Question

What is the general solution of the differential equation? : `dy/dx + (2x)/(x^2+1)y=1/(x^2+1)`

Answer 1

`y = x/(x^2+1) + C/(x^2+1)`

Explanation:

We have:

`dy/dx + (2x)/(x^2+1)y=1/(x^2+1)` ..... [A]

This is a first order linear differential equation of the form:

`(d zeta)/dx + P(x) zeta = Q(x)`

We solve this using an Integrating Factor

`I = exp( \ int \ P(x) \ dx )`
`\ \ = exp( int \ (2x)/(x^2+1) ) \ dx )`
`\ \ = exp( ln|x^2+1| )`
`\ \ = exp( ln(x^2+1) )` as `x^2+1 gt 0`
`\ \ = x^2+1`

And if we multiply the DE [A] by this Integrating Factor, `I`, we will (by virtue of the IF) have a perfect product differential;

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

`:. d/dx((x^2+1)y) = 1`

Which is now a trivial separable DE, so we can "separate the variables" to get:

`(x^2+1)y = int \ dx`

And integrating gives us:

`(x^2+1)y = x + C`

Which we can rearrange to get:

`y = x/(x^2+1) + C/(x^2+1)`

Answer 2

`y(x) = (x+ C)/(x^2 + 1)`

Explanation:

GIven: `dy/dx + (2x)/(x^2 + 1)y = 1/(x^2 + 1)`

I am going to make a slight change of notation and mark it as equation [1]:

`y'(x) + (2x)/(x^2 + 1)y(x) = 1/(x^2 + 1)" [1]"`

Equation [1] is in the form:

`y'(x) + P(x)y(x)=Q(x)`

where `P(x) = (2x)/(x^2+1)` and `Q(x)=1/(x^2+1)`

This type of equation is known to have an integrating factor:

`mu(x) = e^(intP(x)dx`

Substitute for `P(x)`

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

Integrate:

`mu(x) = e^(ln(x^2+1))`

Simplify:

`mu(x) = x^2+1`

Multiply both sides of the equation [1] by `mu(x)`:

`(x^2 + 1)y'(x) + (2x)y(x) = 1`

Set up both sides for integration:

`int((x^2 + 1)y'(x) + (2x)y(x))dx = intdx`

We do not actually integrate the left side integrates, because multiplication by the integrating factor assures that the left side integrates to `mu(x)y(x)`. We do integrate right side but the integral is trivial:

`(x^2 + 1)y(x) = x + C`

Solve for `y(x)`:

`y(x) = (x+ C)/(x^2 + 1)`

Answer 3

`y = (x+C_3)/(x^2+1)`

Explanation:

This differential equation is completely equivalent to

`(x^2+1)(dy)/(dx)+2x y = 1` because `x^2+1 > 0 forall x in RR`

The homogeneous part of this linear differential equation is

`(x^2+1)y'_h+2x y_y = 0` which is separable

`(2x)/(x^2+1) dx = -dy_h/(y_h)` and integrating

`log_e(x^2+1) = -log_e y_h + C_1` then

`x^2+1=C_2/y_h` or

`y_h = C_2/(x^2+1)`

Substituting now `y_p = (C_2(x))/(x^2+1)` into the complete equation we obtain

`C_2'(x) = 1` and then

`C_2(x) = x + C_3`

Now the complete solution is obtained as `y = y_h+y_p` and then

`y = (x+C_3)/(x^2+1)`