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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find a particular solution of the differential equation \((x-y)(dx+dy) = dx - dy\),given that \( y=\;-1\),when \( x=\;0\)

\[(Hint:put\; x-y=\;t)\]

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1 Answer

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  • A linear differential equation of the form $\large\frac{dy}{dx}$$=F(x,y)$ where $F(x.y)$ is of the form $g(x).h(y)$ where $g(x)$ is a function of $x$ and $h(y)$ is a function of $y$,then the equation is variable separable type.Such equations can be solved by separating the variables.
Step 1:
Given :$(x-y)(dx+dy)=dx-dy$
On expanding and rearranging we get,
$x(dx-dy)-y(dx-dy)=dx-dy$
$(x-y+1)dy=(1-x+y)dx$
$\large\frac{dy}{dx}=\frac{(1-x+y)}{(x-y+1)}$
Step 2:
Let $(x-y)=t$ then $1-\large\frac{dy}{dx}=\frac{dt}{dx}$ or $\large\frac{dy}{dx}=$$1-\large\frac{dt}{dx}$
$1-\large\frac{dt}{dx}=\frac{(1-t)}{(1+t)}$
$-\large\frac{dt}{dx}=\frac{(1-t)}{(1+t)}$$-1$
$-\large\frac{dt}{dx}=\frac{-2t}{(1+t)}$
$\large\frac{dt}{dx}=\frac{2t}{(1+t)}$
Separating the variables
$\large\frac{(1+t)dt}{2t}$$=dx$
$\large\frac{dt}{2t}+\frac{dt}{2}$$=dt$
Step 3:
Integrating on both sides we get,
$\large\frac{1}{2}$$\log t+\large\frac{1}{2}$$t=t+C$
substituting for t we get,
$\large\frac{1}{2}$$\log (x-y)+\large\frac{1}{2}$$(x-y)=(x-y)+C$
$\large\frac{1}{2}$$\log (x-y)=\large\frac{(x+y)}{2}$$+C$
$\log(x-y)=(x+y)+2C$
Step 4:
To evaluate for $C$ let us substitute the given values of $x=0$ and $y=-1$
$\log(1)=-1+k$(where $2C=k$)
$k=1$
$\log(x-y)=(x+y)+1$
$\log|x-y|=x+y+1$
This is the required solution.
answered Jul 30, 2013 by sreemathi.v
 

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