Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# Find the Equation of the 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. - Mathematics

Sum

Find the equation of the 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.

#### Solution

According to the question,

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

$\Rightarrow \frac{dy}{dx} - y = x$

$\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}$

$P = - 1$

$Q = x$

Now,

$I . F . = e^{- \int dx} = e^{- x}$

So, the solution is given by

$y \times I . F . = \int Q \times I . F . dx + C$

$\Rightarrow y e^{- x} = x\int e^{- x} dx - \int\left[ \frac{d}{dx}\left( x \right)\int e^{- x} dx \right]dx + C$

$\Rightarrow y e^{- x} = - x e^{- x} + \int e^{- x} dx + C$

$\Rightarrow y e^{- x} = - x e^{- x} - e^{- x} + C$

Since the curve passes throught the origin, it satisfies the equation of the curve.

$\Rightarrow 0 e^0 = - 0 e^0 - e^0 + C$

$C = 1$

Putting the value of C in the equation of the curve, we get

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

$\Rightarrow y e^{- x} + x e^{- x} + e^{- x} = 1$

$\Rightarrow \left( y + x + 1 \right) e^{- x} = 1$

$\Rightarrow \left( x + y + 1 \right) = e^x$

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 12 Maths
Chapter 9 Differential Equations
Q 16 | Page 414
RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 74 | Page 147