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# 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. - CBSE (Commerce) Class 12 - Mathematics

ConceptMethods of Solving First Order, First Degree Differential Equations Linear Differential Equations

#### Question

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 Solution for Mathematics Textbook for Class 12 (2018 to Current)
Chapter 9: Differential Equations
Q: 16 | Page no. 414

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Solution 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. Concept: Methods of Solving First Order, First Degree Differential Equations - Linear Differential Equations.
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