¡) Ans:
The given differential equation is
ysin2xdx=(1+y2+cos2x)dy
⟹ysin2xdx−(1+y2+cos2x)dy=0
Which is of the form Mdx+Ndy=0 .
Where M=ysin2x and N=−(1+y2+cos2x)
∂y∂M=∂y∂ysin2x=sin2x∂x∂N=∂x∂−(1+y2+cos2x)=−(2cosx×−sinx)
=2sinxcos=sin2x
As ∂y∂M=∂x∂N ,Hence the given differential equation is an exact differential equation.
\therefore \ The solution of the given differential equation is
∫y=const.Mdx+∫(only those therms of N which do not contains x)dy
=∫y=const.ysin2x+∫−(1+y2)dy
=y∫sin2xdx−∫(1+y2)dy
=−2ycos2x−y−3y3+c
Where "c" is the integration constant.
¡¡)ans :
The given differential equation is
(xy2−x2)dx+(3x2y2+x2y−2x3+y2)dy=0
Which is of the form Mdx+Ndy=0
Here M=(xy2−x2) and N=(3x2y2+x2y−2x3+y)
∴∂y∂M=2xy and ∂x∂N=6xy2+2xy−6x2
As ∂y∂M=∂x∂N
Therefore ,the given differential equation is not exact.
But M1(∂x∂N−∂y∂M)=6 is a is a constant function , which may be regarded as function of y i,e ϕ(y), then the integration factor of given differential equation is
I.F.=e∫6.dy=e6y
Multiplying by I.F. ,the given differential equation becomes
e6y(xy2−x2)dx+e6y(3x2y2+x2y−2x3+y2)=0..........(1)
Which can be shown to be exact.
Therefore the required solution is
∫y=const.e6y(xy2−x2)dx+∫e6yy2dy=I1+I2(say)
Where I1=∫y=const.e6y(xy2−x2)dx,I2=∫e6yy2dy
I1=e6y(2y2x2−3x3).
I2=y2∫e6ydy−∫(dydy2(∫e6ydy))dy by using by part.
=6y2e6y−∫2y×6e6ydy
=6y2e6y−31y∫e6ydy+31∫(dydy(∫e6ydy))dy by using by parts.
=6y2e6y−181ye6y+1081e6y .
Therefore the required solution of the given differential equation is
=I1+I2
=e6y(2y2x2−3x3)+6y2e6y−181ye6y+1081e6y+k
Where k is the integration constant.
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