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प्रश्न
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.
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उत्तर
`(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)`
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