Question

Solve the given initial-value problem. Give the largest interval / over which the solution is defined.

$\mathbf{x}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{xy}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{e}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

So, the largest interval l on which the solution is defined is $(0,\infty )$. And the solution is $y=\frac{\ln x}{(x+1)}+\frac{e}{(x+1)}$for the given initial problem.

Step-by-step solution

Form of first order differential equation

A first order linear differential equation is a differential equation of the form $y\text{'}+p\left(x\right)y=q\left(x\right)y\text{'}+p\left(x\right)y=q\left(x\right)y\text{'}+p\left(x\right)y=q\left(x\right)$.

Evaluation

Consider the initial value problem

$x(x+1)\frac{dy}{dx}+xy=1,y\left(e\right)=1$

Equation (1) is a first order linear differential equation. Rewrite the equation so it is in standard form.

To solve this, it is required to multiply an integration factor both sides

$$\begin{gathered} \mu \left(x\right)=e^{\int \frac{1}{x+1}dx} \\ =e^{\ln (x+1)} \\ =x+1 \end{gathered}$$

Find the solution

Multiply this on both sides of (1)

$$\begin{gathered} \frac{d}{dx}\left[\right(x+1\left)y\right]=\frac{(x+1)}{x(x+1)} \\ (x+1)y=\int \frac{1}{x}dx \\ y=\frac{\ln x}{(x+1)}+\frac{c}{(x+1)} \end{gathered}$$

Finding the value of e

Substitute the initial condition y(e) = 1 in this,

$$\begin{gathered} 1=\frac{\ln e}{(e+1)}+\frac{c}{(e+1)} \\ \left(e+1\right)=1+c \\ e=c \end{gathered}$$

Use ne = 1.

The solution to the initial value problem is$y=\frac{\ln x}{(x+1)}+\frac{e}{(x+1)}$.

Observe that when x=-1, the solution does not exist.

The solution is defined for all $x\in (0,\infty )$due to the ln x.