Question

Find the general solution of the differential equation:

`(x^2 + 1) dy/dx + 2xy = sqrt(x^2 + 4)`

Answer 1

Given equation is

`(x^2 + 1) dy/dx + 2xy = sqrt(x^2 + 4)`

`\implies dy/dx + (2x)/((x^2 + 1)) xx y = sqrt(x^2 + 4)/((x^2 + 1))`

It is of the form,

`dy/dx + Py` = Q

∴ I.F. = `e^(int P.dx)`

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

Let x² + 1 = t

2x dx = dt

∴ I.F. = `e^(int dt/t)`

= e^{log t}

= t

= x² + 1

General solution is given as

(I.F.)y = `int (I.F.) Q dx + C`

`\implies` (x² + 1)y = `int (x^2 + 1) sqrt(x^2 + 4)/((x^2 + 1)) dx + C` 

`\implies` (x² + 1)y = `int sqrt(x^2 + 4)  dx + C`

`\implies` (x² + 1)y = `int sqrt((x)^2 + 2^2)  dx + C`

`\implies` (x² + 1)y = `x/2 (sqrt(x^2 + 4) + 4/2 log |x + sqrt(x^2 + 4)|) + C`

`\implies` (x² + 1)y = `x/2 sqrt(x^2 + 4) + 2 log |x + sqrt(x^2 + 4) | + C`

is the required solution of given differential equation.