Question

Solve the differential equation ` (1 + x2) dy/dx+y=e^(tan^(−1))x.`

Answer 1

`(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`