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  • Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 17 maths textbook solution

Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 17 maths textbook solution

Answers (1)

Answer: proved.

Given: Slope at any point =y+2 x

To find: We have to show that the equation of curve which pass through the origin isy+2(x+1)=2 e^{2 x}

Hint: Use the linear differential equation i.e. \frac{d y}{d x}+P y=Q

Solution: Slope at any point =y+2 x

        \text { i.e. } \frac{d y}{d x}=y+2 x

        =\frac{d y}{d x}-y=2 x

It is a linear differential equation comparing it with \frac{d y}{d c}+P y=Q

        P=-1, Q=2 x

Integrating factor (I f)=e^{\int P d x}

        \begin{aligned} &=>\text { I.F. }=e^{\int(-1) d x} \\\\ &=>\text { I.F. }=e^{-x} \end{aligned}

Solution of the equation is given by

        \begin{aligned} &=y \times \text { I.F. }=\int Q(\text { I.F. }) d x+C \\\\ &=>y e^{-x}=\int(2 x)\left(e^{-x}\right) d x+C \\\\ &=>y e^{-x}=2 \int(x)\left(e^{-x}\right) d x+C \end{aligned}

        =>y e^{-x}=2\left[x \int e^{-x}-\int\left(\frac{d x}{d x} \int e^{-x} d x\right) d x\right]+C[Using integration by parts]

        =>y e^{-x}=2\left(-x e^{-x}+\int 1 e^{-x} d x\right)+C

        =>y e^{-x}=2\left(-x e^{-x}-e^{-x}\right)+C

        \begin{aligned} &=>y e^{-x}=e^{-x}\left(-2 x-2+C e^{x}\right) \\\\ &=>y=-2 x-2+C e^{x} \\\\ &=>y+2(x+1)=C e^{x} \ldots(i) \end{aligned}

If it passes through origin

        \begin{aligned} &=0+2(0+1)=C e^{0} \\\\ &=>C=2 \end{aligned}

Now equation (i) becomes

        =y+2(x+1)=2 e^{x}

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