By using the substitution y=vx, find the particular solution of the differential equation 2xy (dy/dx)=y^2-x^2 when y=4 and x=2. Express y in terms of x.?

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By using the substitution y=vx, find the particular solution of the differential equation 2xy (dy/dx)=y^2-x^2 when y=4 and x=2. Express y in terms of x.?

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Answer 1 · 2018-03-15T11:16:36

$y = x \sqrt{\frac{10}{x} - 1}$ $y = x sqrt(10/x-1)$

Explanation:

We use the substitution $y = v x$ $y=vx$ . This means that

$\frac{d y}{d x} = v + x \frac{d v}{d x}$ $dy/dx =v+x (dv)/dx$

Now, the given equation is

$2 x y (\frac{d y}{d x}) = y^{2} - x^{2}$ $2xy (dy/dx)=y^2-x^2$
or
$\frac{d y}{d x} = \frac{1}{2} (\frac{y}{x} - \frac{x}{y})$ $dy/dx=1/2(y/x-x/y)$

so that

$v + x \frac{d v}{d x} = \frac{1}{2} (v - \frac{1}{v}) \Rightarrow$ $v+x (dv)/dx =1/2(v-1/v) implies$
$x \frac{d v}{d x} = - \frac{1}{2} (\frac{1}{v} + v)$ $x (dv)/dx =-1/2(1/v+v)$
which can be rewritten in the form

$2 \frac{v}{v^{2} + 1} d v = - \frac{d x}{x} \Rightarrow$ $2 v/(v^2+1) dv=-(dx)/x implies$

$\frac{d (v^{2} + 1)}{v^{2} + 1} = - \frac{d x}{x}$ $(d(v^2+1))/(v^2+1) =-dx/x$

which readily integrates to

$ln (v^{2} + 1) = - ln x + C$ $ln(v^2+1) = -ln x + C$

The given initial condition is $y = 4$ $y=4$ at $x = 2$ $x=2$ . This means that at $x = 2$ $x=2$ , $v = \frac{4}{2} = 2$ $v=4/2=2$ . Substituting this in the solution we have obtained so far, we get

$ln 5 = - ln 2 + C$ $ln 5=-ln 2 +C$

giving the value of the constant of integration as $ln 10$ $ln 10$ .

The solution is thus

$(v^{2} + 1) = \frac{10}{x}$ $(v^2+1) = 10/x$

or

$\frac{y^{2}}{x^{2}} + 1 = \frac{10}{x}$ $y^2/x^2+1 =10/x$

or

$y = x \sqrt{\frac{10}{x} - 1}$ $y = x sqrt(10/x-1)$

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By using the substitution y=vx, find the particular solution of the differential equation 2xy (dy/dx)=y^2-x^2 when y=4 and x=2. Express y in terms of x.?

1 Answer

Explanation:

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