Question

Find the equation of a 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 coordinates of the point.

Answer 1

tangent to the curve at any point (x,y) is equal to the sum of the coordinates of the point.
Let F(x,y) be the curve passing through the origin.
At point(x,y) the slope of the curve will be \(\frac {dy}{dx}\).

According to the given information:

\(\frac {dy}{dx}\) = x+y

⇒\(\frac {dy}{dx}\)-y = x

This is a linear differential equation of the form:

\(\frac {dy}{dx}\)+py = Q (where p=-1 and Q=x)

Now, I.F. = e∫(Q×I.F.)dx + C

⇒ye^-x = ∫xe^-x dx + C      ….....(1)

Now, ∫xe^-xdx = x∫e^-x dx - ∫[\(\frac {d}{dx}\)(x).∫e^-x dx]dx

= -xe^-x-∫-e^-x dx

= -xe^-x+(-e^-x)

= -e^-x(x+1)

Substituting in equation(1), we get:

ye^-x = -e^-x(x+1) + C

⇒Y = -(x+1) + C^ex

⇒x+y+1 = C^ex      …….(2)

The curve passes through the origin.

Therefore,equation(2) becomes:

C = 1

Substituting C=1 in equation(2), we get:

⇒x+y+1 = e^x

Hence, the required equation of curve passing through the origin is x+y+1 = e^x