Consider the homogeneous differential eq...

Consider the homogeneous differential equation `(dy)/(dx)=f((y)/(x))` By putting `(y)/(x)=v`, the equation reduces to `(dv)/(f(v)-v)=(dx)/(x)`. Integrating we get `int(dv)/(f(v)-v)=int(dx)/(x)+c`
Solution of `x(x-y)dy+y^(2)dx=0` is-

`y^(2)=c^(2)e^((2y)/(x))`

`y=ce^((2y)/(x))`

`y^(2)=ce^((y)/(2x))`

`y^(2)=c^(2)e^((y^(2))/(x))`

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Consider the homogeneous differential equation (dy)/(dx)=f((y)/(x)) By putting (y)/(x)=v , the equation reduces to (dv)/(f(v)-v)=(dx)/(x) . Integrating we get int(dv)/(f(v)-v)=int(dx)/(x)+c Solution of (x+ycos""(y)/(x))dx=xcos""(y)/(x)dy is

`cos""(y)/(x)=log|x|+c`

`sin""(x)/(y)=log|y|+c`

`sin""(y)/(x)=log|xc|`

`cos""(y)/(x)=log|yc|`

`x^(2)-y^(2)+2xy=c`

`x^(2)-y^(2)+xy=c`

`x^(2)-y^(2)-2xy=c`

`x^(2)-y^(2)-xy=c`

`2^(-x)-2^(-y)=c`

`2^(-x)+2^(-y)=c`

`2^(x)+2^(y)=c`

`2^(x)-2^(y)=c`