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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

Answers (1)

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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