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

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

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