Question

A curve passes through the origin and the slope of the tangent to the curve at any point (x) is equal to the sum of the coordinates of the point. Find the equation of the curve.

Answer 1

Formula :

i) \(\int\) 1 dx = x

ii) \(\int\) u. v dx = u. \(\int\)  n dx - \(\int\) \((\frac{du}{dx}.\int v\, dx)\) dx

iii) \(\int\) e^kx dx = \(\frac{e^{kx}}{k}\)

iv) \(\frac{d}{dx}\) (xⁿ) = nx^n-1

v) General solution :

For the differential equation in the form of

\(\frac{dy}{dx}\, +Py\, =Q\)

General solution is given by,

y. (I.F.) = \(\int\) Q. (I.F) dx + x

Where, integrating factor,

I.F. = \(e^{\int \,p\,dx}\)

The slope of the tangent to the curve \(=\frac{dy}{dx}\)

The slope of the tangent to the curve is equal to the sum of the coordinates of the point.

\(\therefore \frac{dy}{dx} = x + y\)

Therefore differential equation is

Equation (1) is of the form

\(\frac{dy}{dx} \, + Py\, =Q\)

where, P = -  1 and Q = x

Therefore, integrating factor is

General solution is

Let, u=x and v= e^-x

Substituting I in eq(2),

Dividing above equation by e^-x,

Therefore, general solution is

y + x + 1 = c.e^x

The curve passes through origin , therefore the above equation satisfies for x=0 and y=0,

∴ 0 + 0 + 1 = c.e⁰

∴ c = 1

Substituting c in general solution,

∴ y + x + 1 = e^x

Therefore, equation of the curve is

y + x + 1 = e^x