Question

Find the general solution of the differential equation
$\boxed{\color{green} \displaystyle \rm \frac{dy}{dx} = x + y + xy + 1}$
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Answer 1

$\huge \pink{ \pmb{ \bf{ \frak{answer}}}}$

$\sf {Given: - \: \: \: (x + y + 1) \frac{dy}{dx} = 1}$

$\sf { \frac{dx}{dy} = x + y + + 1}$

which is a linear differential equation with x as dependent variable:-

${\bold{ \red{ \star \: { \boxed{Here, \:  P=−1  \: ; Q=y+1}}}}}</p><p></p><p>$

$\sf \boxed{Integrating \: \: Factor \:  I.F \: </p><p></p><p> = {e}^{pdy} }$

$\bold{I.F. \: = \: {e}^{∫ - dy</p><p></p><p>} </p><p>}</p><p>$

Solution is given by :-

$\sf \: x {e}^{ - y} = ∫</p><p> {e}^{ - y} (y + 1)dy</p><p>$

$\sf \: x {e}^{ - y} = ∫</p><p></p><p> {ye}^{ - y} dy + ∫</p><p></p><p> {e}^{ - y} dy$

$\implies \sf \: {xe}^{ - y} = - y {e}^{ - y} + ∫</p><p></p><p> {e}^{ - y} dy - {e}^{ - y} + k$

$\implies \sf \: x {e}^{ - y} = - y {e}^{ - y} - {e}^{ - y} - {e}^{ - y} + k$

$\sf \implies \: {xe}^{ - y} = - {ye}^{ - y} - {2e}^{ - y} + k$

$\sf \implies \: \: x = {ke}^{y} - (y + 2)$

$\green{ \pmb{ \bf {\frak{hope \: its \: help \: u}}}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given differential equation is

$\rm \: \dfrac{dy}{dx} = x + y + xy + 1$

can be rewritten as

$\rm \: \dfrac{dy}{dx} = 1 + x + y + xy$

can be re-arranged as

$\rm \: \dfrac{dy}{dx} = (1 + x) + (y + xy)$

$\rm \: \dfrac{dy}{dx} = (1 + x) + y(1 + x)$

$\rm \: \dfrac{dy}{dx} = (1 + x)(1 + y)$

On separating the variables, we get

$\rm \: \dfrac{dy}{y + 1} = (1 + x)dx$

On integrating both sides, we get

$\rm \:\displaystyle\int\rm \dfrac{dy}{y + 1} = \displaystyle\int\rm (1 + x)dx$

$\rm \: log |y + 1| = \dfrac{ {x}^{2} }{2} + x + c$

Hence, The general solution of differential equation

$\rm \: \dfrac{dy}{dx} = x + y + xy + 1 \: is \: \\ \: \\ \red{ \rm \: log |y + 1| = \dfrac{ {x}^{2} }{2} + x + c }$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$