1.

Solve the differential equation `2x^2(dy)/(dx)-2xy+y^2=0`, `y^2(e)=e`

Answer» `2x^2dy/dx-2xy+y^2=0`
`2x^2*1/y^2dy/dx-2x/y+1=0`
`2x^2(-dz/dx)-2xz+1=0`
`2x^2dz/dx+2xz=1`
`dz/dx+1/x^2=1/(2x^2)`
`d/dx{ze^(int1/xdx)}=1/(2x^2)*e^(1/xdx`
`d/dx(2x)=1/(2x^2)*x`
`2x=int 1/(2x)dx`
`=1/2log|x|+c`
`x/y=1/2log|x|+c`
`e/(y(e))=1/2log|e|+c`
`e/(y(e))=1/2+c`
`c=e/(y(e))-1/2`
`c=pmsqrte-1/2`
`x/y=1/2log|x|+c`
`c=pmsqrte-1/2`.


Discussion

No Comment Found