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Magazine Chapter 9: Problem 20
\(15-20\) Solve the initial-value problem. $$\left(x^{2}+1\right) \frac{d y}{d x}+3 x(y-1)=0, \quad y(0)=2$$
Short Answer
Expert verified
The solution is \(y(x) = 1 + \frac{1}{(x^2 + 1)^{3/2}}\).
Step by step solution
01
Identify the type of differential equation
The equation given is \((x^{2}+1) \frac{dy}{dx} + 3x(y-1) = 0\). It is a first-order linear differential equation in the standard form \(a(x)\frac{dy}{dx} + b(x)y = c(x)\). In this case, \(a(x) = x^2 + 1\), \(b(x) = 3x\), and \(c(x) = 3x\).
02
Divide by the coefficient of \(\frac{dy}{dx}\)
Divide the entire equation by \(x^2 + 1\) to make it easier to solve: \(\frac{dy}{dx} + \frac{3x}{x^2 + 1}y = \frac{3x}{x^2 + 1}\).
03
Find the integrating factor
The integrating factor \(\mu(x)\) for a linear differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\) is \(\mu(x) = e^{\int P(x) \, dx}\). Here, \(P(x) = \frac{3x}{x^2 + 1}\), so the integrating factor is \(\mu(x) = e^{\int \frac{3x}{x^2 + 1} \, dx}\).
04
Integrate to find \(\mu(x)\)
To compute \(\int \frac{3x}{x^2 + 1} \, dx\), use the substitution \(u = x^2 + 1\), giving \(du = 2x \, dx\). This transforms the integral into \(\frac{3}{2}\int \frac{1}{u} \, du = \frac{3}{2} \ln|u| + C\). Substituting back, we have \(\mu(x) = (x^2 + 1)^{3/2}\).
05
Solve the differential equation using the integrating factor
Multiply the differential equation by the integrating factor: \((x^2 + 1)^{3/2} \frac{dy}{dx} + 3x(x^2 + 1)^{1/2}y = 3x(x^2 + 1)^{1/2}\). This is equivalent to \(\frac{d}{dx}((x^2 + 1)^{3/2}y) = 3x(x^2 + 1)^{1/2}\).
06
Integrate to find \(y(x)\)
Integrate both sides w.r.t. \(x\): \((x^2 + 1)^{3/2}y = \int 3x(x^2 + 1)^{1/2} \, dx\). By substitution \(u = x^2 + 1\), and \(du = 2x \, dx\), this becomes \(\frac{3}{2}\int u^{1/2} \, du = \frac{1}{2}u^{3/2} + C\). Substitute back to get \((x^2 + 1)^{3/2}y = (x^2 + 1)^{3/2} + C\).
07
Apply the initial condition
Using the initial condition \(y(0) = 2\), substitute \(x = 0\) and \(y = 2\) into \((x^2 + 1)^{3/2}y = (x^2 + 1)^{3/2} + C\) to find \(C\): \((0^2 + 1)^{3/2} \times 2 = (0^2 + 1)^{3/2} + C\), resulting in \(2 = 1 + C\) so \(C = 1\).
08
Write the final solution
Substitute \(C = 1\) back into the expression: \((x^2 + 1)^{3/2} y = (x^2 + 1)^{3/2} + 1\), thus \(y(x) = 1 + \frac{1}{(x^2 + 1)^{3/2}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial-Value Problem
An initial-value problem is a type of differential equation accompanied by a condition that specifies the value of the unknown function at a given point.
In our exercise, the equation is given as \[(x^2 + 1) \frac{dy}{dx} + 3x(y-1) = 0,\] and the initial condition is \(y(0)=2\).
The initial condition enables us to find a specific solution from the general solution of the differential equation. This problem combines the differential equation with this additional piece of information, which helps us to determine the particular solution that satisfies both the differential equation and the initial condition.
In our exercise, the equation is given as \[(x^2 + 1) \frac{dy}{dx} + 3x(y-1) = 0,\] and the initial condition is \(y(0)=2\).
