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

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Hint- We will try to separate both the terms of $x\& y$. In that case it will be easy to integrate separately.

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

Before solving the differential equation, first let us rearrange the given equation by taking some common terms.

\[

\Rightarrow \dfrac{{dy}}{{dx}} = 1 + x + y + xy \\

\Rightarrow \dfrac{{dy}}{{dx}} = 1\left( {1 + x} \right) + y\left( {1 + x} \right) \\

\Rightarrow \dfrac{{dy}}{{dx}} = \left( {1 + x} \right)\left( {1 + y} \right) \\

\]

Now, let us separate the like terms together on either side of the equation.

\[ \Rightarrow \dfrac{{dy}}{{\left( {1 + y} \right)}} = \left( {1 + x} \right)dx\]

Now, integrating both the sides

\[ \Rightarrow \int {\dfrac{{dy}}{{\left( {1 + y} \right)}}} = \int {\left( {1 + x} \right)dx} \]

As we know that

\[\left[ {\because \int {\dfrac{{dx}}{x} = \ln x} } \right]\& \left[ {\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right]\]

So using the above formula and by solving the integral, we get

\[ \Rightarrow \ln \left( {y + 1} \right) = \dfrac{{{x^2}}}{2} + x + c\]

As we know by the property of natural logarithm

$

\ln x = y \\

\Rightarrow x = {e^y} \\

$

So using this in the above equation, we have

\[

\Rightarrow y + 1 = {e^{\dfrac{{{x^2}}}{2} + x + c}} \\

\Rightarrow y = {e^{\dfrac{{{x^2}}}{2} + x + c}} - 1 \\

\]

Hence, the solution of the given equation is\[y = {e^{\dfrac{{{x^2}}}{2} + x + c}} - 1\]

Note- To solve any differential equation, rearranging of the equation in the correct form at the beginning is a very basic step. Re-arrangement should be made in such a way as the terms on L.H.S. and R.H.S. must contain different variables. $\ln $ in the solution represents natural logarithm which means logarithm with base $e$.

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

Before solving the differential equation, first let us rearrange the given equation by taking some common terms.

\[

\Rightarrow \dfrac{{dy}}{{dx}} = 1 + x + y + xy \\

\Rightarrow \dfrac{{dy}}{{dx}} = 1\left( {1 + x} \right) + y\left( {1 + x} \right) \\

\Rightarrow \dfrac{{dy}}{{dx}} = \left( {1 + x} \right)\left( {1 + y} \right) \\

\]

Now, let us separate the like terms together on either side of the equation.

\[ \Rightarrow \dfrac{{dy}}{{\left( {1 + y} \right)}} = \left( {1 + x} \right)dx\]

Now, integrating both the sides

\[ \Rightarrow \int {\dfrac{{dy}}{{\left( {1 + y} \right)}}} = \int {\left( {1 + x} \right)dx} \]

As we know that

\[\left[ {\because \int {\dfrac{{dx}}{x} = \ln x} } \right]\& \left[ {\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right]\]

So using the above formula and by solving the integral, we get

\[ \Rightarrow \ln \left( {y + 1} \right) = \dfrac{{{x^2}}}{2} + x + c\]

As we know by the property of natural logarithm

$

\ln x = y \\

\Rightarrow x = {e^y} \\

$

So using this in the above equation, we have

\[

\Rightarrow y + 1 = {e^{\dfrac{{{x^2}}}{2} + x + c}} \\

\Rightarrow y = {e^{\dfrac{{{x^2}}}{2} + x + c}} - 1 \\

\]

Hence, the solution of the given equation is\[y = {e^{\dfrac{{{x^2}}}{2} + x + c}} - 1\]

Note- To solve any differential equation, rearranging of the equation in the correct form at the beginning is a very basic step. Re-arrangement should be made in such a way as the terms on L.H.S. and R.H.S. must contain different variables. $\ln $ in the solution represents natural logarithm which means logarithm with base $e$.

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