1.

Find the particular solution of the differentialequation `(3x y+y^2)dx+(x^2=x y)dy=0;for x=1, y=1.`

Answer» We may write the given differential equation as
`(dy)/(dx)=-(3xy+y^(2))/(x^(2)+xy)`……………(i)
On dividing the Nr and Dr of RHS of (i) by `x^(2)`, we get
`(dy)/(dx)=-{(3(y/x)+(y/x)^(2))/(1+y/x)}=f(y/x)`.
So, the given differential equation is homogeneous.
Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)` in (i), we get
`v+x(dv)/(dx)=-(3vx^(2)+v^(2)x^(2))/(x^(2)+vx^(2))=-(3v+v^(2))/(v+1)`
`rArr x(dv)/(dx)={-(3v+v^(2))/(v+1)-v}=-(2(v^(2)+2v))/(v+1)`
`rArr (v+1)/(v^(2)+2v)dv=-2/xdx`
`rArr 1/2int(2(v+1))/(v^(2)+2v)dv+int2/xdx=log|C_(1)|`
`rArr log|x^(2)sqrt(v^(2)+2v)|=log|C_(1)|`
`rArr log|xsqrt(y^(2)+2xy)|=log|C_(1)| [therefore v=y/x]`.
`rArr xsqrt(y^(2)+2xy)=+-|C_(1)|`
`rArr x^(2)(y^(2)+2xy)=C,` Where `C=C_(1)^(2)`................(ii)
`therefore x^(2)(y^(2)+2xy)=3` is the required solution.


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