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Q. 48

Expert-verifiedFound in: Page 570

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 $y\left(x\right)=-1+(1+c){e}^{\frac{1}{2}(x+1{)}^{2}}$

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

Calculating, we get

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

Integrating, we get

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

Calculating, we get

$c=-1+A\phantom{\rule{0ex}{0ex}}A=1+c\phantom{\rule{0ex}{0ex}}y\left(x\right)=-1+(1+c){e}^{\frac{1}{2}(x+1{)}^{2}}$

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