Question

Solution of the differential equation $xdx + ydy = x(xdy - ydx)$ is

• A. $\frac{y+1}{\sqrt{x^2 + y^2}} + C = 0$
• B. $\frac{y-1}{\sqrt{x^2 + y^2}} + C = 0$
• C. $\frac{y+1}{\sqrt{x^2 - y^2}} + C = 0$
• D. $\frac{y-1}{\sqrt{x^2 - y^2}} + C = 0$

Answer 1

Answer: (A)

$\frac{y+1}{\sqrt{x^2 + y^2}} + C = 0$

Answer 2

The given differential equation is: $xdx + ydy = x(xdy - ydx)$

This equation can be simplified using polar coordinates. Let $x = r\cos\theta$ and $y = r\sin\theta$. From these substitutions, we have the following standard differential relations:

1. $x^2 + y^2 = r^2$ Differentiating both sides: $2xdx + 2ydy = 2rdr \implies xdx + ydy = rdr$.
2. The term $xdy - ydx$ can be related to $d\theta$. $xdy - ydx = (r\cos\theta)(dr\sin\theta + r\cos\theta d\theta) - (r\sin\theta)(dr\cos\theta - r\sin\theta d\theta)$ $= rdr\sin\theta\cos\theta + r^2\cos^2\theta d\theta - rdr\sin\theta\cos\theta + r^2\sin^2\theta d\theta$ $= r^2(\cos^2\theta + \sin^2\theta)d\theta = r^2d\theta$.

Now, substitute these into the original differential equation: $rdr = (r\cos\theta)(r^2d\theta)$ $rdr = r^3\cos\theta d\theta$

Assuming $r \neq 0$, we can divide both sides by $r$: $dr = r^2\cos\theta d\theta$

Now, separate the variables $r$ and $\theta$: $\frac{dr}{r^2} = \cos\theta d\theta$

Integrate both sides: $\int \frac{dr}{r^2} = \int \cos\theta d\theta$ $\int r^{-2} dr = \int \cos\theta d\theta$ $-\frac{1}{r} = \sin\theta + K$ (where $K$ is the constant of integration)

Rearrange the terms: $\frac{1}{r} + \sin\theta = -K$ Let $C = -K$ (since $K$ is an arbitrary constant, $-K$ is also an arbitrary constant). $\frac{1}{r} + \sin\theta = C$

Finally, substitute back the Cartesian coordinates: $r = \sqrt{x^2+y^2}$ $\sin\theta = \frac{y}{r} = \frac{y}{\sqrt{x^2+y^2}}$

So the solution becomes: $\frac{1}{\sqrt{x^2+y^2}} + \frac{y}{\sqrt{x^2+y^2}} = C$ $\frac{1+y}{\sqrt{x^2+y^2}} = C$

This can be written as $\frac{y+1}{\sqrt{x^2+y^2}} - C = 0$. If we redefine the constant as $-C$, it matches option A. $\frac{y+1}{\sqrt{x^2+y^2}} + C' = 0$ (where $C' = -C$).

The given differential equation $xdx + ydy = x(xdy - ydx)$ is most efficiently solved by transforming it into polar coordinates. By substituting $x = r\cos\theta$, $y = r\sin\theta$, we use the identities $xdx + ydy = rdr$ and $xdy - ydx = r^2d\theta$. The equation simplifies to $rdr = r\cos\theta (r^2d\theta)$, which further reduces to $\frac{dr}{r^2} = \cos\theta d\theta$. Integrating both sides yields $-\frac{1}{r} = \sin\theta + C'$. Substituting back $r=\sqrt{x^2+y^2}$ and $\sin\theta = \frac{y}{\sqrt{x^2+y^2}}$ gives $\frac{1+y}{\sqrt{x^2+y^2}} = C$, which matches option A.