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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the equation of 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 tangent to the curve at that point by 5

$\begin{array}{1 1} y=x-4-2e^x \\ y=x-4+2e^x \\ y=x+4-2e^x \\ y=-x+4-2e^x \end{array}$

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  • Slope of the curve is $\large\frac{dy}{dx}$
  • To solve a first order linear differential equation of the form $\large\frac{dy}{dx}$$ + Py = Q$
  • (i) First write the equation in the form $\large\frac{dy}{dx}$$ + Py = Q$
  • (ii) Find the integral Factor (I.F) =$ e^{\int Pdx}$
  • Write the solution as y(I.F) = integration of Q. (I.F) dx + C
Step 1:
According to the given information, let F(x,y) be the curve and dy/dx be the slope to the curve.
Hence $\large\frac{dy}{dx }$$ + 5 = x + y$
Let us rewrite the equation as $\large\frac{dy}{dx}$$ -y = x + 5$
The equation is of the form $\large\frac{dy}{dx}$$ + Py = Q$
Here $P = -1$ and $Q = x + 5$
Step 2:
To find the integrating factor
$\int -dx=-x$
Hence $I.F = e^{-x}$
The required solution is $ye^{-x} = \int(x+5).e^{-x}.dx + C$
$\int(x+5) e^{-x }dx$ can be done by parts.
Let $u = x+5$, hence $du = dx$ and $dv = e^{-x}dx ;v = -e^{-x}$
$\int(x+5)e^{-x} = (x+5)(-e^{-x}) -\int(-e^{-x})dx$
$\qquad\qquad\;\;=-(x+5)e^{-x}+\int e^{-x}dx$
Hence the required solution is $ye^{-x }= -(x+5)e^{-x} - e^{-x} + C$
$ye^-x = (5-x)e^{-x} - e^{-x} + C $
$ye^-x = (4-x)e^{-x} + C$
Step 3:
To evaluate $C$, let us substitute the values of $x$ and $y$ in the above equation.
$2e^0 = (4-0)e^0 + C$
$C = - 2$
Substituting for $C$ we get,
$ye^{-x} = (4 - x) e^{-x} - 2$
dividing throughout by $e^{-x}$ we get,
$y = (4-x) - \large\frac{2}{e^{-x}}$
$y = x - 4 - 2e^x$ is the required solution.
answered Jul 30, 2013 by sreemathi.v

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