Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 30

Answer: $x+y=e^{x}-1$

Given: The slope of tangent at each point of a curve is equal to the sum of the co-ordinate of the point i.e. slope of tangent at $(x, y)=x+y$

To find: The curves that pass through the origin.

Hint: As  $\frac{d y}{d x}+P y=Q$ then it is a linear differential equation.

Solution: Slope of tangent at $(x, y)=x+y$

\begin{aligned} &=\frac{d y}{d x}=x+y \\\\ &=\frac{d y}{d x}-y=x \end{aligned}

It is a linear differential equation

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

$=P=-1, Q=x$

$=I f=e^{\int \mathrm{P} d x}$

\begin{aligned} &=I f=e^{-\int 1 d x} \\\\ &=I f=e^{-x} \end{aligned}

It is a linear differential equation

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

\begin{aligned} &=y(I f)=\int \mathrm{Q}(I f) d x+C \\\\ &=y\left(e^{-x}\right)=\int x e^{-x} d x+C \end{aligned}

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

\begin{aligned} &=y e^{-x}=-x e^{-x}-\int e^{-x} d x+C \\\\ &=y e^{-x}=-x e^{-x}+e^{-x}+C \end{aligned}

Dividing by $e^{-x}$ on both sides

$=y=-x-1+C e^{x} \ldots(i)$

As it passes through origin $(x, y)=(0,0)$

\begin{aligned} &=0=0-1+C e^{0} \\\\ &=C=1 \end{aligned}

Substituting C in equation (i)

\begin{aligned} &=y=-x-1+e^{x} \\\\ &=x+y=e^{x}-1 \end{aligned}

Hence, equation of curve is found