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

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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}


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