#### Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 43

Answer: $r=r_{0} e^{\frac{-t^{2}}{2}}$

Hint: Separate the terms of x and y and then integrate them.

Given: $\frac{d r}{d t}=-r t, r(0)=r_{0}$

Solution:

\begin{aligned} &\frac{d r}{d t}=-r t \\\\ &\frac{d r}{r}=-t d t \end{aligned}

Integrating both sides

\begin{aligned} &\Rightarrow \int \frac{d r}{r}=-\int t d t \\\\ &\Rightarrow \log |r|=-\frac{t^{2}}{2}+c \end{aligned}                        ...............(1)

Given that $r(0)=r_{0} \text { i.e. at } t=0, r=r_{0}$

Put in (1) $\log r_{0}=0+c \Rightarrow c=\log r_{0}$

Put in (1) we get

\begin{aligned} &\log |r|=-\frac{t^{2}}{2}+\log r_{0} \Rightarrow \log |r|-\log r_{0}=-\frac{t^{2}}{2} \\\\ &\Rightarrow \log \frac{r}{r_{0}}=-\frac{t^{2}}{2} \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} \log _{e} a=x \\ \Rightarrow a=e^{x} \end{array}\right]} \\\\ &\Rightarrow \frac{r}{r_{0}}=e^{\frac{-t^{2}}{2}} \\\\ &\Rightarrow r=r_{0} e^{\frac{-t^{2}}{2}} \end{aligned}