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Find the particular solution of the diff...

Find the particular solution of the differential equation: `(1+e^(2x))dy+(1+y^2)e^x dx=0,` given that `y=1,\ ` when `x=0.`

Text Solution

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`(1+e^(2x))dy+(1+y^(2))e^(x)dx=0`
or `(dy)/(1+y^(3))+(e^(x)dx)/(1+e^(2x))=0`
Integrating both sides, we get
`int(dy)/(1+y^(2))+int(e^(x)dx)/(1+e^(2x))=C`
`Now, tan^(-1)y+tan^(-1)(e^(x))=C`…………..(1)
`therefore tan^(-1)+tan^(-1)=C`
or `C=pi/2`
Substituting `C=pi/2`, in equation(1), we get
`tan^(-1)y+tan^(-1)(e^(x))=pi/2`
This is the required solution of the given differential equation.
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Knowledge Check

  • The solution of the differential equation x(e^(2y)-1)dy + (x^2-1) e^y dx=0 is

    A
    `e^y + e^(-y) =log x - x^2/2+c`
    B
    `e^y-e^(-y) = log x - x^2/2+c `
    C
    `e^y + e^(-y) = log x + x^2/2 +c`
    D
    `e^y - e^(-y) = log x +x^2/2 + c `

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