For each of the differential equations given in find the general … - Vidyadip

Question

For each of the differential equations given in find the general solution:
$(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=1$

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Answer

$(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+\text{y}}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=\text{x}+\text{y}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}-\text{x}=\text{y}$
This is a linear differential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{px}=\text{Q}\ (\text{where p}=-1\ \text{and}\ \text{Q}=\text{y})$
$\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{\int-\text{dy}}=\text{e}^{-\text{y}}.$
The general solution of the given differential equation is given by the relation,
$\text{x(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dy}+\text{C}$
$\Rightarrow\text{xe}^{-\text{y}}=\int(\text{y}\cdot\text{e}^{-\text{y}})\text{dy}+\text{C}$
$\Rightarrow\text{xe}^{-\text{y}}=\text{y}\cdot\int\text{e}^{-\text{y}}\text{dy}-\int\Big[\frac{\text{d}}{\text{dy}}(\text{y})\int\text{e}^{-\text{y}}\text{dy}\Big]\text{dy}+\text{C}$
$\Rightarrow\text{xe}^{-\text{y}}=\text{y}(-\text{e}^{-\text{y}})-\int(-\text{e}^{-\text{y}})\text{dy}+\text{C}$
$\Rightarrow\text{xe}^{-\text{y}}=-\text{ye}^{-\text{y}}+\int\text{e}^{-\text{y}}\text{dy}+\text{C}$
$\Rightarrow\text{xe}^{-\text{y}}=-\text{ye}^{-\text{y}}-\text{e}^{-\text{y}}+\text{C}$
$\Rightarrow\text{x}=-\text{y}-1+\text{Ce}^{\text{y}}$
$\Rightarrow\text{x}+\text{y}+1=\text{Ce}^{\text{y}}$

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