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Find the equation of a curve passing through (2,1) if the slope of the tangent to the curve at any point (x,y) is $\Large \frac{x^2+y^2}{2xy}.$

1 Answer

  • A linear differential equation of the form $\large\frac{dy}{dx}=$$f(x,y)$ is a homogenous differential equation. Such equation can be solved by substituting $y=vx$ and $\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
  • $\int \large \frac{dx}{x+a}$$=\log |x+a|+c$
  • Slope of the tangent to the curve is $\large\frac{dy}{dx}$
Step 1:
Given : The slope of the tangent to the curve is
$\large\frac{x^2+y^2}{2xy}$ we know slope of the tangent is $\large\frac{dy}{dx}$
Hence $\large\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$
Clearly this is a homologous differential equation
This can be solved by putting $y=vx$ and $\large\frac{dy}{dx} $$=v+x \large\frac{dv}{dx}$
$v+x \large\frac{dv}{dx}$$=\large\frac{x^2+v^2x^2}{2x(vx)}$$=\large\frac{x^2(1+v^2)}{2vx^2}$
$=>v+x \large\frac{dv}{dx}=\frac{1+v^2}{2v}$
$x \large\frac{dv}{dx}=\frac{1+v^2}{2v}$$-v$
$=>x \large\frac{dv}{dx}=\frac{1-v^2}{2v}$
Step 2:
Now seperating the variables we get
Let us integrate on both sides,
$\int \large\frac{2v}{1-v^2}$$dv=\int \large\frac{dx}{x}$
Put $ 1-v^2=t$ on differentiating we get,
Substituting this we get,
$-\int \large\frac{dt}{t}=\int \frac{dx}{x}$
Integrating ob both sides we get,
$-\log |t|=\log x+\log c$
$=>-\log |t|^{-1}=\log cx$
Substituing for t
$=> \large\frac{1}{1-v^2}$$=cx$
Substituting for $v=\large\frac{y}{x}$
$=> \large\frac{1}{1-\Large\frac{y^2}{x^2}}$$=cx$
=>$ \large\frac{x^2}{x^2-y^2}$$=cx$
or $x=c(x^2-y^2)$
Step 3:
It is given that the curve passes through the point (2,1)
Now substitute 2 for x and 1 for y
$=>2=3c\qquad or \qquad c=2/3$
Substituting the value of c we get,
$x=\large\frac{2}{3} $$(x^2-y^2)$
or $3x=2(x^2-y^2)$ is the required solution


answered May 13, 2013 by meena.p

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