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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Solve: $(x+y)(dx-dy)=dx+dy.$[Hint:Substitute x+y=z after separating dx and dy]

$\begin{array}{1 1}(A)\;c\;e^{\large y+x}=x+y \\(B)\;c\;e^{\large x-y}=x+y \\(C)\;c\;e^{\large y-x}=x+y \\(D)\;c\;e^{\large y-x}=x-y \end{array} $

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Toolbox:
  • A linear differential equation of the form.$ \large\frac{dy}{dx}$$=(x+y)$ can be solved by substituting $x+y=z$ and $\large\frac{dy}{dx}=\frac{dz}{dx}$$-1$
  • Then the variable can be seperated and integrated.
  • $ \int \large\frac{dx}{x}$$=\log x$
Given $(x+y)(dx-dy)=dx+dy$
This can be written as
$xdx-xdy+ydx-ydy=dx+dy$
=> $ dx(x+y-1)=dy(x+y+1)$
(ie)$dx(x+y-1)=dy(x+y+1)$
=>$\large\frac{dy}{dx}=\large\frac{x+y-1}{x+y+1}$-----(1)
Let $x+y=z$ on differentiating w.r.t x we get
$1+\large\frac{dy}{dx}=\frac{dz}{dx}$
or $ \large\frac{dy}{dx}=\frac{dz}{dx}-1$
Now substituting in equ (1) we get,
$\large\frac{dz}{dx}$$-1=\large\frac{z-1}{z+1}$
=>$\large\frac{dz}{dx}=\frac{z-1}{z+1}$$+1$
$=\large\frac{z-1+z+1}{z-1}$
$\large\frac{dz}{dx}=\frac{2z}{z-1}$
On seperating the variables we get
$\bigg(\large\frac{z-1}{2z}\bigg)$$dz=dx$
=>$\large\frac{1}{2}\bigg(\frac{z-1}{z}\bigg)$$dz=dx$
On integrating we get
$\large\frac{1}{2}\bigg[\int \frac{z}{z} $$dz-\int \large\frac{1}{z}$$dz\bigg]=\int dx$
=>$\large\frac{1}{2}\bigg[\int $$dz-\int \large\frac{1}{z}$$dz\bigg]=\int dx$
$\large\frac{1}{2}\bigg[$$z-\log z\bigg]=x+c$
$z-\log z=2x+c$
Now substituting for $z=(x+y)$
$x+y-\log(x+y)=2x+c$
$y=x+\log (x+y)+c$
or $y-x=\log _e (x+y)$
$c\;e^{\large y-x}=x+y$
answered May 9, 2013 by meena.p
 
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