Q. 52 - Calculus

Question

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

$\frac{dy}{dx}=\frac{y+1}{x^2+1}, y\left(0\right)=1$

Step-by-Step Solution

Answer

On solving, we get $y\left(x\right)=2e^{{\tan }^{-1}x}-1$

Step 1. Given information

Given expression $\frac{dy}{dx}=\frac{y+1}{x^2+1}, y\left(0\right)=1$

Step 2: Using the variable separable method

Calculating, we get

$$\begin{gathered} \int \frac{1}{y+1}dy=\int \frac{1}{x^2+1}dx \\ \ln |y+1|={\tan }^{-1}x+C \\ y+1=e^{{\tan }^{-x}+c} \\ y=-1+A{e}^{{\tan }^{-1}x} \end{gathered}$$

Step 3: Substitute x = 0 y = 1 and calculate

Calculating, we get

$$\begin{gathered} 1=-1+A \\ A=2 \\ y\left(x\right)=2e^{{\tan }^{-1}x}-1 \end{gathered}$$