The initial condition enables us to find a specific solution from the general solution of the differential equation. This problem combines the differential equation with this additional piece of information, which helps us to determine the particular solution that satisfies both the differential equation and the initial condition.
Integrating Factor
The integrating factor is a crucial tool in solving first-order linear differential equations. It simplifies the equation into a form that is easier to solve.
For an equation in the form \[\frac{dy}{dx} + P(x)y = Q(x),\] we find the integrating factor by calculating \[\mu(x) = e^{\int P(x) \, dx}.\]
In our exercise, after reformatting the equation, we identified \(P(x) = \frac{3x}{x^2 + 1}\). By integrating \(\int \frac{3x}{x^2 + 1} \, dx\), we found the integrating factor: \[\mu(x) = (x^2 + 1)^{3/2}.\]
This factor is then used to make the differential equation easily integrable by multiplying through each part, transforming the left side into an exact derivative.
For an equation in the form \[\frac{dy}{dx} + P(x)y = Q(x),\] we find the integrating factor by calculating \[\mu(x) = e^{\int P(x) \, dx}.\]
In our exercise, after reformatting the equation, we identified \(P(x) = \frac{3x}{x^2 + 1}\). By integrating \(\int \frac{3x}{x^2 + 1} \, dx\), we found the integrating factor: \[\mu(x) = (x^2 + 1)^{3/2}.\]
This factor is then used to make the differential equation easily integrable by multiplying through each part, transforming the left side into an exact derivative.
Substitution Method
Substitution is a method used to simplify integrals, especially when dealing with complex or non-standard forms.
In this exercise, substitution facilitated solving the integral necessary for finding the integrating factor. We set \(u = x^2 + 1\), which implies that \(du = 2x \, dx\).
This changes the integral \(\int \frac{3x}{x^2 + 1} \, dx\) into \(\frac{3}{2}\int \frac{1}{u} \, du\).
The method of substitution greatly simplifies the process, as it reduces the complicated integral into a more recognizable form that is easier to work with.
In this exercise, substitution facilitated solving the integral necessary for finding the integrating factor. We set \(u = x^2 + 1\), which implies that \(du = 2x \, dx\).
This changes the integral \(\int \frac{3x}{x^2 + 1} \, dx\) into \(\frac{3}{2}\int \frac{1}{u} \, du\).
The method of substitution greatly simplifies the process, as it reduces the complicated integral into a more recognizable form that is easier to work with.
Differential Equation Solution
Solving a differential equation involves finding a function or functions that satisfy the given equation.
In this case, we use the integrating factor method to transform and integrate the equation, leading to a solution that fits the initial condition.
After applying the integrating factor, the equation \(\frac{dy}{dx} + \frac{3x}{x^2 + 1}y = \frac{3x}{x^2 + 1}\) was converted into \(\frac{d}{dx}((x^2 + 1)^{3/2}y) = 3x(x^2 + 1)^{1/2}\).
Integrating both sides with respect to \(x\), and substituting back, we arrive at the solution: \[y(x) = 1 + \frac{1}{(x^2 + 1)^{3/2}}.\]
This satisfies both the differential equation and the initial condition \(y(0) = 2\). The complexity lies in carefully following the steps and understanding each part's impact on the equation as a whole.
In this case, we use the integrating factor method to transform and integrate the equation, leading to a solution that fits the initial condition.
After applying the integrating factor, the equation \(\frac{dy}{dx} + \frac{3x}{x^2 + 1}y = \frac{3x}{x^2 + 1}\) was converted into \(\frac{d}{dx}((x^2 + 1)^{3/2}y) = 3x(x^2 + 1)^{1/2}\).
Integrating both sides with respect to \(x\), and substituting back, we arrive at the solution: \[y(x) = 1 + \frac{1}{(x^2 + 1)^{3/2}}.\]
This satisfies both the differential equation and the initial condition \(y(0) = 2\). The complexity lies in carefully following the steps and understanding each part's impact on the equation as a whole.
