Question

Solve the given differential equations. $$ \left(u^{2}+1\right) \frac{d v}{d u}+4 u v=3 u $$

Short answer

The solution to the given differential equation \(\left(u^{2} +1\right)\frac{dv}{du} + 4uv = 3u\) is given by: \(v(u) = \frac{\int 3u(u^2+1)du + C}{(u^2 + 1)^2}\).

Step-by-step solution

Identify the given equation as a first-order linear differential equation

The given differential equation is in the form: \[ \left(u^{2} +1\right)\frac{dv}{du} + 4uv = 3u \] This equation is linear in \(v\) and its first derivative \(\frac{dv}{du}\). Thus, this is a first-order linear differential equation.

Rewrite the given equation in the standard form

The standard form of a first-order linear differential equation is: \[ \frac{dv}{du} + P(u)v = Q(u) \] In our case, we have: \[ \left(u^{2} +1\right)\frac{dv}{du} + 4uv = 3u \] Divide both sides of the equation by \(\left(u^2 + 1\right)\), to get: \[ \frac{dv}{du} + \frac{4u}{u^2 + 1}v = \frac{3u}{u^2 + 1} \] Now we have the equation in the standard form with \(P(u) = \frac{4u}{u^2+1}\) and \(Q(u) = \frac{3u}{u^2+1}\).

Find the integrating factor

The integrating factor is obtained by calculating the following: \[ \mathrm{IF} = e^{\int P(u)du} \] In our case, we calculate: \[ \mathrm{IF} = e^{\int \frac{4u}{u^2+1}du} \] To compute this integral, use the substitution \(x = u^2 + 1\), which gives \(dx = 2udu\): \[ \mathrm{IF} = e^{\int \frac{2}{x}dx} = e^{\ln x^2} = x^2 = (u^2 + 1)^2 \] So, the integrating factor \(\mathrm{IF}\) = \((u^2 + 1)^2\).

Multiply the differential equation by the integrating factor and solve for \(v(u)\)

Multiply both sides of the standard-form equation by the integrating factor \((u^2 +1)^2\): \[ (u^2+1)^2 \frac{dv}{du} + 4u(u^2+1)v = 3u(u^2+1) \] The left side of the equation is now the derivative of a product: \[ \frac{d}{du}((u^2 + 1)^2v) = 3u(u^2+1) \] Integrate both sides with respect to \(u\): \[ \int\frac{d}{du}((u^2 + 1)^2v) du = \int 3u(u^2+1)du \] This simplifies to: \[ (u^2 + 1)^2v = \int 3u(u^2+1)du + C \] Now, we just need to find the integral of \(\int 3u(u^2+1)du\) and then divide by \((u^2+1)^2\) to get the solution for \(v(u)\).

Key concepts

Integrating Factor Method

The integrating factor method is a powerful tool commonly used to solve first-order linear ordinary differential equations (ODEs). This technique turns a non-exact equation into an exact one by multiplying an integrating factor. To determine the integrating factor (IF), we typically perform the integral \( e^{\int P(u)du} \) where \( P(u) \) is the coefficient of \( v \) in the standard form \( \frac{dv}{du} + P(u)v = Q(u) \) of the differential equation.

The calculated integrating factor is then multiplied with each term of the standard form, which allows us to express the left side of the equation as the derivative of a product. The uniqueness of the integrating factor lies in its tailor-made form for each equation, ensuring the left side becomes a straightforward derivative.

In the exercise provided, the ODE was first rewritten into the standard form to identify \( P(u) = \frac{4u}{u^2+1} \) and thus find the integrating factor \( (u^2 + 1)^2 \). By multiplying this integrating factor across, the ODE was turned into an easily integrable form, showcasing how the integrating factor simplifies the process of finding the solution to an ODE.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that involve functions of one independent variable and their derivatives. They are termed 'ordinary' to distinguish them from partial differential equations, which involve partial derivatives of functions of multiple independent variables. First-order linear ODEs have the particular form \( \frac{dv}{du} + P(u)v = Q(u) \) and are solved for a function \( v \) that depends on a single variable \( u \).

These equations are ubiquitous in the mathematical modeling of natural phenomena, including physics, engineering, biology, and economics. Solving ODEs is crucial as the solutions often reveal the behavior of the modeled systems over time. The method to solve an ODE depends on its type. For first-order linear ODEs, the integrating factor method serves as a standard approach.

Integration by Substitution

The integration by substitution is a technique that simplifies complex integrals by substituting part of the integral with a new variable. This method is analogous to the chain rule for differentiation, and it's typically used when an integral contains a composite function.

To apply this method, one chooses a substitution that simplifies the integral's form, allowing easier computation. In the example problem, the substitution \( x = u^2 + 1 \) was chosen, creating a simpler integrand and leading to an easier antiderivative to find. Once the integration is performed with the new variable, the results are then substituted back in terms of the original variable. This demonstrates how integration by substitution not only simplifies integrals but is also invaluable in solving differential equations where direct integration is not straightforward.