Q. 41 - Calculus

Question

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52

$\frac{dy}{dx}=\frac{x}{1+x^2}$

Step-by-Step Solution

Answer

The solution of the initial -value problem  $\frac{dy}{dx}=\frac{x}{1+x^2} as y=\frac{1}{2}\ln |1+x^2|+4$

Step 1. Given information

The given initial value problem $\frac{dy}{dx}=\frac{x}{1+x^2}.....................\left(1\right)$

Step 2. Use antidifferentiation and/or separation of variables to solve each of the initial-value

Note that the differential equation in (1) does not contain the independent variable at all, so technically the variables have already been separated. Hence, the differential equation can be solved by antidifferentiating. Thus, the solution of the differential equation involved in the initialvalue problem is given by

$$\begin{gathered} \int dy=\int \frac{x}{1+x^2}dx \\ y=\frac{1}{2}\int \frac{2x}{1+x^2}dx \\ =\frac{1}{2}\int \frac{1}{u}du (1+x^2=u,2xdx=du) \\ =\frac{1}{2}\ln \left|u\right|+C \end{gathered}$$

Replace the variable u back in terms of variable x to get

$y=\frac{1}{2}\ln |1+x^2|+C$

Now, use the given initial condition $y\left(0\right)=4$ , that is take $x=0,y=4$ in the above result and evaluate the constant C.

$$\begin{gathered} 4=\frac{1}{2}\ln 1+C \\ C=4 \end{gathered}$$

Substitute this value of the constant C in the solution of the differential equation and obtain the solution of the initial -value problem  $\frac{dy}{dx}=\frac{x}{1+x^2} as y=\frac{1}{2}\ln |1+x^2|+4$