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  • Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (xi) textbook solution.

Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (xi) textbook solution.

Answers (1)

Answer :  y \log x=-\frac{2}{x}(1+\log x)+c

Hint :  you integrate by integrating x^{n}

Given :  x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x

Solution : \text { put in form } \frac{d y}{d x}+P y=Q

x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x

Divide bot side by (x \log x)

\begin{aligned} &\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \log x \times \frac{1}{x \log x}\\ &\frac{d y}{d x}+\left(\frac{1}{x \log x}\right) y=\frac{2}{x^{2}} \end{aligned}               ...(i)

By comparing (i) with \frac{d y}{d x}+P y=Q

\begin{aligned} &P=\frac{1}{x \log x} \quad \text { and } Q=\frac{2}{x^{2}} \\ &\text { I. } F=e^{\int P d x} \\ &\text { I. } F=\int e^{\frac{1}{x \log x} d x} \\ &\text { Let } t=\log x \end{aligned}

\begin{aligned} &d t=\frac{1}{x} d x \\ &I . F=e^{\int \frac{1}{t} d t} \\ &I \cdot F=e^{\log |t|} \\ &\text { I.F }=|t|=\log |x| \end{aligned}

\begin{gathered} y \times \text { I.F }=\int Q \times \text { I.F } d x+c \\ y \times \log x=\int \frac{2}{x^{2}} \log x d x+c \end{gathered}                                ....(ii)

           \text { Let } I=2 \int \log x \cdot x^{-2} d x

\begin{aligned} &I=2\left[\log x \int x^{-2} d x-\int\left\{\frac{1}{x} \int x^{-2} d x\right\} d x\right] \\ &I=2\left[\log x \frac{x^{-1}}{-1}-\int \frac{1}{x} \frac{x^{-1}}{-1} d x\right] \end{aligned}

\begin{aligned} &=2\left(-\log x \frac{1}{x}+\int \frac{1}{x^{2}} d x\right) \\ &I=2\left(\frac{-1}{x} \log x-\frac{1}{x}\right)=\frac{-2}{x}(1+\log x) \end{aligned}

Now enq (ii) becomes

\begin{aligned} &y \log x=I+C \\ &y \log x=\frac{-2}{x}(1+\log x)+c \end{aligned}

 

Posted by

infoexpert23

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