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Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (v)

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Answer:  \begin{aligned} &y=\frac{1}{2} \log x+\frac{c}{\log x}\\ & \end{aligned}

Give: x \log x) \frac{d y}{d x}+y=\log x\\

Hint: Using   \int \frac{1}{x} d x\\

Explanation:  (x \log x) \frac{d y}{d x}+y=\log x\\

Divide by  x \log x

\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\frac{1}{x \log x} \text { and } Q=\frac{1}{x} \end{aligned}


The integrating factor  If  of this differential equation is


 \begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \frac{1}{x \log x} d x} \quad\left[\log x=t, \frac{1}{x} d x=d t\right] \\ &=e^{\int \frac{1}{t} d t} \\ &=e^{\log t} \end{aligned}

\begin{aligned} &=e^{\log (\log x)} \quad\left[e^{\log x}=x\right] \\ &=\log x \end{aligned}

Hence, the solution of the differential equation is

 \begin{aligned} &y(I f)=\int \text { QIfd } x+C \\ &=y \log x=\int \frac{1}{x} \log x d x+C \\ &=y \log x=\frac{\log ^{2} x}{x}+C \quad\left[\int x d x=\frac{x^{2}}{2}\right] \\ &=y=\frac{1}{\log x}\left(\frac{\log ^{2} x}{2}+C\right) \\ &=y=\frac{\log ^{2} x}{2 \log x}+\frac{C}{\log x} \\ &=y=\frac{\log x}{2}+\frac{C}{\log x} \end{aligned}

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