Chapter 7.1, Problem 15E

To determine

To solve the initial value problem explicitly.

Answer to Problem 15E

The solution of differential equation is $y=\frac{1}{x}+\frac{1}{x^3}+12x-11$ .

Explanation of Solution

Given information:

It is given that

  $\frac{dy}{dx}=-\frac{1}{x^2}-\frac{3}{x^4}\text{+12 and y}=3\text{ when }x=1$

Calculation:

Differential equation is

  $\frac{dy}{dx}=-\frac{1}{x^2}-\frac{3}{x^4}\text{+12}$

Now,

  $\begin{array}{l}dy=\left(-\frac{1}{x^2}-\frac{3}{x^4}\text{+12}\right)dx\\ \text{integrating both side}\\ {\displaystyle \int dy}={\displaystyle \int \left(-\frac{1}{x^2}-\frac{3}{x^4}\text{+12}\right)dx}\\ y={\displaystyle \int \left(-\frac{1}{x^2}\right)dx-{\displaystyle \int \frac{3}{x^4}dx+{\displaystyle \int 12dx}}}\\ y=\frac{1}{x}+\frac{1}{x^3}+12x+C\\ \text{intial condition, y}=3\text{ when }x=1\\ 3=1+1+12+C\\ C=3-14\\ C=-11\end{array}$

Therefore, the solution of differential equation is

  $y=\frac{1}{x}+\frac{1}{x^3}+12x-11$