Question

Find the equation of curve passing through the origin given that the slope of the tangent to the curve at any point \((x,y)\) is equal to the sum of the coordinates of the point .

Answer 1

Answer:

$\text{The equation of the required curve is }\bf\:log(x+y+1)=x$

Step-by-step explanation:

Given:

$\text{Slope of the tangent = x+y}$

$\frac{dy}{dx}=x+y$

$\boxed{\begin{minipage}{4cm} Take,\:u=x+y\\ \\\frac{du}{dx}=1+\frac{dy}{dx}\\ \\ \implies\:\frac{dy}{dx}=\frac{du}{dx}-1 \end{minipage}}$

$\implies\frac{du}{dx}-1=u$

$\implies\frac{du}{dx}=u+1$

$\implies\frac{du}{u+1}=dx$

$\text{Integrating }$

$\int{\frac{du}{u+1}}=\int{dx}$

$log(u+1)=x+c$

$log(x+y+1)=x+c$

since the curve passes through (0,0),

$log(0+0+1)=0+c$

$\implies\:c=0$

$\therefore\text{The equation of the required curve is }\bf\:log(x+y+1)=x$

Answer 2

Answer:

Equation of the curve is,

$log(x+y+1)=x$

Step-by-step explanation:

In the question,

We have been given that the equation of the slope is,

Slope = Sum of the co-ordinates at that point

So,

$\frac{dy}{dx}=x+y$

Now,

Let us say,

x + y = t

So,

On differentiating it w.r.t 'x' we get,

$\frac{dt}{dx}=1+\frac{dy}{dx}\\\frac{dy}{dx}=\frac{dt}{dx}-1\\\frac{dt}{dx}-1=t\\\frac{dt}{dx}=t+1\\\frac{dt}{t+1}=dx$

Now, on integrating the given equation we get,

$log(t+1)=x+c\\log(x+y+1)=x+c$

Now, we know that curve is passing through the origin i.e. points (0, 0).

So, on putting we get,

$log(0+0+1)=0+c\\c=log1=0$

So, the equation of the curve is given by,

$log(x+y+1)=x$