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

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

Answers (1)

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}

 

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