`y'=arctan(x/2)` Solve the differential equation.

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Given ,

`y'= tan^(-1) (x/2)`

we have to get the y

so ,

=> `y = int (tan^(-1) (x/2)) dx`

By Applying the integration by parts we get the solution

so,

let` u=tan^(-1) (x/2) => u'= (tan^(-1) (x/2) )'`

let` t= x/2 `

=> `u' =(du/dt)*(dt/dx) = (d ((tan^(-1)(t))/(dt))(d/dx (x/2))`

=`(1/(t^2+1))*1/2`

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Given ,

`y'= tan^(-1) (x/2)`

we have to get the y

so ,

=> `y = int (tan^(-1) (x/2)) dx`

By Applying the integration by parts we get the solution

so,

let` u=tan^(-1) (x/2) => u'= (tan^(-1) (x/2) )'`

let` t= x/2 `

=> `u' =(du/dt)*(dt/dx) = (d ((tan^(-1)(t))/(dt))(d/dx (x/2))`

=`(1/(t^2+1))*1/2`

=`(1/((x/2)^2+1))*1/2`

=`(4/(x^2+4))*1/2`

=`2/(x^2+4)`

 

and `v'=1 =>v =x`

now by Integration by parts ,

`int uv' dx= uv-int u'v dx`

so , now

`int (tan^(-1) (x/2))dx`

`= xtan^(-1) (x/2) - int 2x/(x^2+4)dx`

let `x^2+4 = q`

=> `2x dx= dq`

so ,

`int (2x)/(x^2+4)dx = int 1/q dq = ln(q)+c =ln(x^2+4)+c`

 

= ` x(tan^(-1) (x/2)) -ln(x^2+4)+C`

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