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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 22

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 22

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Answer: 3(x+3 y)=2\left(1-e^{3 x}\right)

Given: Equation of slope x+3 y-1

To find: We have to show that the equation of the curve which passes through the origin.

Hint: Use linear differential equation to solve i.e. \frac{d y}{d x}+P y=Q  and its solution y \times I f=\int(Q \times I f) d x   whereIf=e^{\int Pdx}

Solution: Equation of slope =x+3 y-1

                                          =\frac{d y}{d x}=x+3 y=1

                                          =\frac{d y}{d x}-3 y=x-1        [It is linear differential equation]

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

        \begin{aligned} &P=-3, Q=x-1 \\\\ &=I f=e^{\int P d x} \\\\ &=e^{\int-3 d x} \end{aligned}

        \\\begin{aligned} &=e^{-3 \int d x} \\\\ &=e^{-3 x} \end{aligned}

Solution of the given equation is

        \begin{aligned} &=y(I f)=\int Q(I f) d x+C \\\\ &=y\left(e^{-3 x}\right)=\int(x-1)\left(x^{-3 x}\right) d x+C \\\\ &=y e^{-3 x}=(x-1) \int e^{-3 x} d x-\int\left[\frac{d}{d x}(x-1) \int e^{-3 x} d x\right] d x+C \end{aligned}[using integration by parts]

        \begin{aligned} &=y e^{-3 x}=(x-1)\left(-\frac{1}{3} e^{-3 x}\right)-\int\left(-\frac{e^{-8 x}}{3}\right) d x+C \\\\ &=y e^{-3 x}=-\left(\frac{x-1}{3}\right) e^{-3 x}-\frac{1}{3} \int e^{-3 x} d x+C \end{aligned}

        =y e^{-3 x}=-\left(\frac{x-1}{3}\right) e^{-3 x}-\frac{1}{3}\left(\frac{e^{-s x}}{3}\right)+C

Takinge^{-3 x} common on both sides,

        =y=-\frac{x}{3}+\frac{1}{3}-\frac{1}{9}+\frac{c}{e^{-3 x}}

        \begin{aligned} &=y=-\frac{x}{3}+\frac{3-1}{9}+C e^{3 x} \\\\ &=y=-\frac{x}{3}+\frac{2}{9}+C e^{3 x} \ldots(i) \end{aligned}

As it passes through origin

        \begin{aligned} &=0=-\frac{0}{3}+\frac{2}{9}+C e^{3(0)} \\\\ &=C=-\frac{2}{9} \end{aligned}

Now, substituting in equation (i)

        \begin{aligned} &=y=-\frac{x}{3}+\frac{2}{9}-\frac{2}{9} e^{3 x} \\\\ &=y=\frac{-3 x+2-2 e^{8 x}}{9} \\\\ &=9 y=-3 x+2-2 e^{3 x} \end{aligned}

        \begin{aligned} &=9 y+3 x=2\left(1-e^{3 x}\right) \\\\ &=3(3 y+x)=2\left(1-e^{3 x}\right) \end{aligned}

Hence, required equation of curve is found.

 

 

Posted by

infoexpert26

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