrated.

The integrating factor technique is especially powerful because it transforms differential equations into a solvable form through the process of integration, leading us directly toward finding a general solution.

Particular Solution

A particular solution of a differential equation satisfies both the equation itself and the initial condition given. The initial condition is a specific point where the solution must pass, often provided as \( y(x_0) = y_0 \). Once we have the general solution, we apply the initial condition to find the specific values of any constants involved.

• For our particular problem, the general solution is \( y(x) = Ce^{4x} \).
• We use the initial condition \( y(0) = 1 \) to determine \( C \).
• Plugging these into the equation, we solve \( 1 = Ce^{0} \), leading us to find \( C = 1 \).

Thus, the particular solution for our initial value problem is \( y(x) = e^{4x} \), which means that this specific solution meets the criteria of the differential equation under the initial condition constraint.