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

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