Question

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

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

Short answer

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

Step-by-step solution

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 variableu back in terms of variablex 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 Cin the solution of the differential equation and obtain the solution of the initial -value problem $\frac{dy}{dx}=\frac{x}{1+x^2}asy=\frac{1}{2}\ln |1+x^2|+4$