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  • Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 subquestion (ii)

Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 subquestion (ii)

Answers (1)

Answer:  x+y+\log x=1

Give:  x \frac{d y}{d x}-y=\log x, y(1)=0

Hint: Using  \begin{aligned} &\int \frac{1}{x} d x\\ & \end{aligned}

Explanation: x \frac{d y}{d x}-y=\log x

Divide by x

=\frac{d y}{d x}+\left(\frac{-1}{x}\right) y=\frac{\log x}{x} 

This is a first order linear differential equation of the form

 \begin{aligned} &\frac{d y}{d x}+P x=Q \\ & \end{aligned}

 P=\frac{-1}{x} \text { and } Q=\frac{\log x}{x}

The integrating factor If  of this differential equation is

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

Hence, the solution of different equation is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \frac{1}{x}=\int \frac{\log x}{x} \frac{1}{x} d x+C \\ &=\frac{y}{x}=\int \frac{\log x}{x} \frac{1}{x} d x+C \ldots(i) \end{aligned}

We have  \begin{aligned} &\int \frac{\log x}{x} \frac{1}{x} d x \\ & \end{aligned}

Put  x=t \Rightarrow x=e^{t}

Using integration by parts

\begin{aligned} &=t \frac{e^{-t}}{-1}-\int(1) \frac{e^{-t}}{-1} d t \\ &=-t e^{-t}+e^{-t}+C \\ &=-\frac{t}{e^{t}}+\frac{1}{e^{t}}+C \\ &=-\frac{\log x}{x}+\frac{1}{x}+C \end{aligned}

Substituting in i

 =\frac{y}{x}=-\frac{\log x}{x}+\frac{1}{x}+C

Multiplying by x

 =y=1-\log x+C \ldots(i i)

Given  y\left ( 1 \right )= 0

\begin{aligned} \text { When } x &=1, y=0 \\ &=0=1-0+C \quad[\log 0=1] \\ &=C=-1 \end{aligned}

Substituting in ii

 \begin{aligned} &=y=1-\log x-1 x \\ &=y=1-\log x-x \\ &=y++\log x=1 \end{aligned}

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