Question

In problems 1–24Find the general solution of the given differential equation. Give the largest interval I over which the general solution is defined. Determine whether there are any transient terms in the general solution.

$(x+2{)}^2\frac{dy}{dx}=5-8y-4xy$

Short answer

The general solution is$y=\frac{5}{3}(x+2{)}^{-1}+C(x+2{)}^{-4},-2<x<\infty ,\text{solution}\text{is}\text{Transient}$

Step-by-step solution

Step 1: Concept of Linear Differential Equation

$\frac{dy}{dx}+py=q$, where p and q can be constants or variables.

Converting the given equation into standard form

After subtracting 5 from both sides, the left hand side is simplified to$-4y(x+2)$

$\frac{dy}{dx}+\frac{4y}{x+2}=\frac{5}{{(x+2)}^2}$

Find the integrating factor

$$\begin{gathered} {}^{\int \left(\frac{4}{x+2}\right)dx} \\ (x+2{)}^4 \end{gathered}$$

Finding the General Solution

Multiply the right side by the integrating factor and take the integral.

$\frac{d}{dx}(x+2{)}^{4}y=\int 5(x+2{)}^{2}dx$

Solution to the integral on the right side,

$(x+2{)}^{4}y=\frac{5}{3}(x+2{)}^3+c$

Now we are simplifying

$y=\frac{5}{3}(x+2{)}^{-1}+C(x+2{)}^{-4}$

Entire Solution is Transient $-2<x<\infty$

Hence, the general solution is $y=\frac{5}{3}(x+2{)}^{-1}+C(x+2{)}^{-4},-2<x<\infty ,\text{solution}\text{is}\text{Transient}$