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Expert-verified Found in: Page 570 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29-52.$\frac{dy}{dx}=xy+x+y+1,y\left(0\right)=c$

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

See the step by step solution

## Step 1. Given information

Given the expression $\frac{dy}{dx}=xy+x+y+1,\phantom{\rule{0ex}{0ex}}y\left(0\right)=c$

## Step 2: Take common and use variable separable method

Calculating, we get

$\frac{dy}{dx}=x\left(y+1\right)+1\left(y+1\right)\phantom{\rule{0ex}{0ex}}=\left(x+1\right)\left(y+1\right)$

Integrating, we get

$\int \frac{1}{y+1}dy=\int \left(x+1\right)dx\phantom{\rule{0ex}{0ex}}\mathrm{ln}|y+1|=\frac{1}{2}\left(x+1{\right)}^{2}+C\phantom{\rule{0ex}{0ex}}y+1={e}^{\frac{1}{2}\left(x+1{\right)}^{2}+C}\phantom{\rule{0ex}{0ex}}y=-1+A{e}^{\frac{1}{2}\left(x+1{\right)}^{2}}$

## Step 3: Substitute x=0,y=c in the equation and solve

Calculating, we get

$c=-1+A\phantom{\rule{0ex}{0ex}}A=1+c\phantom{\rule{0ex}{0ex}}y\left(x\right)=-1+\left(1+c\right){e}^{\frac{1}{2}\left(x+1{\right)}^{2}}$ ### Want to see more solutions like these? 