अवकल समीकरण `e^(-x)(y+1)dy+(cos^(2)x-sin2x)y(dx)=0` का हल है, जब `y = 1` तथा `x = 0` है

Options

A. `y+logy+e^(x)cos^(2)x=2`
B. `log(y+1)+e^(x)cos^(2)x=1`
C. `y+logy=e^(x)cos^(2)x`
D. `(y+1)+e^(x)cos^(2)x=2`

Answer

A

Explanation

दी गई अवकल समीकरण
`e^(-x)(y+1)dy=-y(cos^(2)x-sin2x)dx`
`implies ((y+1))/(y)dy=-(cos^(2)x-sin2x)e^(x)dx`
`implies (1+(1)/(y))dy=-e^(x)(cos^(2)x-sin2x)dx`
समाकलन करने पर,
`y+logy=-e^(x)cos^(2)x+C`
y = 1 तथा x = 0 रखने पर,
`implies1+log1=-e^(0).cos^(2)0+C`
`implies 1=-1+CimpliesC=2`
अतः अभीष्ट हल `y+logy=-e^(x)cosx+2`
`implies y+logy+e^(x)cos^(2)x=2`