Question

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(y d y+\left(x y^{2}-8 x\right) d x=0\) \(y=3\) when \(x=1\)

Short answer

y = √(8 + e^{-x^2 + 1})

Step-by-step solution

Rewrite the Differential Equation

Starting with the given differential equation: y dy + (xy^2 - 8x) dx = 0First, we need to separate the variables y and x.

Separate Variables

Move all terms involving y to one side and all terms involving x to the other side: y dy = - (xy^2 - 8x) dxDivide both sides by y (assuming y ≠ 0):dy/y = (-xy^2 + 8x)/y dxSimplify it to:dy/y = (-x y + 8x/y) dxSince this will overcomplicate our process, instead we should just separate and integrate directly:(y dy)/(y^2 - 8) = - x dx.

Integrate Both Sides

Integrate both sides with respect to their respective variables:∫ (y dy)/(y^2 - 8) = ∫ -x dxFor the left side, let u = y^2 - 8, then du/2 = y dy:∫ (1/2) du/u = ∫ -x dxIntegrate both sides to obtain:(1/2) ln|y^2 - 8| = - (1/2) x^2 + C, where C is the constant of integration.

Simplify the Equation

Multiply both sides by 2 to simplify:ln|y^2 - 8| = - x^2 + 2CLet K = 2C, a new constant, this results in:ln|y^2 - 8| = - x^2 + K

Solve for y

Solve for y in terms of x:|y^2 - 8| = e^{-x^2 + K} y^2 - 8 = ± e^{-x^2 + K} y^2 = 8 ± e^{-x^2 + K}y = ±√(8 ± e^{-x^2 + K})

Apply Initial Conditions

Use the initial condition y = 3 when x = 1 to find K:3^2 = 8 ± e^{-1 + K}9 = 8 ± e^{-1 + K}1 = ± e^{-1 + K}So e^{-1 + K} = 1 or -1 (negative value is not valid for exponential terms): e^{-1 + K} = 1-1 + K = 0K = 1

Write the Particular Solution

Substitute K back into the general solution:y = ±√(8 ± e^{-x^2 + 1})Therefore, the particular solution is:y = √(8 + e^{-x^2 + 1})

Key concepts

Differential Equations

A differential equation is an equation that involves an unknown function and its derivatives. Differential equations are essential in modeling various physical and natural phenomena, such as heat conduction, fluid flow, and population growth.
There are many types of differential equations, but a common method to solve them involves finding a general solution that includes an arbitrary constant. This general solution can be adjusted to meet specific initial or boundary conditions.
In this exercise, we are dealing with a first-order differential equation. This means it involves only the first derivative of the unknown function. We use the separation of variables technique to solve it, which simplifies this equation by isolating the variables on different sides of the equation.

Initial Conditions

Initial conditions are specific values or conditions given at the start of a problem. They help us find a unique solution to a differential equation from the general solution.
In this exercise, the initial condition is given as y=3 when x=1. This allows us to determine the constant in the general solution of our differential equation.
Here's how it's used: After integrating the separated differential equation, we get an expression with an arbitrary constant ‘C’. We then substitute the initial condition values into this equation to solve for ‘C’.
Once we have determined the constant, we can rewrite our general solution as a specific solution that satisfies the given initial condition.

Integration Techniques

Integration is the process of finding the integral of a function, which is a fundamental technique in solving differential equations.
In this exercise, we integrate both sides of our separated differential equation.
To integrate the left side, we use a substitution method where we set u = y^2 - 8. This substitution simplifies the integral, allowing us to integrate easily.
The right side involves a straightforward integration: integrating -x with respect to x, which gives us -(1/2)x^2.
Combining the results of these integrations gives us an expression with a constant of integration, which we later determine using the initial conditions provided.