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

93. Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangents to the curve at that point by 5.

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Vishal Baghel | Contributor-Level 10

4 months ago

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

$∴ x + y = \frac{d y}{d x} + 5$

$\frac{d y}{d x} - y = x - 5$ Which has form $\frac{d y}{d x} + P x = Q$

$w h e r e, P = 1 & Q = x - 5$

$∴ I . F = e^{\int P d x} = e^{\int - 1 d x} = e^{- x}$

Thus the solution has the form

$\begin{array}{l} y e^{- x} = \int (x - 5) e^{- x} d x + c \\ = \int x e^{- x} d x - \int 5 e^{- x} d x + c \\ = y e^{- x} = I + 5 e^{- x} + c \\ w h e r e, I = \int x e^{- x} d x \\ = x \int e^{- x} d x - \int \frac{d}{d x} x \int e^{- x} d x d x \\ = - x e^{- x} + \int e^{- x} d x \\ = - x e^{- x} - e^{- x} \end{array}$

$\begin{array}{l} ∴ y e^{- x} = - x e^{- x} - e^{- x} + 5 e^{- x} + c \\ = y e^{- x} = - x e^{- x} + 4 e^{- x} + c \\ = y = - x + 4 + \frac{c}{e^{- x}} \\ = y + x = 4 + c e^{x} \end{array}$

Given, the curve passes through (0,2) so y=2 when x=0

$\begin{array}{l} 2 + 0 = 4 c e^{0} \\ 2 - 4 = c \\ c = - 2 \end{array}$

$∴$ The particular solution is

$y + x = 4 - 2 e^{x}$

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93. Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangents to the curve at that point by 5.

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