Chapter 7.1, Problem 13E

To determine

To solve the initial value problem explicitly.

Answer to Problem 13E

The solution of differential equation is $u=x^7-x^3+5x-4$

Explanation of Solution

Given information:

It is given that

  $\frac{du}{dx}=7x^6-3x^2\text{+5 and u}=1\text{ when }x=1$

Calculation:

Differential equation is

  $\frac{du}{dx}=7x^6-3x^2\text{+5}$

Now,

  $\begin{array}{l}du=\left(7x^6-3x^2+5\right)dx\\ \text{integrating both side}\\ {\displaystyle \int du}={\displaystyle \int \left(7x^6-3x^2+5\right)dx}\\ u={\displaystyle \int 7x^{6}dx-{\displaystyle \int 3x^{2}dx+{\displaystyle \int 5dx}}}\\ u=x^7-x^3+5x+C\\ \text{intial condition, u}=1\text{ when }x=1\\ 1={\left(1\right)}^7-{\left(1\right)}^3+5\left(1\right)+C\\ 1=1-1+5+C\\ C=1-5\\ C=-4\end{array}$

Therefore, the solution of differential equation is

  $u=x^7-x^3+5x-4$