Question

: Show that the solution of the differential equation: = − − − − − is of the form: + + ( − ) = + , When = and =

Answer 1

`y = \tan(x + (x^2)/(2))`

Explanation:

Poorly formatted question; The complete question requires that we prove that
`y=\tan(x+(x\²)/(2))`

When

`(dy)/(dx) =1+xy\²+x+y\²` and
`y(0)=0`

We have:

`(dy)/(dx) =1+xy\²+x+y\²`

Rewrite as:

`(dy)/(dx) =1+x+xy\²+y\²`

Factorize

`(dy)/(dx) = (1+x)+y\²(x+1)`

Rewrite as:

`(dy)/(dx) = (1+x)+y\²(1+x)`

Factor out 1 + x

`(dy)/(dx) = (1+y\²)(1+x)`

Multiply both sides by
`(dx)/(1 + y^2)`

`(dy)/(1+y\²) = (1+x)dx`

Integrate both sides

`\int (dy)/(1+y\²) = \int (1+x)dx`

Rewrite as:

`\int (1)/(1+y\²) dy = \int (1+x)dx`

Integrate the left-hand side

`\int (1)/(1+y\²) dy = \tan^(-1)y`

Integrate the right-hand side

`\tan^(-1)y = x + (x^2)/(2) + c`

`y(0)=0` implies that:
`(x,y) = (0,0)`

So:

`\tan^(-1)y = x + (x^2)/(2) + c` becomes

`\tan^(-1)(0) = 0 + (0^2)/(2) + c`

This gives:

`0 = 0 +0 + c`

`0 =c`

`c = 0`

The equation
`\tan^(-1)y = x + (x^2)/(2) + c` becomes

`\tan^(-1)y = x + (x^2)/(2) + 0`

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

Take tan of both sides

`y = \tan(x + (x^2)/(2))` --- Proved