Q. 48 - Calculus

Question

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29-52.

$\frac{dy}{dx}=xy+x+y+1, y\left(0\right)=c$

Step-by-Step Solution

Answer

On solving, we get $y\left(x\right)=-1+(1+c)e^{\frac{1}{2}(x+1{)}^2}$

Step 1. Given information

Given the expression $$\begin{gathered} \frac{dy}{dx}=xy+x+y+1, \\ y\left(0\right)=c \end{gathered}$$

Step 2: Take common and use variable separable method

Calculating, we get

$$\begin{gathered} \frac{dy}{dx}=x(y+1)+1(y+1) \\ =(x+1)(y+1) \end{gathered}$$

Integrating, we get

$$\begin{gathered} \int \frac{1}{y+1}dy=\int (x+1)dx \\ \ln |y+1|=\frac{1}{2}(x+1{)}^2+C \\ y+1=e^{\frac{1}{2}(x+1{)}^2+C} \\ y=-1+A{e}^{\frac{1}{2}(x+1{)}^2} \end{gathered}$$

Step 3: Substitute x = 0 , y = c in the equation and solve

Calculating, we get

$$\begin{gathered} c=-1+A \\ A=1+c \\ y\left(x\right)=-1+(1+c)e^{\frac{1}{2}(x+1{)}^2} \end{gathered}$$