Question

Find the particular solution of the differential equation `(1+x^2)dy/dx=(e^(mtan^-1 x)-y)` , give that y=1 when x=0.

Answer 1

`(1+x^2)dy/dx=(e^(mtan^-1 x)-y)`

`=>dy/dx=(e^(mtav^-1 x))/(1+x^2)-y/(1+x^2)`

`=>dy/dx+y/(1+x^2)=e^(m tan^-1 x)/(1+x^2)`

`P=1/(1+x^2), Q=e^(m tan^-1 x)/(1+x^2)`

`I.F=e^(intPdx)`

`=e^(int(1/(1+x^2))dx)`

`=e^(tan^-1 x)`

Thus the solution is

`ye^(intPdx)=intQe^(intPdx)dx`

`=>yxxe^(tan^-1 x)=inte^(m tan^-1 x)/(1+x^2) .e^(tan^-1 x)dx`

`=>yxxe^(tan^-1 x)=inte^((m+1) tan^-1 x)/(1+x^2)dx ...............(i)`

`inte^((m+1) tan^-1 x)/(1+x^2)dx.............(ii)`

`Let (m+1)tan^-1 x = z`

`(m+1)/(1+x^2)dx=dz`

`dx/(1+x^2)=(dz)/(m+1)`

Substituting in (ii),

`1/(m+1)inte^z dz`

`=e^z/(m+1)`

`=e^((m+1)tan^-1 x)`

Substituting in (i),

`=>y xx e^(tan^-1 x)=e^((m+1)tan^-1 x)/(m+1)+C..........(iii)`

Putting y=1 and x=1, in the above equation,

`=>yxxe^(tan^-1 1)=e^((m+1)tan^-1 x)/(m+1)+C`

`=>1 xx e^(pi/4) = e^((m+1)tan^-1 pi/4)/(m+1)+C`

`=>C= e^((m+1)tan^-1 pi/4)/(m+1)- e^(pi/4)`

Particular solution of the D.E. is `yxxe^(tan^-1x)=e^((m+1)tan^-1 x)/(m+1)+e^((m+1)tan^-1 pi/4)/(m+1)- e^(pi/4)`