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

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

Answers (1)

Answer : c y=\log \left|\frac{y}{x}\right|-1

Hint : You must know the rules of solving differential equation and integration

Given : y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0

Solution : y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0

               \begin{aligned} &d y\left(x \log \left(\frac{y}{x}\right)-2 x\right)=-y d x\\ &\frac{d y}{d x}=\frac{-y}{x \log \left(\frac{y}{x}\right)-2 x}\\ &\frac{d y}{d x}=\frac{-y}{-x\left(2-\log \left(\frac{y}{x}\right)\right)}\\ &\frac{d y}{d x}=\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)} \end{aligned}

Step : 2

Putting f(x, y)=\frac{d y}{d x} \text { and finding }(f(\lambda x, \lambda x))

           \begin{gathered} f(x, y)=\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)} \\ \qquad \begin{array}{c} f(\lambda x, \lambda y)=\frac{\frac{\lambda y}{\lambda x}}{2-\log \left(\frac{\lambda y}{\lambda x}\right)} \\ =\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)} \\ f(\lambda x, \lambda y)=f(x, y) \end{array} \end{gathered}

Step : 3

Solve by y=vx

Put y=vx

Diff w.r.t.x

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

Put value of \frac{d y}{d x} \text { and } y=v x \text { in eqn(i) }

\begin{aligned} &\frac{d y}{d x}=\frac{\frac{y}{x}}{2-\log \frac{y}{x}} \\ &x \frac{d v}{d x}=\frac{v}{2-\log v}-v \\ &x \frac{d v}{d x}=\frac{v-2 v+v \log v}{2-\log v} \end{aligned}

\begin{aligned} &x \frac{d v}{d x}=\frac{v \log v-v}{2-\log v} \\ &\frac{2-\log v}{v \log v-v} d v=\frac{d x}{x} \end{aligned}

Integrate both sides

\begin{aligned} &\int \frac{2-\log v}{v \log v-v} d v=\int \frac{d x}{x} \\ &\int \frac{1+1-\log v}{v(\log v-1)} d v=\log x+\log c \\ &\int \frac{1}{v(\log v-1)} d v-\int \frac{1}{v} d v=\log x+\log c \\ &\int \frac{d v}{v(\log v-1)}-\log v=\log x+\log c \end{aligned}

Put t=\log v-1

dt = \frac{1}{v}dv

So our equation

\int \frac{d t}{t}-\log v=\log x+\log c

Put value of t

\begin{aligned} &\log (\log v-1)-\log v=\log x+\log c \\ &\log (\log v-1)=\log x+\log c+\log v \\ &\log (\log v-1)=\log (x c v) \\ &\text { put } v=\frac{y}{x} \end{aligned}

\begin{aligned} &\log \left(\log \frac{y}{x}-1\right)=\log x c \frac{y}{x} \\ &\log \left(\log \frac{y}{x}-1\right)=\log c y \\ &\log \frac{y}{x}-1=c y \\ &\Rightarrow c y=\log \frac{y}{x}-1 \end{aligned}

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