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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the equation of 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 .

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1 Answer

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Toolbox:
  • $\large\frac{dy}{dx}$ is the slope of the curve.
  • If the given equation is a first order linear equation, to solve this:
  • (i) Write the given equation in the form of $\large\frac{dy}{dx}$$ + Py = Q$
  • (ii) Find the integrating factor (I.F) = $e^{\int Pdx}$.
  • (iii) Write the solution as y(I.F) = integration of $Q(I.F) dx + C$
Step 1:
With the given information:
Let the curve be $F(x,y)$ and $\large\frac{dy}{dx}$ be the slope.
Hence the equation is $\large\frac{dy}{dx }$$ = x + y$
$\large\frac{dy}{dx}$$ - y = x$
Hence the equation is a first order linear equation, where $P = - 1$ and $Q = x$
$\int Pdx = \int-dx = - x$
Hence the $ I.F = e^{-x}$
Hence the required solution is $ye^{-x} =\int x. e^{-x} dx + C$
Step 2:
$ye^{-x} =\int x. e^{-x} dx + C$ can be done by parts.
Let $u = x$, hence $du = dx$ and $dv = e^{-x}dx ;v = -e^{-x}$
$\int ye^{-x} = x(-e^{-x}) - \int- e^{-x}dx$
$\qquad\;\;\;= -xe^{-x} +\int e^{-x}dx$
$\qquad\;\;\;= -(xe^{-x}) + (-e^{-x}) + C$
$\qquad\;\;\;= -e^{-x(x + 1)} + C$
Hence the required solution is $ye^{-x}=-e^{-x(x+1)}+C$
Step 3:
To evaluate $C$ let us substitute the given values of $x$ and $y$
It is given that the curve passes through the origin, hence $x = 0$ and $y = 0$
$0 = - e^{0(0+1)} + C$
$C = 1$
Hence the solution is $ye^{-x} = -e^{-x(x + 1)} + 1$
divide throughout by $e^{-x}$
$y=-(x+1)+\large\frac{1}{e^{-x}}$
$x+y+1=e^x$
This is the required solution.
answered Jul 31, 2013 by sreemathi.v
 

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