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प्रश्न
Solve the differential equation:
`(xdy - ydx) ysin(y/x) = (ydx + xdy) xcos(y/x)`.
Find the particular solution satisfying the condition that y = π when x = 1.
उत्तर १
`(xdy - ydx)ysin(y/x) = (ydx + xdy)xcos(y/x)`.
`\implies (xdy - ydx) (sin(y/x))/(cos(y/x)) = ((ydx + xdy)x)/y`
`\implies ((xdy - ydx))/x^2 tan(y/x) = ((ydx + xdy)x)/(x^2y)`
`\implies ((xdy - ydx))/x^2 tan(y/x) = ((ydx + xdy))/(xy)`
`\implies tan(y/x)*d(y/x) = (d(xy))/(xy)`,
which is variable separable form where variables are `y/x` and xy
Integrating both sides,
`\implies int tan y/x * d(y/x) = int(d(xy))/(xy)`
`\implies log|sec(y/x)| = log(xy) + logc`
`\implies sec(y/x) = c(xy)`.
Given, when x = 1, y = π `\implies` c = `-1/π`.
Hence required particular solution is `sec(y/x) = -1/π(xy)`.
उत्तर २
`(xdy - ydx) ysin(y/x) = (ydx + xdy) xcos(y/x)`.
`\implies dy/dx = ((xy)cos(y/x) + y^2sin(y/x))/((xy)sin(y/x) - x^2cos(y/x))`
`\implies dy/dx = (1 + (y/x)tan(y/x))/(tan(y/x) - 1/((y/x))`,
which is homogenous differential equation of type `dy/dx = f(y/x)`.
Let y = vx
`\implies dy/dx = v + x(dv)/dx`
`\implies v + x(dv)/dx = (1 + vtanv)/(tanv - 1/v)`
`\implies (tan v - 1/v)dv = 2dx/x`,
which is variable separable form where variables are v and x
Integrating both sides,
`\implies int(tanv - 1/v)dv = 2intdx/x`
`\implies log|secv| - log|v| = 2log|x| + logc`
`\implies log|secv| = log|v.x^2.c|`
`\implies secv = c(xy)`
`\implies sec(y/x) = c(xy)`.
Given, when x = 1, y = π `\implies` c = `-1/π`.
Hence required particular solution is `sec(y/x) = -1/π(xy)`.
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