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

Given: Slope at any point $=y+2 x$

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

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

$=y+2(x+1)=2 e^{x}$