Question

Find the general solution to the differential equation.\((1+x) y^{\prime}=(x+2)(y-1)\)

Short answer

The general solution is \( y = C_1(x+1)e^{x} + 1 \).

Step-by-step solution

Rearrange the Equation

Given the differential equation \[(1+x)y^{\prime}=(x+2)(y-1)\]we rearrange it to separate the differentials by dividing both sides by \((y-1)\): \[\frac{y^{\prime}}{y-1} = \frac{(x+2)}{(1+x)}\].

Substitute for Simplicity

Let's make a substitution to simplify the equation. Set \( u = y - 1 \). Thus, \( y = u + 1 \) and \( y' = u' \).Substituting this into the rearranged equation gives:\[ \frac{u^{\prime}}{u} = \frac{x + 2}{1 + x} \].

Separation of Variables

Separate the variables:\[ \frac{du}{u} = \frac{(x + 2)}{(1 + x)}dx \].

Integrate Both Sides

Integrate both sides: \[ \int \frac{du}{u} = \int \frac{x+2}{1+x}dx \].The left side integrates to \( \ln |u| \).The right side simplification:\[ \int \left( 1 + \frac{1}{x+1} \right) dx = \int 1 \ dx + \int \frac{1}{x+1} \ dx = x + \ln|x+1| \].Thus, the integration gives:\[ \ln |u| = x + \ln|x+1| + C \] where \(C\) is the constant of integration.

Solve for the Original Variable

Substitute back \( u = y - 1 \):\[ \ln |y-1| = x + \ln|x+1| + C \].Exponentiating both sides, we get:\[ |y-1| = e^{x}e^{\ln|x+1|}e^{C} \].This simplifies to:\[ |y-1| = C_1(x+1)e^{x} \] where \(C_1 = e^{C} \). So,\[ y - 1 = C_1(x+1)e^{x} \].The general solution is:\[ y = C_1(x+1)e^{x} + 1 \], where \(C_1\) can be positive or negative.

Key concepts

General Solution

A general solution for a differential equation provides a family of solutions that includes arbitrary constants. This means it isn't just one particular answer, but many potential ones depending on specific conditions. In the given equation, \[(1+x)y^{\prime}=(x+2)(y-1)\], we aim to derive a formula that solves it under various conditions by introducing constants. The general solution is not constrained by initial conditions or boundaries, which differentiates it from a particular solution. By successfully separating, integrating, and substituting variables, we arrive at a general form like \[y = C_1(x+1)e^{x} + 1\]. This form reflects all potential solutions applicable, with \(C_1\) being any real number.

Separation of Variables

Separation of variables is a straightforward and powerful technique to solve differential equations. By isolating each variable on opposite sides of the equation, we can simplify the process of integration.Consider the equation after substitution:\[\frac{du}{u} = \frac{(x + 2)}{(1 + x)}dx\]. Here, separating the variables involves getting all terms with \(u\) on one side and \(x\) terms on the other. This way, each side of the equation depends on only one variable. This allows for the integration of each side independently. It's like tackling an algebra problem one part at a time. Once the variables are separated effectively, we can integrate directly and find relationships between the variables intuitive. This method is typically one of the first taught in calculus, owing to its conceptual clarity and ease of use.

Integration

Integration is the process of finding a function given its derivative. It's like reversing differentiation, and it allows us to solve equations by finding the original function.In our exercise, integration appears when we rearrange the separated variables:\[\int \frac{du}{u} = \int \frac{x+2}{1+x}dx\].The left side integrates to \[\ln|u|\], while the right side involves breaking down the fraction into simpler components:\[\int \left(1 + \frac{1}{x+1}\right) dx\].Each part can be integrated separately to get \[x + \ln|x+1|\]. With the integration constant \(C\), these equations represent a broad range of possible answers, considering all shifts or stretches of one solution to fit others, embodying the concept of a general solution.

Substitution Method

The substitution method is a clever tool for simplifying complex equations, often making seemingly unsolvable problems manageable. In our exercise, we use a substitution to tackle the differential equation.We set \(u = y - 1\), which simplifies replacement to \(y = u + 1\). By rephrasing \(y'\) as \(u'\), we transform the equation into a simpler form:\[\frac{u^{\prime}}{u} = \frac{x + 2}{1 + x}\]. This new equation offers simpler terms for separation and subsequent integration. The substitution reduces the complexity of the original problem, making it easier to apply separation of variables and integration methods. In essence, substitution acts as a translator between the original complex formula and a simpler, more solvable form.