Question

Solve the following differential equation
${\displaystyle \frac{dy}{dx}}=1+x+y+xy$

Answer 1

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

Given equation: ${\displaystyle \frac{dy}{dx}}=1+x+y+xy$
Before solving the differential equation, first let us rearrange the given equation by taking some common terms.
$$\begin{gathered} \Rightarrow {\displaystyle \frac{dy}{dx}}=1+x+y+xy \\ \Rightarrow {\displaystyle \frac{dy}{dx}}=1\left(1+x\right)+y\left(1+x\right) \\ \Rightarrow {\displaystyle \frac{dy}{dx}}=\left(1+x\right)\left(1+y\right) \end{gathered}$$
Now, let us separate the like terms together on either side of the equation.
$$\Rightarrow {\displaystyle \frac{dy}{\left(1+y\right)}}=\left(1+x\right)dx$$
Now, integrating both the sides
$$\Rightarrow \int {\displaystyle \frac{dy}{\left(1+y\right)}}=\int \left(1+x\right)dx$$
As we know that
$$\left[\because \int {\displaystyle \frac{dx}{x}}=\ln x\right]\mathrm{\&}\left[\int x^{n}dx={\displaystyle \frac{x^{n+1}}{n+1}}\right]$$
So using the above formula and by solving the integral, we get
$$\Rightarrow \ln \left(y+1\right)={\displaystyle \frac{x^2}{2}}+x+c$$
As we know by the property of natural logarithm
$$\begin{gathered} \ln x=y \\ \Rightarrow x=e^y \end{gathered}$$
So using this in the above equation, we have
$$\begin{gathered} \Rightarrow y+1=e^{{\displaystyle \frac{x^2}{2}}+x+c} \\ \Rightarrow y=e^{{\displaystyle \frac{x^2}{2}}+x+c}-1 \end{gathered}$$
Hence, the solution of the given equation is$$y=e^{{\displaystyle \frac{x^2}{2}}+x+c}-1$$

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 $e$.