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प्रश्न
Find the integrating factor of the differential equation.
`((e^(-2^sqrtx))/sqrtx-y/sqrtx)dy/dx=1`
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उत्तर
We have
`((e^(-2^sqrtx))/sqrtx-y/sqrtx)dy/dx=1`
`dy/dx=(e^(-2^sqrtx))/sqrtx-y/sqrtx`
`=>dy/dx+(1/sqrtx)y=(e^(-2^sqrtx))/sqrtx`
It is in the form dy/dx+Py=Q, where P and Q are the constants or functions of x.
Thus, the integrating factor of the given differential equation is
`e^(intPdx)=e^(int1/sqrtxdx)`
`=e^(intx^(-1/2)dx)=e^(x^((1/2))/(1/2))`
`=e^(2sqrtx)`
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