#### Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 23

Answer: $y+1=2 e^{\frac{x^{2}}{2}}$

Given: slope $=x+y x$

Hint: x-coordinate is abscissa and y-coordinate is ordinate

Solution:

According to ques,

\begin{aligned} &\frac{d y}{d x}=x+y x \\\\ &\frac{d y}{d x}=x(1+y) \\\\ &\frac{d y}{1+y}=x d x \end{aligned}

Integrating both sides w.r.t x

$\int \frac{d y}{1+y}=\int x d x$

$\log |1+y|=\frac{x^{2}}{2}+c \quad \text {........(i) } \quad\left[\int \frac{d x}{x}=\log x+c\right]$

Since, the curve passes through (0,1)

\begin{aligned} &\log |1+1|=\frac{0^{2}}{2}+c \\\\ &\mathrm{c}=\log 2 \end{aligned}

Putting the value of c in eqn (i)

\begin{aligned} &\log |1+y|=\frac{x^{2}}{2}+\log 2 \\\\ &\log |1+y|-\log 2=\frac{x^{2}}{2} \quad\quad\quad\quad\left[\log m-\log n=\log _{n}^{m}\right] \end{aligned}

$\begin{gathered} \log \frac{1+y}{2}=\frac{x^{2}}{2} \\\\ \frac{1+y}{2}=e^{\frac{x^{2}}{2}} \\\\ y+1=2 e^{\frac{x^{2}}{2}} \end{gathered}$