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Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

92. 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 coordinates of that point.

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Vishal Baghel | Contributor-Level 10

4 months ago

We know the slope of tangent to curve is $\frac{d y}{d x}$ .

$∴$ $\frac{d y}{d x} = x + y$

$= \frac{d y}{d x} - y = x$ which has form $\frac{d y}{d x} + P y = Q$

So, $P = - 1 & Q = x$

$∴ I . F = e^{\int P d x} = e^{\int - d x} = e^{- x}$

Thus the solution is of the form .

$\begin{array}{l} y \times e^{- x} = \int x . e^{- x} d x + c \\ = x \int e^{- x} d x - \int \frac{d x}{d x} \int e^{- x} d x d x + c \\ = - x e^{- x} + \int e^{- x} d x + c \\ = y e^{- x} = - x e^{- x} - e^{- x} + c \\ = y = - x - 1 + \frac{c}{e^{- x}} \\ = y + x + 1 = c e^{x} \end{array}$

Given, the curve passes through origin (0,0) i.e, $y = 0, w h e n, x = 0$

$0 + 0 + 1 = c e^{0} = c = 1$

$∴$ Thus, equation of the curve is

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

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92. 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 coordinates of that point.

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