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  • Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 20 maths

Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 20 maths

Answers (1)

Answer:  y=\frac{x}{x+1}(x+\log x-1)

Given: x(x+1) \frac{d y}{d x}-y=x(x+1)

To find: We have to show that the curve which satisfies x(x+1) \frac{d y}{d x}-y=x(x+1)  and passes through \left ( 1,0 \right )

Hint: Use linear differential equation to solve.

Solution: x(x+1) \frac{d y}{d x}-y=x(x+1)

Dividing by  x(x+1)

        =\frac{d y}{d x}-\frac{y}{x(x+1)}=1        [This is a linear differential equation]

Comparing with \frac{d y}{d x}+P y=Q   we get

        P=\frac{1}{x(x+1)}, Q=1

\text { Now If }=e^{\int P d x}

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

        \begin{aligned} &=e^{-\log \left(\frac{x}{x+1}\right)} \\\\ &=e^{\log \left(\frac{x}{x+1}\right)^{-1}} \quad\quad\quad\left[e^{-\log x}=\log x^{-1}\right] \end{aligned}

        =\left(\frac{x}{x+1}\right)^{-1} \quad \quad \quad\left[e^{\log x}=x\right]

        =\frac{x+1}{x}

So, the solution is given by

        \begin{aligned} &=y(I f)=\int Q(I f) d x+C \\\\ &=y \times\left(\frac{x+1}{x}\right)=\int \frac{x+1}{x} d x+C \\\\ &=\left(\frac{x+1}{x}\right) y=\int \frac{x}{x} d x+\int \frac{1}{x} d x+C \end{aligned}

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

Since curve passes through the point \left ( 1,0 \right ) it satisfies the equation of the curve

        =\left(\frac{1+1}{1}\right) 0=1+\log 1+C

        =0=1+0+C \quad[\log 1=0]

        =C=-1

Substituting value of C in equation (i) we get

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

Hence the required equation for the curve is found.

 

Posted by

infoexpert26

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