Question #3e972

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Answer 1 · 2018-02-08T09:19:38

#dy/dx=(y+1+x^2)/(x-x^2(siny))#

#ydx - xdy + (1+x^2)dx + x^2(siny)dy =0#

#ydx + (1+x^2)dx =xdy- x^2(siny)dy#

#dx[y+1+x^2]=dy[x-x^2(siny)]#

#dy/dx=(y+1+x^2)/(x-x^2(siny))#

Answer 2 · 2018-02-08T10:15:34

# y=x(x-cosy+C)-1#.

Let us rewrite the given Diff. Eqn. as,

#(1+x^2)dx+x^2sinydy=xdy-ydx#.

Dividing by #x^2#, we have,

#((1+x^2)/x^2)dx+sinydy=(xdy-ydx)/x^2#.

The Right Member of the eqn. is note-worthy :

Observe that, #d(y/x)=(xdy-ydx)/x^2#.

Utilising this, we find that the given eqn. is,

# ((1+x^2)/x^2)dx+sinydy=d(y/x)#, which looks like,

separable variable type. So, Integrating term-wise,

#int((1+x^2)/x^2)dx+intsinydy+C=intd(y/x)#.

#:. int(1/x^2+1)dx-cosy+C=y/x, i.e., #

# -1/x+x-cosy+C=y/x, or,#

# y=x(x-cosy+C)-1,# is the desired General Solution!