Question

Find the particular solution of the differential equation $\dfrac{dy}{dx}=1+x+y+xy$ , given that $y=0$ , when $x=1$.

Answer 1

Hint: Rearrange the terms of the equation to get all the terms related to x on one side of the equation and all the terms related to y to the other side of the equation. Take common while rearranging the terms to make the process easier. Now integrate both the sides and use the identity that $\int{\dfrac{dy}{1+y}=\log \left( 1+y \right)+c}$ to reach the solution of the general equation. For finding the constant, use the condition given in the question.

Complete step by step answer:
Let us start the solution to the above question by rearranging the terms, such that the terms related to x on one side of the equation and all the terms related to y to the other side of the equation.
$\dfrac{dy}{dx}=1+x+y+xy$
Now, we will take y common from the last two terms of the right-hand side of the equation. On doing so, we get
$\dfrac{dy}{dx}=1\left( 1+x \right)+y\left( 1+x \right)$
$\Rightarrow \dfrac{dy}{dx}=\left( 1+y \right)\left( 1+x \right)$
$\Rightarrow \dfrac{dy}{1+y}=dx\left( 1+x \right)$
Now, we will integrate both sides of the equation. On doing so, we get
$\int{\dfrac{dy}{1+y}}=\int{dx\left( 1+x \right)}$

Now, we know that $\int{\dfrac{dy}{1+y}=\log \left( 1+y \right)+c}$ and $\int{\left( 1+x \right)dx}=\int{dx+\int{xdx=}}x+\dfrac{{{x}^{2}}}{2}+c$ . So, our integral becomes:
$\log \left( 1+y \right)+{{c}_{1}}=x+\dfrac{{{x}^{2}}}{2}+{{c}_{2}}$
$\Rightarrow \log \left( 1+y \right)=x+\dfrac{{{x}^{2}}}{2}+{{c}_{2}}-{{c}_{1}}$
$\Rightarrow \log \left( 1+y \right)=x+\dfrac{{{x}^{2}}}{2}+c........(i)$
Now, as it is given that $y=0$ when $x=1$ . We will put this in the equation and get the constant c. On doing so, we get
$\Rightarrow \log \left( 1+0 \right)=1+\dfrac{{{1}^{2}}}{2}+c$
We know that log1=0.
$0=\dfrac{3}{2}+c$
$\Rightarrow c=-\dfrac{3}{2}$
Now, we will substitute the value of c in equation (i). On doing so, we get
$\log \left( 1+y \right)=x+\dfrac{{{x}^{2}}}{2}-\dfrac{3}{2}$

Hence, the answer to the above question is $\log \left( 1+y \right)=x+\dfrac{{{x}^{2}}}{2}-\dfrac{3}{2}$ .

Note: The general mistake that a student makes in a problem related to integrations is that they forget about the constant term, which makes the equation that we get after integration is completely wrong. Also, you need to learn all the formulas related to integration and differentiation, as they are used very often in such questions. Identities related to logarithmic and exponential functions are also important.