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प्रश्न
Solve the differential equation ` (1 + x2) dy/dx+y=e^(tan^(−1))x.`
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उत्तर
`(1 + x2) dy/dx+y=e^(tan^(−1))x`
`=>dy/dx+y/(1+x^2)=e^(tan^(−1)x)/(1+x^2)`
Comparing it with the standard equation, f'(x)+yP=Q
Now, integrating factor, `I.F. = e^(∫Pdx)=e^(int1/(1+x^2)dx)=e^(tan^(-1)x)`
`ye^(∫P(x)dx)=∫Q(x)e^(∫P(x)dx)dx+C`
`∴ ye^(tan^(−1) x)=∫(e^(tan^(−1) x))/(1+x^2)e^(tan^(−1) x)dx+C .....(1)`
`Let I=∫(e^(tan^(−1) x))/(1+x^2)e^(tan^(−1) x)dx`
Putting `e^(tan^(−1) x)=t`
`∫1/(1+x^2)e^(tan^(−1) x)dx=dt`
`∴ I=∫t dt `
`I=t^2/2`
`⇒I=((e^(tan^(−1) x))^2)/2`
Considering (1), we get:
`ye^(tan^(−1) x)=(e^(tan^(−1) x)/2)^2+C`
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