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Chapter 2: Problem 11

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

Short Answer

Expert verified
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

01

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: $$ y^{\prime} - 4y = 0 $$
02

Find the integrating factor

For the linear part, \(y^{\prime} - 4y = 0\), we will find the integrating factor. The integrating factor is given by $$ \mu(x) = e^{\int{-4}dx} = e^{-4x} $$
03

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 $$ (e^{-4x}y)^{\prime} = 0 $$
04

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.
05

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: $$ (Ce^{4x})^{\prime} - 4(Ce^{4x}) = \frac{48x}{(Ce^{4x})^2} $$
06

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: $$ C = 1 $$
07

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: $$ y(x) = e^{4x} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

We've shown that if \(p\) and \(f\) are continuous on \((a, b)\) then every solution of $$ y^{\prime}+p(x) y=f(x) $$ on \((a, b)\) can be written as \(y=u y_{1},\) where \(y_{1}\) is a nontrivial solution of the complementary equation for \((\mathrm{A})\) and \(u^{\prime}=f / y_{1} .\) Now suppose \(f, f^{\prime}, \ldots, f^{(m)}\) and \(p, p^{\prime}, \ldots, p^{(m-1)}\) are continuous on \((a, b),\) where \(m\) is a positive integer, and define $$ \begin{array}{l} f_{0}=f \\ f_{j}=f_{j-1}^{\prime}+p f_{j-1}, \quad 1 \leq j \leq m \end{array} $$ Show that $$ u^{(j+1)}=\frac{f_{j}}{y_{1}}, \quad 0 \leq j \leq m $$

In Exercises \(1-17\) determine which equations are exact and solve them. $$ \left(3 y \cos x+4 x e^{x}+2 x^{2} e^{x}\right) d x+(3 \sin x+3) d y=0 $$

In Exercises \(1-17\) determine which equations are exact and solve them. $$ \left(e^{x}\left(x^{2} y^{2}+2 x y^{2}\right)+6 x\right) d x+\left(2 x^{2} y e^{x}+2\right) d y=0 $$

Suppose all second partial derivatives of \(M=M(x, y)\) and \(N=N(x, y)\) are continuous and \(M d x+N d y=0\) and \(-N d x+M d y=0\) are exact on an open rectangle \(R\). Show that \(M_{x x}+M_{y y}=N_{x x}+N_{y y}=0\) on \(R\)

Solve the initial value problem. $$ y^{\prime}=\frac{y^{2}-3 x y-5 x^{2}}{x^{2}}, \quad y(1)=-1 $$
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