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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 18

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 18

Answers (1)

Answer: y e^{-3 x}=\left(-\frac{2}{3} x-\frac{2}{9}\right) e^{-3 x}+\frac{26}{9} e^{-3}

Given: Tangent makes an angle (2 x+3 y) with x-axis

To find: We have to show that the equation of curve which pass through (1,2)

Hint: Find the slope of the tangent then solve using linear differential equation.

Solution: Slope of tangent t=\tan \theta and tangent makes angle (2 x+3 y)

        =>\frac{d y}{d x}=\tan \left[\tan ^{-1}(2 x+3 y)\right]

        =>\frac{d y}{d x}=2 x+3 y

        =>\frac{d y}{d x}-3 y=2 x

It is a linear differential equation

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

        =P=-3, Q=2 x

        \begin{aligned} &\text { If }=e^{\int P d x} \\\\ &=>I f=e^{-\int 3 d x} \\\\ &=>I f=e^{-3 x} \end{aligned}

Solution of equation is given by

        =y(I f)=\int Q(I f) d x+C \\

        =>y\left(e^{-3 x}\right)=\int 2 x e^{-3 x} d x+C

Using integration by parts we get

        \begin{aligned} &=>y\left(e^{-3 x}\right)=2\left[x \int e^{-3 x}-\int\left(\frac{d x}{d x} \int e^{-3 x} d x\right) d x\right]+C \\\\ &=>y\left(e^{-3 x}\right)=2\left[-\frac{x e^{-3 x}}{3}-\frac{1}{3} \int e^{-3 x} d x\right]+C \end{aligned}

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

Taking e^{-3 x} common on both sides

        =>y=-\frac{2}{3} x-\frac{2}{9}+C \ldots(i)

If it passes through \left ( 1,2 \right )

        \begin{aligned} &=>2=-\frac{2}{3} \times 1-\frac{2}{3}+C e^{3} \\\\ &=>C=\frac{26}{9} e^{-3} \end{aligned}

So equation (i) becomes

        =>y e^{-3 x}=\left(-\frac{2}{3} x-\frac{2}{9}\right) e^{-3 x}+\frac{26}{9} e^{-3}

 

Posted by

infoexpert26

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